CHAPTER 5 - CMOS AMPLIFIERS
|
|
- Jewel Hunt
- 6 years ago
- Views:
Transcription
1 CMOS Analog Circuit Design Page 5.0 Chapter Outline 5. Inverters 5. Differential Amplifiers 5.3 Cascode Amplifiers 5.4 Current Amplifiers 5.5 Output Amplifiers 5.6 HighGain Architectures Goal CHAPTER 5 CMOS AMPLIFIERS To develop an understanding of the amplifier building blocks used in CMOS analog circuit design. Design Hierarchy Functional blocks or circuits (Perform a complex function) Chapter 5 Blocks or circuits (Combination of primitives, independent) Subblocks or subcircuits (A primitive, not independent) Fig. 5.0 CMOS Analog Circuit Design Page 5.0 Illustration of Hierarchy in Analog Circuits for an Op Amp Operational Amplifier Biasing Circuits Input Differential Amplifier Second Gain Stage Output Stage Current Source Current Mirrors Current Sink Source Coupled Pair Current Mirror Load Inverter Current Sink Load Source Follower Current Sink Load Fig. 5.0
2 CMOS Analog Circuit Design Page 5. SECTION 5. CMOS INVERTING AMPLIFIERS Characterization of Amplifiers Amplifiers will be characterized by the following properties: Largesignal voltage transfer characteristics Largesignal voltage swing limitations Smallsignal, frequency independent performance Gain Input resistance Output resistance Smallsignal, frequency response Other properties Noise Power dissipation Etc. CMOS Analog Circuit Design Page 5. Inverters The inverting amplifier is an amplifier which amplifies and inverts the input signal. The inverting amplifier generally has the source on ac ground or the commonsource configuration. Various types of inverting CMOS amplifiers: I D I D I D V GG I D v IN I D vout v IN v IN v IN v IN Active NMOS Load Inverter Active PMOS Load Inverter Depletion NMOS Load Inverter We will consider: Active PMOS Load Inverter (active load inverter) Current Source Load Inverter Pushpull Inverter Current Source Load Inverter Pushpull Inverter Fig. 5.
3 CMOS Analog Circuit Design Page 5.3 Voltage Transfer Characteristic of the Active Load Inverter ID (ma) v IN =5.0Vv IN =4.5V vin =4.0V 0.5 K J v IN =3.5V I 0.4 H v IN =3.0V G 0.3 F E v IN =.0V v IN =.5V D C A,B 0.0 v IN =.0V A B 4 C 0 0 v IN The boundary between active and saturation operation for is v DS v GS V TN v IN 0.7V vout3 v IN =.5V 5V W = µm L µm I D W v IN = µm L µm D cutoff saturated E F saturated active G H I J K Fig. 5. CMOS Analog Circuit Design Page 5.4 LargeSignal Voltage Swing Limits of the Active Load Inverter Maximum output voltage, (max): (max) V TP (ignores subthreshold current influence on the MOSFET) Minimum output voltage, (min): Assume that is nonsaturated and that V T = V T = V T. v DS v GS V TN v IN 0.7V The current through is i D = β (v GS V T )v DS v DS (V DD V T )( ) () and the current through is i D = β (v SG V T ) = β ( V T ) = β ( V T ) Equating these currents gives the minimum as, (min) = V T V T (β /β )
4 CMOS Analog Circuit Design Page 5.5 SmallSignal Midband Performance of the Active Load Inverter The development of the smallsignal model for the active load inverter is shown below: S=B v IN I D g m v gs G D=D=G v in g m v gs S=B r ds r ds R out v out v in v out g m v in rds g m v out r ds Sum the currents at the output node to get, g m v in g ds v out g m v out g ds v out = 0 Solving for the voltage gain, v out /v in, gives v out g m v = in g ds g ds g g m m g = K' N W L / m K' P L W The smallsignal output resistance can also be found from the above by letting v in = 0 to get, R out = g ds g ds g m g m Fig. 5.3 CMOS Analog Circuit Design Page 5.6 Frequency Response of the Active Load Inverter Incorporation of the parasitic capacitors into the smallsignal model (x is connected to v out ): C gs x C bd C M C gd C gd V in C bd V out C L V in gm V in R out C out V out Fig. 5.4 If we assume the input voltage has a small source resistance, then we can write the following: g V out (s) m R out s db V in (s) = z 0log0(g m R out ) s p z 0dB log 0 f where p ω 3dB g m = g m, p = R out (C out C M ), and z = g m C M and R out = [g ds g ds g m ] /g m, C M = C gd, and C out = C bd C bd C gs C L If p < z, then the 3dB frequency is approximately equal to [R out (C out C M )]. Fig. 5.4A
5 CMOS Analog Circuit Design Page 5.7 Example 5. Performance of an Active ResistorLoad Inverter Calculate the outputvoltage swing limits for = 5 volts, the smallsignal gain, the output resistance, and the 3 db frequency of active load inverter if (W /L ) is µm/ µm and W /L = µm/ µm, C gd = 00fF, C bd = 00fF, C bd = 00fF, C gs = 00fF, C L = pf, and I D = I D = 00µA, using the parameters in Table 3.. Solution From the above results we find that: (max) = 4.3 volts (min) = 0.48 volts Smallsignal voltage gain =.9V/V R out = 9.7 kω including g ds and g ds and 0 kω ignoring g ds and g ds z =.0x0 9 rads/sec p = 64.x0 6 rads/sec. Thus, the 3 db frequency is 0. MHz. CMOS Analog Circuit Design Page 5.8 Voltage Transfer Characteristic of the Current Source Inverter ID (ma) v IN =5.0Vv IN =4.5V vin =4.0V 0.5 v IN =3.5V 0.4 v IN =3.0V 0.3 vout v IN =.5V.5V KJIH F E v IN =.0V 0. G 0. D v C IN =.5V A,B 0.0 v IN =.0V A B C D 4 active 3 saturated Regions of operation for the transistors: 5V W = µm L µm I D W v IN = L µm µm saturated active E F G H I J K 0 0 v 3 4 IN 5 : v DS v GS V Tn v IN 0.7V : v SD v SG V Tp V GG V Tp 3.V Fig. 5.5
6 CMOS Analog Circuit Design Page 5.9 LargeSignal Voltage Swing Limits of the Current Source Load Inverter Maximum output voltage, (max): (max) Minimum output voltage, (min): Assume that is nonsaturated. The minimum output voltage is, (min) = (min) = ( V T ) This result assumes that v IN is taken to. β V GG V T β V T CMOS Analog Circuit Design Page 5.0 SmallSignal Midband Performance of the Current Source Load Inverter SmallSignal Model: S=B V GG v IN I D r ds G D=D v in g m v gs r ds S=B=G Midband Performance: v out g m v = in g ds g = K' N W ds L / I D λ λ A max log A v R out v out v in v out g m v in rds r ds Ι D!!! and R out = g ds g ds Fig. 5.5B I D (λ λ ) A max 0 A max 00 A max 000 Weak inversion Strong inversion I D 0.µA µa 0µA 00µA ma 0mA Fig. 5.6
7 CMOS Analog Circuit Design Page 5. Frequency Response of the Current Source Load Inverter Incorporation of the parasitic capacitors into the smallsignal model (x is connected to V GG ): C gs x C bd C M C gd C gd V in C bd V out C L V in gm V in R out C out V out Fig. 5.4 If we assume the input voltage has a small source resistance, then we can write the following: g V out (s) m R out s V in (s) = z s p where g m = g m, p = R out (C out C M ), and z = g m C M and R out = g ds g and C ds out = C gd C bd C bd C L C M = C gd Therefore, if p < z, then the 3 db frequency response can be expressed as g ds g ds ω 3dB ω = C gd C gd C bd C bd C L CMOS Analog Circuit Design Page 5. Example 5. Performance of a CurrentSink Inverter The performance of a currentsink CMOS inverter is to be examined. The currentsink inverter is shown in Fig Assume that V SG W = µm, L = µm, W = µm, L = µm, = 5 volts, V GG v IN = 3 volts, and the parameters of Table 3. describe and. Use I D v the capacitor values of Example 5. (C gd = C gd ). Calculate the OUT outputswing limits and the smallsignal performance. V Solution GG To attain the output signalswing limitations, we treat Fig. 5.7 as a current source CMOS inverter with PMOS parameters for the Figure 5.7 Current sink CMOS inverter. NMOS and NMOS parameters for the PMOS and use NMOS equations. Using a prime notation to designate the results of the current source CMOS inverter which exchanges the PMOS and NMOS model parameters, we get (max) = 5V and (min) = (50.7) = 0.74V In terms of the current sink CMOS inverter, these limits are subtracted from 5V to get (max) = 4.6V and (min) = 0V. To find the small signal performance, we must first calculate the dc current. The dc current, I D, is I D = K N W L (V GG V TN ) = 0 (30.7) = 9µA v out /v in = 9.V/V, R out = 38. kω, and f 3dB =.78 MHz.
8 CMOS Analog Circuit Design Page 5.3 Voltage Transfer Characteristic of the PushPull Inverter ID (ma) v v IN =5.0V IN =4.5V v IN =4.0V v IN =3.5V.0 v IN =0.5V 0.8 v IN =.0V v IN =.5V v IN =.0V v IN =3.0V v IN =.5V v IN =.5V F 0. H I G v IN =3.0V E v IN =.0V v IN =3.5V v IN =4.5V D v IN =.5V C J,K A,B 5 A B C v IN =.0V vout 4 3 D v IN E active saturated 5V W = µm L µm I D W = L µm µm F saturated active Note the railtorail output voltage swing G H I J K 0 0 v 3 4 IN 5 Regions of operation for and : : v DS v GS V T v IN 0.7V : v SD v SG V T v IN V T v IN 0.7V Fig. 5.8 CMOS Analog Circuit Design Page 5.4 SmallSignal Performance of the PushPull Amplifier 5V C M v in v out v in g m v in r ds g m v in r ds C out v out Fig. 5.9 Smallsignal analysis gives the following results: and v out v = (g m g m ) in g ds g = (/I D ) ds R out = g ds g ds z = g m g m C M = g m g m C gd C gd (g ds g ds ) p = C gd C gd C bd C bd C L If z > p, then g ds g ds ω 3dB = C gd C gd C bd C bd C L K' N (W /L ) K' P (W /L ) λ λ
9 CMOS Analog Circuit Design Page 5.5 Example 5.3 Performance of a PushPull Inverter The performance of a pushpull CMOS inverter is to be examined. Assume that W = µm, L = µm, W = µm, L = µm, = 5 volts, and use the parameters of Table 3. to model and. Use the capacitor values of Example 5. (C gd = C gd ). Calculate the outputswing limits and the smallsignal performance assuming that I D = I D = 300µA. Solution The output swing is seen to be from 0V to 5V. In order to find the small signal performance, we will make the important assumption that both transistors are operating in the saturation region. Therefore: v out 57µS 45µS v = in µs 5µS = 8.6V/V R out = 37 kω f 3dB =.86 MHz and z = 399 MHz CMOS Analog Circuit Design Page 5.6 Noise Analysis of Inverting Amplifiers Noise model: Approach: e n * Noise Free MOSFETs e out e n e eq v in v in * * Noise Free MOSFETs e out Fig. 5.0.) Assume a meansquare inputvoltagenoise spectral density e n in series with the gate of each MOSFET. (This step assumes that the MOSFET is the common source configuration.).) Calculate the outputvoltagenoise spectral density, e out (Assume all sources are additive). 3.) Refer the outputvoltagenoise spectral density back to the input to get equivalent input noise e eq. 4.) Substitute the type of noise source, /f or thermal.
10 CMOS Analog Circuit Design Page 5.7 Noise Analysis of the Active Load Inverter.) See model to the right..) e out = e n g m g en m 3.) e eq = e n g m g e n m e n Up to this point, the type of noise is not defined. /f Noise Substituting e KF n = fc ox WLK = B fwl, into the above gives, e eq (/f) = B fw / L K' B K' B L / L To minimize /f noise,.) Make L >>L,.) increase the value of W and 3.) choose as a PMOS. Thermal Noise Substituting e n = 8kT 3g into the above gives, m e eq (th) = 8kT( η ) 3[K' (W/L) I ] / W L K' / / L W K' To minimize thermal noise, maximize the gain of the inverter. e n * Noise Free MOSFETs e out e n e eq v in v in * * Noise Free MOSFETs e out Fig. 5.0 CMOS Analog Circuit Design Page 5.8 Noise Analysis of the Active Load Inverter Continued When calculating the contribution of e n to e out, it was assumed that the gain was unity. To verify this assumption consider the following model: e n * v gs g m v gs r ds r ds eout _ We can show that, Fig. 5. e out e = g m (r ds r ds ) n g m (r ds r ds )
11 CMOS Analog Circuit Design Page 5.9 Noise Analysis of the Current Source Load Inverting Amplifier Model: Noise e n Free MOSFETs * V GG e e n out e eq v in v in * * The outputvoltagenoise spectral density of this inverter can be written as, e out = (g m r out ) e n (g m r out ) e n or e eq = e n (g m r out ) g m g m e n (g m r out ) e n = e n e n Noise Free MOSFETs e out Fig. 5.. This result is identical with the active load inverter. Thus the noise performance of the two circuits are equivalent although the smallsignal voltage gain is significantly different. CMOS Analog Circuit Design Page 5.0 Noise Analysis of the PushPull Amplifier Model: v in e n * e n * Noise Free MOSFETs e out Fig The equivalent inputvoltagenoise spectral density of the pushpull inverter can be found as g m e n g m e n e eq = g m g m g m g m If the two transconductances are balanced (g m = g m ), then the noise contribution of each device is divided by two. The total noise contribution can only be reduced by reducing the noise contribution of each device. (Basically, both and act like the load transistor and input transistor, so there is no defined input transistor that can cause the noise of the load transistor to be insignificant.)
12 CMOS Analog Circuit Design Page 5. Summary of CMOS Inverting Amplifiers Inverter pchannel active load inverter nchannel active load inverter AC Voltage Gain gm gm gm gmgmb AC Output Resistance gm gmgmb Bandwidth (CGB=0) gm CBDCGSCGSCBD gmgmb CBDCGDCGSCBS Equivalent, inputreferred,meansquare noise voltage g m e n e n g m e n e n g m g m Current source load inverter gm gdsgds gdsgds gdsgds CBDCGDCDGCBD g m e n e n g m nchannel depletion load inverter ~ g m gmb gmbgdsgds gmbgdsgds CBDCGDCGSCBD e n e n g m g m PushPull inverter (gmgm) gdsgds gdsgds gdsgds CBDCGDCGSCBD e n g m e n g m g m g m g m g m Inverting configurations we did not examine. CMOS Analog Circuit Design Page 5. SECTION 5. DIFFERENTIAL AMPLIFIERS What is a Differential Amplifier? A differential amplifier is an amplifier that amplifies the difference between two voltages and rejects the average or common mode value of the two voltages. Symbol for a differential amplifier: v v v OUT Fig. 5.A Differential and common mode voltages: v and v are called singleended voltages. They are voltages referenced to ac ground. The differentialmode input voltage, v ID, is the voltage difference between v and v. The commonmode input voltage, v IC, is the average value of v and v. v ID = v v and v IC = v v v = v IC 0.5v ID and v = v IC 0.5v ID v ID = A VD v ID ± A VC v IC = A VD (v v ) ± A v v VC v IC v ID Fig. 5.B where A VD = differentialmode voltage gain A VC = commonmode voltage gain
13 CMOS Analog Circuit Design Page 5. Differential Amplifier Definitions Common mode rejection rato (CMRR) A VD CMRR = A VC CMRR is a measure of how well the differential amplifier rejects the commonmode input voltage in favor of the differentialinput voltage. Input commonmode range (ICMR) The input commonmode range is the range of commonmode voltages over which the differential amplifier continues to sense and amplify the difference signal with the same gain. Typically, the ICMR is defined by the commonmode voltage range over which all MOSFETs remain in the saturation region. Output offset voltage (V OS (out)) The output offset voltage is the voltage which appears at the output of the differential amplifier when the input terminals are connected together. Input offset voltage (V OS (in) = V OS ) The input offset voltage is equal to the output offset voltage divided by the differential voltage gain. V OS = V OS (out) A VD CMOS Analog Circuit Design Page 5.3 Transconductance Characteristic of the Differential Amplifier Consider the following nchannel differential amplifier (sometimes called a sourcecoupled pair): v I ID Bias M4 i D i D v G v G v GS v GS M3 I SS V Bulk Fig. 5. Where should bulk be connected? Consider a pwell, CMOS technology, D G S S G D n n p n n n pwell nsubstrate Fig. 5.3.) Bulks connected to the sources: No modulation of V T but large common mode parasitic capacitance..) Bulks connected to ground: Smaller common mode parasitic capacitors, but modulation of V T. If the technology is nwell CMOS, there is no choice. The bulks must be connected to ground.
14 CMOS Analog Circuit Design Page 5.4 Transconductance Characteristic of the Differential Amplifier Continued Defining equations: v ID = v GS v GS = i D / β i D / β and I SS = i D i D Solution: i D = I SS I SS βv ID I SS β v 4 ID 4I SS which are valid for v ID < (I SS /β) /. Illustration of the result: / and i D = I SS I SS βv ID I SS β v 4 ID 4I SS / i D /I SS i 0.6 D 0.4 i D 0. v ID (I SS /ß)0.5 Fig. 5.4 Differentiating i D (or i D ) with respect to v ID and setting V ID =0V gives g m = di D dv (V ID = 0) = (βi SS /4) / = K' I SS W / ID 4L (half the g m of an inverting amplifier) CMOS Analog Circuit Design Page 5.5 Voltage Transfer Characteristic of the Differential Amplifier In order to obtain the voltage transfer characteristic, a load for the differential amplifier must be defined. We will select a current mirror load as illustrated below. v GS v G µm µm M3 i D3 µm µm V Bias i D µm µm µm µm M4 i D4 i OUT µm i D µm v GS I SS v G M5 Fig. 5.5 Note that output signal to ground is equivalent to the differential output signal due to the current mirror. The shortcircuit, transconductance is given as g m = di OUT dv (V ID = 0) = (βi SS ) / = K' I SS W / ID L
15 CMOS Analog Circuit Design Page 5.6 Voltage Transfer Function of the Differential Amplifer with a Current Mirror Load vout (Volts) Regions of operation of the transistors: is saturated when, V IC = V M4 active M4 saturated saturated active v ID (Volts) Fig. 5.6 v DS v GS V TN V S V IC 0.5v ID V S V TN V IC V TN where we have assumed that the region of transition for is close to v ID = 0V. M4 is saturated when, v SD4 v SG4 V TP V SG4 V TP V SG4 V TP The regions of operations shown on the voltage transfer function assume I SS = 00µA. Note: V SG4 = V TP = V TP = 4V CMOS Analog Circuit Design Page 5.7 Differential Amplifier Using pchannel Input MOSFETs V Bias v G v SG i D i D3 M3 M5 IDD v SG i D i OUT i D4 v G v M4 OUT Fig. 5.7
16 CMOS Analog Circuit Design Page 5.8 Input Common Mode Range (ICMR) ICMR is found by setting v ID = 0 and varying v IC until one of the transistors leaves the saturation region. Highest Common Mode Voltage Path from G through and M3 to : V IC (max) =V G (max) =V G (max) = V SG3 V DS (sat) V GS or V IC (max) = V SG3 V TN Path from G through and M4 to : V IC (max) = V SD4 (sat) V DS (sat) V GS = V SD4 (sat) V TN V IC (max) = V SG3 V TN Lowest Common Mode Voltage V IC (min) = V DS5 (sat) V GS = V DS5 (sat) V GS v GS v G µm µm M3 i D3 µm µm V Bias i D µm µm µm µm M4 i D4 i OUT µm i D µm v GS I SS v G M5 Fig. 5.5 where we have assumed that V GS = V GS during changes in the input common mode voltage. CMOS Analog Circuit Design Page 5.9 Example 5. Calculation of the WorstCase Input CommonMode Range of the nchannel Input, Differential Amplifier Assume that varies from 4 to 6 volts and that V SS = 0, and use the values of Table 3. under worstcase conditions to calculate the input commonmode range of nchannel input differential amplifier with a current mirror load. Assume that I SS is 00 µa, W /L = W /L = 5, W 3 /L 3 = W 4 /L 4 =, and V DS5 (sat) = 0. V. Include worstcase variation in K' in calculations. Solution If varies 5 ± volts, then the upper commonmode voltage is, V IC (max) = 4 50µA 45µA/V = =. volts The lower commonmode voltage is, V IC (min) = µA 99µA/V = =.50 volts which gives a worstcase input commonmode range of 0.7 volts with a nominal 5volt power supply. Reducing by several volts more will result in a worstcase commonmode range of zero. We have assumed in this example that all bulksource voltages are zero.
17 CMOS Analog Circuit Design Page 5.0 SmallSignal Analysis of the DifferentialMode of the Differential Amplifier A requirement for differentialmode operation is that the differential amplifier is perfectly balanced. VDD v id M3 i D3 i D M4 i D4 i out i D v out G G v id v g v g C D=G3=D3=G4 C3 r ds S=S r ds i 3 r g r ds5 i 3 ds3 m3 gmv gs g m v gs S3 D=D4 r ds4 C S4 v out M5 I SS G G v D=G3=D3=G4 id V Bias i v 3 gs v gs C3 g m v gs r ds rds3 g m3 gm v gs i 3 C S=S=S3=S4 Differential Transconductance: Assume that the output of the differential amplifier is an ac short. rds D=D4 i out = g m g m3 r p g m3 r p v gs g m v gs g m v gs g m v gs = g md v id where g m = g m = g md, r p = r ds r ds3 and i' out designates the output current into a short circuit. r ds4 C i out ' v out Fig. 5.8 It can be shown that the current mirror causes this requirement to be invalid because the drain loads are not matched. However, we will continue to use the assumption regardless. See Problem 5.6. CMOS Analog Circuit Design Page 5. SmallSignal Analysis of the DifferentialMode of the Differential Amplifier Continued r out = = r g ds g ds r ds4 ds4 Differential Voltage Gain: A v = v out g md v = id g ds g ds4 If we assume that all transistors are in saturation and replace the small signal parameters of g m and r ds in terms of their largesignal model equivalents, we achieve A v = v out v = (K' I SS W /L ) / id (λ λ 4 )(I SS /) = K' W / λ λ 4 I SS L I SS Note that the smallsignal gain is inversely proportional to the square root of the bias current! Example: If W /L = µm/µm and I SS = 50µA (0µA), then A v (nchannel) = 46.6V/V (04.3V/V) v out v in A v (pchannel) = 3.4V/V (70.7V/V) r out = = g ds g ds4 5µA 0.09V = 0.444MΩ (.MΩ) Weak Inversion µa Stong Inversion log(i Bias ) ALA075
18 CMOS Analog Circuit Design Page 5. Common Mode Analysis for the Current Mirror Load Differential Amplifier The current mirror load differential amplifier is not a good example for common mode analysis because the current mirror rejects the common mode signal. M3 M3M4 M4 v out 0V v ic VBias M5 Fig. 5.8A Total common Common mode Common mode mode Output = output due to due to v ic M3M4 path output due to path Therefore The common mode output voltage should ideally be zero. Any voltage that exists at the output is due to mismatches in the gain between the two different paths. CMOS Analog Circuit Design Page 5.3 SmallSignal Analysis of the CommonMode of the Differential Amplifier The commonmode gain of the differential amplifier with a current mirror load is ideally zero. To illustrate the commonmode gain, we need a different type of load so we will consider the following: M3 M4 M3 M4 M3 v o vo v o v o v o M4 v o v v v id v id V Bias I SS M5 v ic I SS M5x V Bias I SS v ic DifferentialMode Analysis: Differentialmode circuit General circuit Commonmode circuit Fig. 5.9 and v o v id g m g m3 v o v g m id g m4 Note that these voltage gains are half of the active load inverter voltage gain.
19 CMOS Analog Circuit Design Page 5.4 SmallSignal Analysis of the CommonMode of the Differential Amplifier Continued CommonMode Analysis: v gs g m v gs v ic r ds5 rds r ds3 g m3 v o Fig. 5.0 Assume that r ds is large and can be ignored (greatly simplifies the analysis). v gs = v g v s = v ic g m r ds5 v gs Solving for v gs gives v ic v gs = g m r ds5 The singleended output voltage, v o, as a function of v ic can be written as v o v = g m [r ds3 (/g m3 )] (g m /g m3 ) ic g m r ds5 g m r g ds5 ds5 g m3 CommonMode Rejection Ratio (CMRR): CMRR = v o /v id v o /v ic = g m /g m3 g ds5 /g = g m3 m r ds5 How could you easily increase the CMRR of this differential amplifier? CMOS Analog Circuit Design Page 5.5 Frequency Response of the Differential Amplifer (Differential Mode) Back to the current mirror load differential amplifier: VDD C gs3 C gs4 M3 M4 C bd3 C bd4 v id C gd C gd4 C bd C bd M5 C gd v out C L G G v D=G3=D3=G4 id i v 3 gs v gs C3 gmv gs gm3 gmv gs i 3 C S=S=S3=S4 D=D4 i out ' rds rds4 C v out Fig. 5.0B V Bias Ignore the zeros that occur due to C gd, C gd and C gd4. C = C gd C bd C bd3 C gs3 C gs4, C = C bd C bd4 C gd C L and C 3 = C gd4 If C 3 0, then we can write g m V out (s) g ds g g m3 V ds4 g m3 sc gs (s) V gs (s) ω s ω where ω = g ds g ds4 C gm3 If we further assume that C >> g ds g ds4 C then the frequency response of the differential amplifier reduces to V out (s) V id (s) g m g ω ds g ds4 s ω (A more detailed analysis will be made later)
20 CMOS Analog Circuit Design Page 5.6 An Intuitive Method of Small Signal Analysis Small signal analysis is used so often in analog circuit design that it becomes desirable to find faster ways of performing this important analysis. Intuitive Analysis (or Schematic Analysis) Technique:.) Identify the transistor(s) that convert the input voltage to current (these transistors are called transconductance transistors)..) Trace the currents to where they flow into an equivalent resistance to ground. 3.) Multiply this resistance by the current to get the voltage at this node to ground. 4.) Repeat this process until the output is reached. Simple Example: R g m v in v o gm v o v out v in R Fig. 5.0C v o = (g m v in ) R v out = (g m v o )R v out = (g m R g m R )v in CMOS Analog Circuit Design Page 5.7 Intuitive Analysis of the CurrentMirror Load Differential Amplifier VDD M3 V Bias g m v id.) i = 0.5g m v id and i = 0.5g m v id.) i 3 = i = 0.5g m v id 3.) i 4 = i 3 = 0.5g m v id M5 g m v id M4 r out g m v id g m v id v id v id v id 4.) The resistance at the output node, r out, is r ds r ds4 or g ds g ds4 v out Fig. 5. g m v in g m v in 5.) v out = (0.5g m v id 0.5g m v id )r out = g ds g = ds4 g ds g ds4 v out g m v = in g ds g ds4
21 CMOS Analog Circuit Design Page 5.8 Some Concepts to Help Extend the Intuitive Method of SmallSignal Analysis.) Approximate the output resistance of any cascode circuit as R out (g m r ds )r ds where is a transistor cascoded by..) If there is a resistance, R, in series with the source of the transconductance transistor, let the effective transconductance be g m g m(eff) = g m R Proof: g m (eff)v in g m (eff)v in g m v gs i out v gs v in r ds v in v in r ds Smallsignal model V Bias Fig. 5.A v in v gs = v g v s = v in (g m r ds )v gs v gs = g m r ds g m v in Thus, i out = g m r = g ds m (eff) v in CMOS Analog Circuit Design Page 5.9 Slew Rate of the Differential Amplifier Slew Rate (SR) = Maximum outputvoltage rate (either positive or negative) d It is caused by the relationship, i OUT = C L dt. When i OUT is a constant, the rate becomes a constant. Consider the following currentmirror load, differential amplifiers: VDD M3 v GS v G V Bias i D3 i D M5 M4 i D4 i D i OUT v GS I SS v G C L V Bias v SG v G M3 i D i D3 M5 IDD v SG i D i OUT i D4 M4 C L v G Fig. 5.B Note that slew rate can only occur when the differential input signal is large enough to cause I SS (I DD ) to flow through only one of I the differential input transistors. SS SR = I DD C = L C If C L L = 5pF and I SS = 0µA, the slew rate is SR = V/µs. (For the BJT differential amplifier slewing occurs at ±00mV whereas for the MOSFET differential amplifier it can be ±V or more.)
22 CMOS Analog Circuit Design Page 5.0 Noise Analysis of the Differential Amplifier M5 V Bias e n e n e eq * * i to e n3 e n4 M3 * * M4 V out * M5 V Bias M3 M5 M4 Solve for the total outputnoise current to get, i to = gm e n gm e n gm3 e n3 gm4 e n4 Fig. 5.C This outputnoise current can be expressed in terms of an equivalent input noise voltage, e eq, given as i to = gm e eq Equating the above two expressions for the total outputnoise current gives, e eq = en en g m3 g e m n3 en4 /f Noise ( e n = e n and e n3 = e n4 ): Thermal Noise ( e n = e n and e n3 = e n4 ): e eq(/f) = B P fw L K N B N K L P B P L 3 e eq(th) = 6kT 3[K' (W/L) I ] / W 3 L K' 3 L / 3 W K' CMOS Analog Circuit Design Page 5. CurrentSource Load Differential Amplifier Gives a truly balanced differential amplifier. X M7 v 3 M3 X M4 X I 3 I 4 v4 I Bias v I I X X M6 M5 I 5 X X v Also, the upper input commonmode range is extended. However, a problem occurs if I I 3 or if I I 4. Fig. 5. Current Current I 3 I I 3 I 0 0 V DS <V DS (sat) v DS 0 0 V SD3 <V SD (sat) (a.) I>I3. (b.) I3>I. v DS Fig. 5.3
23 CMOS Analog Circuit Design Page 5. A DifferentialOutput, DifferentialInput Amplifier Probably the best way to solve the current mismatch problem is through the use of commonmode feedback. Consider the following solution to the previous problem. I Bias V CM MC3 Commonmode feedback circuit I C3 MC MCA MC4 I C4 MCB M3 M4 v 3 I 3 I 4 v v 4 Selfresistances of M4 v MB MC5 M5 V SS Fig. 5.4 Operation: Common mode output voltages are sensed at the gates of MCA and MCB and compared to V CM. The current in MC3 provides the negative feedback to drive the common mode output voltage to the desired level. CMOS Analog Circuit Design Page 5.3 Design of a CMOS Differential Amplifier with a Current Mirror Load Design Considerations: Constraints Specifications Power supply Technology Temperature Relationships A v = g m R out ω 3dB = /R out C L Smallsignal gain Frequency response (C L ) ICMR Slew rate (C L ) Power dissipation V IC (max) = V SG3 V TN V IC (min) = V SS V DS5 (sat) V GS = V SS V DS5 (sat) V GS vin V Bias M3 V SS I 5 M5 M4 ALA0 C L v out SR = I SS /C L P diss = ( V SS )xall dc currents flowing from or to V SS
24 CMOS Analog Circuit Design Page 5.4 Design of a CMOS Differential Amplifier with a Current Mirror Load Continued Schematicwise, the design procedure is illustrated as shown: M3 vin Min. ICMR VBias Max. ICMR V SG4 g m R out V SS M5 M4 ALA0 v out C L I 5 I 5 = SR C L, ω 3dB, P diss Procedure:.) Pick I SS to satisfy the slew rate knowing C L or the power dissipation.) Check to see if R out will satisfy the frequency response, if not change I SS or modify circuit 3.) Design W 3 /L 3 (W 4 /L 4 ) to satisfy the upper ICMR 4.) Design W /L (W /L ) to satisfy the gain 5.) Design W 5 /L 5 to satisfy the lower ICMR 6.) Iterate where necessary CMOS Analog Circuit Design Page 5.5 Example 5 Design of a MOS Differential Amplifier with a Current Mirror Load Design the currents and W/L values of the current mirror load MOS differential amplifier to satisfy the following specifications: = V SS =.5V, SR 0V/µs (C L =5pF), f 3dB 00kHz (C L =5pF), a small signal gain of 00V/V,.5V ICMR V and P diss mw. Use the parameters of K N =0µA/V, K P =50µA/V, V TN = 0.7V, V TP = 0.7V, λ N = 0.04V and λ P = 0.05V. Solution.) To meet the slew rate, I SS 50µA. For maximum P diss, I SS 00µA..) f 3dB of 00kHz implies that R out 38kΩ. This gives R out = (λ N λ P )I 38kΩ SS I SS 70µA Thus, pick I SS = 00µA 3.) V IC (max) = V SG3 V TN V =.5 V SG3 0.7 V SG3 =.V = W 3 L = W 4 3 L = 4 (0.5) = 8 g m 4.) 00V/V = g m R out = g ds g = ds4 5.) V IC (min) = V SS V DS5 (sat) V GS.5 =.5 V DS5 (sat) V DS5 (sat) = = !! 50µA 50µA/V (W 3 /L 3 ) 0.7 0µA/V (W /L ) ( ) 50µA = 3.3 W /L W L = W L = 8.4 W 5 L 5 = I SS K N V DS5 (sat) = µA 0µA/V (8.4) 0.7 We probably should increase W /L to reduce V GS and allow a smaller W 5 /L 5. If we choose W /L = 40, then W 5 /L 5 = 9. (Larger than specified gain should be okay.)
25 CMOS Analog Circuit Design Page 5.3 SECTION 5.3 CASCODE AMPLIFIER Why Use the Cascode Amplifier? Can provide higher output resistance and larger gain if the load is also high resistance. It reduces the Miller effect when the driving source has a large source resistance. V GG3 M3 v S V C GG gd R s R S v IN v Fig. 5.3 The Miller effect causes C gd to be increased by the value of (v /v in ) and appear in parallel with the gatesource of causing a dominant pole to occur. The cascode amplifier eliminates this dominant pole by keeping the value of v /v in small by making the value of R to be approximately /g m. CMOS Analog Circuit Design Page 5.3 LargeSignal Characteristics of the Cascode Amplifier v IN =5.0V v IN =4.5V 5V 0.5 v IN =4.0V M3 W v 3 = µm IN =3.5V L 3 µm 0.4 v IN =3.0V I.3V D v IN =.5V 0.3 K W G F = JIH E L µm µm 0. v IN =.0V 3.4V M3 v OUT 0. D W v = µm IN =.5V L v µm C A,B v IN IN =.0V A B C D 4 E M3 active 3 M3 saturated saturated active ID (ma) Fig. 5.3 is saturated when V GG V GS V GS V T vout F G H saturated active I J K 0 0 v 3 4 IN 5 v IN 0.5(V GG V TN ) where V GS = V GS is saturated when V DS V GS V TN V DS V GG V DS V TN V GG V TN M3 is saturated when V GG3 V TP V GG3 V TP
26 CMOS Analog Circuit Design Page 5.33 LargeSignal Voltage Swing Limits of the Cascode Amplifier Maximum output voltage, (max): (max) = Minimum output voltage, (min): Referencing all potentials to the negative power supply (ground in this case), we may express the current through each of the devices, through M3, as i D = β ( V T )v DS v DS β ( V T )v DS i D = β (V GG v DS V T )( v DS ) ( v DS ) and β (V GG v DS V T )( v DS ) i D3 = β 3 ( V GG3 V T3 ) where we have also assumed that both v DS and are small, and v IN =. Solving for by realizing that i D = i D = i D3 and β = β we get, (min) = β 3 β (V DD V GG3 V T3 ) V GG V T V T CMOS Analog Circuit Design Page 5.34 Example 5.3 Calculation of the Minimum Output Voltage for the Simple Cascode Amplifier (a.) Assume the values and parameters used for the cascode configuration plotted in the previous slide on the voltage transfer function and calculate the value of (min). (b.) Find the value of (max) and (min) where all transistors remain in saturation. Solution (a.) Using the previous result gives, (min) = 0.50 volts. We note that simulation gives a value of about 0.75 volts. If we include the influence of the channel modulation on M3 in the previous derivation, the calculated value is 0.6 volts which is closer. The difference is attributable to the assumption that both v DS and are small. (b.) The largest output voltage for which all transistors of the cascode amplifier are in saturation is given as (max) = V SD3 (sat) and the corresponding minimum output voltage is (min) = V DS (sat) V DS (sat). For the cascode amplifier of Fig. 5.3, these limits are 3.0V and.7v. Consequently, the range over which all transistors are saturated is quite small for a 5V power supply.
27 CMOS Analog Circuit Design Page 5.35 SmallSignal Midband Performance of the Cascode Amplifier Smallsignal model: Using nodal analysis, we can write, g m v gs = g m v G D=S D=D3 r ds v in = v v g m v r v gs gs rds ds3 out S=G=G3 Smallsignal model of cascode amplifier neglecting the bulk effect on. C r G D=S ds D=D3 v in g m v in rds C v r g g m v ds3 C 3 v out m Simplified equivalent model of the above circuit. Fig [g ds g ds g m ]v g ds v out = g m v in [g ds g m ]v (g ds g ds3 )v out = 0 Solving for v out /v in yields v out g m (g ds g m ) v = in g ds g ds g ds g ds3 g ds g ds3 g ds3 g m The smallsignal output resistance is, r out = [r ds r ds g m r ds r ds ] r ds3 r ds3 g m g ds3 = K' W L I D λ 3 CMOS Analog Circuit Design Page 5.36 SmallSignal Analysis of the Cascode Amplifier Continued It is of interest to examine the voltage gain of v /v in. From the previous nodal equations we can solve for this gain as, v v in = g m (g ds g ds3 ) g ds g ds g ds g ds3 g ds g ds3 g ds3 g g ds g ds3 m g g m ds3 g m g m g m = If the W/L ratios of and are identical and g ds = g ds3, then v /v in is approximately. Why is this gain instead of? Consider the smallsignal model looking into the source of : R s g m v s The voltage loop is written as, v s = (i g m v s )r ds i r ds3 i i B = i (r ds r ds3 ) g m r ds v s Solving this equation for the ratio of v s to i gives R s = v s i = r ds r ds3 g m r ds W L L W We see that R s is indeed equal to /g m if r ds r ds3. Thus, if g m g m, the voltage gain v /v in. Note that: r ds3 = 0 that R s /g m or r ds3 = r ds that R s /g m or r ds3 r ds g m r ds that R s r ds!!! Principle: The smallsignal resistance looking into the source of a MOSFET depends on the resistance connected from the drain of the MOSFET to ac ground. v s r ds i A r ds3 Fig. 5.34
28 CMOS Analog Circuit Design Page 5.37 Frequency Response of the Cascode Amplifier Smallsignal model (R S = 0): C r G D=S ds D=D3 v in g m v in rds C v r g g m v ds3 C 3 v out m Fig. 5.34A where C = C gd, C = C bd C bs C gs and C 3 = C bd C bd3 C gd C gd3 C L The nodal equations now become: (g m g ds g ds sc sc )v g ds v out = (g m sc )v in and (g ds g m )v (g ds g ds3 sc 3 )v out = 0 Solving for V out (s)/v in (s) gives, V out (s) V in (s) = (g m sc )(g ds g m ) as bs g ds g ds g ds3 (g m g ds g ds ) where a = C 3(g ds g ds g m ) C (g ds g ds3 ) C (g ds g ds3 ) g ds g ds g ds3 (g m g ds g ds ) and C 3 (C C ) b = g ds g ds g ds3 (g m g ds g ds ) CMOS Analog Circuit Design Page 5.38 A Simplified Method of Finding an Algebraic Expression for the Two Poles Assume that a general secondorder polynomial can be written as: P(s) = as bs = s p s p = s p s p p p Now if p >> p, then P(s) can be simplified as P(s) s s p p p Therefore we may write p and p in terms of a and b as p = a and p = a b Applying this to the previous problem gives, p = [g ds g ds g ds3 (g m g ds g ds )] C 3 (g ds g ds g m ) C (g ds g ds3 ) C (g ds g ds3 ) g ds3 C 3 The nondominant root p is given as p = [C 3(g ds g ds g m ) C (g ds g ds3 ) C (g ds g ds3 )] C 3 (C C ) g m C C Assuming that C, C, and C 3 are the same order of magnitude, and that g m is greater than g ds3, then p is smaller than p (closer to the origin). Therefore the approximation of p >> p is valid. Note that there is a righthalf plane zero at z = g m C.
29 CMOS Analog Circuit Design Page 5.39 Driving Amplifiers from a High Resistance Source The Miller Effect Examine the frequency response of a currentsource load inverter driven from a high resistance source: vin R s V GG R s v out V in R R s C s C C gs C V gm V C 3 R3 C 3 = C bd C bd C gd C = C gd R 3 = r ds r ds Fig Assuming the input is I in, the nodal equations are, [G s(c C )]V sc V out = I in and (g m sc )V [G 3 s(c C 3 )]V out = 0 where G = G s (=/R s ), G 3 = g ds g ds, C = C gs, C = C gd and C 3 = C bd C bd C gd. Solving for V out (s)/v in (s) gives V out (s) V in (s) = (sc g m )G G G 3 s[g 3 (C C )G (C C 3 )g m C ](C C C C 3 C C 3 )s or V out (s) V in (s) = g m [ s(c /g m )] G 3 [R (C C )R 3 (C C 3 )g m R R 3 C ]s(c C C C 3 C C 3 )R R 3 s Assuming that the poles are split allows the use of the previous technique to get, g m C p = R (C C ) R 3 (C C 3 ) g m R R 3 C g m R R 3 C and p C C C C 3 C C 3 The Miller effect has caused the input pole, /R C, to be decreased by a value of g m R 3. V out CMOS Analog Circuit Design Page 5.30 How does the Cascode Amplifier Solve the Miller Effect? The dominant pole of the inverting amplifier with a large source resistance was found to be p (inverter) = R (C C )R 3 (C C 3 )g m R R 3 C Now if a cascode amplifier is used, R 3, can be approximated as /g m of the cascoding transistor (assuming the drain sees an r ds to ac ground). p (cascode) = R (C C ) g (C m C 3 )g m R = g C m R (C C ) g (C m C 3 )R C R (C 3C ) Thus we see that p (cascode) >> p (inverter).
30 CMOS Analog Circuit Design Page 5.3 High Gain and High Output Resistance Cascode Amplifier If the load of the cascode amplifier is a cascode current source, then both high output resistance and high voltage gain is achieved. M4 V D=D3 GG4 M3 V GG3 g v m v g mbs v r ds g m3 v 4 g mbs3 v 4 r ds3 out V R out G D=S D4=S3 GG v out v in v in v r g m v ds v 4 r ds4 in G=G3=G4=S=S4 Fig The output resistance is, r out = [r ds r ds g m r ds r ds (η )] [r ds3 r ds4 g m3 r ds3 r ds4 (η 3 )] [g m r ds r ds ] [g m3 r ds3 r ds4 ] r out I.5 D λ λ K' (W/L) Knowing r out, the gain is simply λ 3 λ 4 K' 3 (W/L) 3 A v = g m r out g m {[g m r ds r ds ] [g m3 r ds3 r ds4 ]} K' (W/L) ID λ λ λ 3 λ 4 K' (W/L) K' 3 (W/L) 3 CMOS Analog Circuit Design Page 5.3 Example 5.3 Comparison of the Cascode Amplifier Performance Calculate the smallsignal voltage gain, output resistance, the dominant pole, and the nondominant pole for the lowgain, cascode amplifier and the highgain, cascode amplifier. Assume that I D = 00 microamperes, that all W/L ratios are µm/µm, and that the parameters of Table 3. are valid. The capacitors are assumed to be: C gd = 3.5 ff, C gs = 30 ff, C bsn = C bdn = 4 ff, C bsp = C bdp = ff, and C L = pf. Solution The lowgain, cascode amplifier has the following smallsignal performance: A v = 37.V/V R out = 5kΩ p g ds3 /C 3 =. MHz p g m /(C C ) = 605 MHz. The highgain, cascode amplifier has the following smallsignal performance: A v = 44V/V R out =.40 MΩ p /R out C 3 = 08 khz p g m /(C C ) = 579 MHz (Note that at this frequency, the drain of is shorted to ground by the load capacitance, C L )
31 CMOS Analog Circuit Design Page 5.33 Designing Cascode Amplifiers Pertinent design equations for the simple cascode amplifier. I = K PW 3 L 3 ( V GG3 V TP ) Fig V GG3 V GG V GG = V DS (sat) V GS v IN M3 I (max) = V SD3 (sat) =V I DD K P (W 3 /L 3 ) (min) =V DS (sat) V DS (sat) = I I K N (W /L ) K N (W /L ) I = P diss = (SR) C out A v = g m g ds3 = K N(W /L ) λ P I CMOS Analog Circuit Design Page 5.34 Example 5.33 Design of a Cascode Amplifier The specifications for a cascode amplifier are = 5V, P diss = mw, A v = 50V/V, (max) = 4V, and (min) =.5V. The slew rate with a 0pF load should be 0V/µs or greater. Solution The slew rate requires a current greater than 00µA while the power dissipation requires a current less than 00µA. Let us compromise with a current of 50µA. We will first begin with M3. W 3 I L = 3 K P [ (max)] = 50 50() = 6 From this find V GG3 : V GG3 = V TP I K P (W 3 /L 3 ) = = 3V W Next, L = (A v λ) I K = ( ) (50) N 0 =.73 To design W /L, we will first calculate V DS (sat) and use the (min) specification to define V DS (sat). I V DS (sat) = K N (W /L ) = = 0.8V Subtracting this value from.5v gives V DS (sat) = 0.7V. W L = I K N V DS (sat) = Finally, V GG = V DS (sat) = 5.57 I K N (W /L ) V TN = 0.8V 0.7V 0.7V =.V
32 CMOS Analog Circuit Design Page 5.4 SECTION 5.4 CURRENT AMPLIFIERS What is a Current Amplifier? An amplifier that has a defined outputinput current relationship Low input resistance High output resistance Application of current amplifiers: ii i i i i i o i o A i is RS A i i S RS Current Amplifier RL Current Amplifier R L Singleended input. Differential input. Fig. 5.4 R S >> R in and R out >> R L Advantages of current amplifiers: Currents are not restricted by the power supply voltages so that wider dynamic ranges are possible with lower power supply voltages. 3dB bandwidth of a current amplifier using negative feedback is independent of the closed loop gain. CMOS Analog Circuit Design Page 5.4 Frequency Response of a Current Amplifier with Current Feedback Consider the following current amplifier with resistive negative feedback applied. Assuming that the smallsignal resistance looking into the current amplifier is much less than R or R, R i A i v in v i o = A i (i i ) = A i R i in o Solving for i o gives i o = A i v in A i R v out = R i o = R A i R A v i in A o If A i (s) = s, then v out v = R ω in R A = R R A o s A i (s) = R A o R ω (A o ) A ω 3dB = ω A (A o ) Amplitude The unitygainbandwidth is GB = R R A o A ω o A (A o ) = R R (ω A A o ) = R R R R GB i A o 0dB i R i o Fig. 5.4 v out s ω A (A o ) A o ω A (A o ) ω A GB=(R /R )(A o ω A ) ω
33 CMOS Analog Circuit Design Page 5.43 Current Amplifier using the Simple Current Mirror i in I I i out i in i out R Current Amplifier R in = g R out = m λ I o and A i = W /L W /L. Frequency response: p = (g m g ds ) C C = (g m g ds ) C bd C gs C gs C gd C v R L in g m v in r ds C g m v in rds C 0 3 g m C bd C gs C gs C gd Note that the bandwidth can be almost doubled by including the resistor, R. (R removes C gs from p ) Fig CMOS Analog Circuit Design Page 5.44 Example 5.4 Performance of a Simple Current Mirror as a Current Amplifier Find the smallsignal current gain, A i, the input resistance, R in, the output resistance, R out, and the 3dB frequency in Hertz for the current amplifier of Fig. 5.43(a) if 0I = I = 00µA and W /L = 0W /L = 0µm/µm. Assume that C bd = 0fF, C gs = C gs = 00fF, and C gs = 50fF. Solution Ignoring channel modulation and mismatch effects, the smallsignal current gain, A i = W /L W /L 0A/A. The smallsignal input resistance, R in, is approximately /g m and is R in K N (/)0µA = 46.9µS =.3kΩ The smallsignal output resistance is equal to R out = λ N I = 50kΩ. The 3dB frequency is ω 3dB = 46.9µS 60fF = 80.4x06 radians/sec. f 3dB = 8.7 MHz
34 CMOS Analog Circuit Design Page 5.45 SelfBiased Cascode Current Mirror Implementation of a Current Amplifier i in R I I i out v in M3 M4 v out R in R Current Amplifier Fig g m, R out r ds g m4 r ds4, and A i = W /L W /L CMOS Analog Circuit Design Page 5.46 Example 5.4 Current Amplifier Implemented by the SelfBiased, Cascode Current Mirror Assume that I and I of the selfbiased cascode current mirror are 00µA. R has been designed to give a V ON of 0.V. Thus R = kω. Find the value of R in, R out, and A i if the W/L ratios of all transistors are 8µm/µm. Solution The input resistance requires g m which is = ms R in 000Ω 500Ω =.5kΩ From our knowledge of the cascode configuration, the small signal output resistance should be R out g m4 r ds4 r ds = (00µS)(50kΩ)(50kΩ) = 5MΩ Because V DS = V DS, the smallsignal current gain is A i = W /L W /L = Simulation results using the level model for this example give R in =.497kΩ R out = 64.7MΩ and A i =.000 A/A.
35 CMOS Analog Circuit Design Page 5.47 LowInput Resistance Current Amplifier To decrease R in below /g m requires the use of negative, shunt feedback. Consider the following example. i in I I iout I 3 M3 V GG3 i in i = 0 g m3 v gs3 v in v gs3 r v ds3 gs g m v gs r ds Fig Current Amplifier Feedback concept: Input resistance without feedback r ds. g m g m3 Loop gain g ds g assuming that the resistances of I ds3 and I 3 are very large. R in (no fb.) R in = Loop gain r ds g m r ds g m3 r = ds3 g m g m3 r ds3 Small signal analysis: i in = g m v gs g ds v gs3 and v gs3 = v in v gs = v in (g m3 v gs3 r ds3 ) = v in (g m3 r ds3 ) i in = g m (g m3 r ds3 )v in g ds v in g m g m3 r ds3 v in R in g m g m3 r ds3 CMOS Analog Circuit Design Page 5.48 DifferentialInput, Current Amplifiers Definitions for the differentialmode, i ID, and commonmode, i IC, input currents of the differentialinput current amplifier. i IC i ID i IC i i io Fig i O = A ID i ID ± A IC i IC = A ID (i i ) ± A i i IC Implementations: i i I I I i O i M3 M4 i i M3 M4 i O i i V GG M5 V GG M6 Fig. 5.47
36 CMOS Analog Circuit Design Page 5.49 Summary Current amplifiers have a low input resistance, high output resistance, and a defined outputinput current relationship Input resistances less than /g m require feedback However, all feedback loops have internal poles that cause the benefits of negative feedback to vanish at high frequencies. In addition, these feedback loops can have a slow time constant from a polezero pair. Voltage amplifiers using a current amplifier have high values of gainbandwidth Current amplifiers are useful at low power supplies and for switched current applications CMOS Analog Circuit Design Page 5.5 SECTION 5.5 OUTPUT AMPLIFIERS General Considerations of Output Amplifiers Requirements:.) Provide sufficient output power in the form of voltage or current..) Avoid signal distortion. 3.) Be efficient 4.) Provide protection from abnormal conditions (short circuit, over temperature, etc.) Types of Output Amplifiers:.) Class A amplifiers.) Source followers 3.) Pushpull amplifiers 4.) Substrate BJT amplifiers 5.) Amplifiers using negative shunt feedback
37 CMOS Analog Circuit Design Page 5.5 Class A Amplifiers Current source load inverter: V GG IQ i D i OUT V SS R L I Q i D R L dominates as the load line v IN C L V SS R L V SS I Q R L I Q R L Fig. 5.5 A Class A circuit has current flow in the MOSFETs during the entire period of a sinusoidal signal. Characteristics of Class A amplifiers: Unsymmetrical sinking and sourcing Linear Poor efficiency (peak) P RL R L Efficiency = P = Supply ( V SS )I = Q (peak) R L (peak) = ( V SS ) V DD V SS ( V SS ) R L Maximum efficiency occurs when (peak) = = V SS which gives 5%. CMOS Analog Circuit Design Page 5.53 Specifying the Performance of a Class A Amplifier Output resistance: r out = g ds g = ds (λ λ )I D Current: Maximum sinking current is, I ŌUT = K' W L (V DD V SS V T ) I Q Maximum sourcing current is, I OUT = K' W L (V DD V GG V T ) I Q Requirements: Want r out << R L I OUT > C L SR I OUT > (peak) R L The maximum current will be determined by both the current required to provide the necessary slew rate (C L ) and the current required to provide a voltage across the load resistor (R L ).
38 CMOS Analog Circuit Design Page 5.54 SmallSignal Performance of the Class A Amplifier Although we have considered the smallsignal performance of the Class A amplifier as the current source load inverter, let us include the influence of the load. The modified smallsignal model: The smallsignal voltage gain is: v out v in = g m g ds g ds G L C v in g m v in rds r ds R L The smallsignal frequency response includes: A zero at z = g m C gd and a pole at (g ds g ds G L ) p = C gd C gd C bd C bd C L C vout Fig. 5.5 CMOS Analog Circuit Design Page 5.55 Example 5.5 Design of a Simple ClassA Output Stage Use the values of Table 3. and design the W/L ratios of and so that a voltage swing of ± volts and a slew rate of volt/µs is achieved if R L = 0 kω and C L = 000 pf. Assume that = V SS = 3 volts and V GG = 0 volts. Let the channel lengths be µm and assume that C gd = 00fF. Solution Let us first consider the effects of R L and C L. i OUT (peak) = ±V 0kΩ = ±00µA and CL SR = = 000µA Since the slew rate current is so much larger than the current needed to meet the voltage specification across R L, we can safely assume that all of the current supplied by the inverter is available to charge C L. Using a value of ± ma, W (I OUT I Q ) L = K N ( V SS V TN ) = (5.3) 3µm µm and W I OUT L = K P ( V GG V TP ) = (.3) 5µm µm The smallsignal performance of this amplifier is, A v = 8. V/V (includes R L = 0kΩ) The roots are, Zero = g m /C gd.59ghz and r out = 50kΩ Pole = /[(R L r out )C L )].4kHz
39 CMOS Analog Circuit Design Page 5.56 Broadband Harmonic Distortion The linearity of an amplifier can be characterized by its influence on a pure sinusoidal input signal. Assume the input is, V in (ω) = V p sin(ωt) The output of an amplifier with distortion will be V out (ω) = a V p sin (ωt) a V p sin (ωt)... an V p sin(nωt) Harmonic distortion (HD) for the ith harmonic can be defined as the ratio of the magnitude of the ith harmonic to the magnitude of the fundamental. For example, secondharmonic distortion would be given as HD = a a Total harmonic distortion (THD) is defined as the square root of the ratio of the sum of all of the second and higher harmonics to the magnitude of the first or fundamental harmonic. Thus, THD can be expressed as THD = [a a3... an] / a The distortion of the class A amplifier is good for small signals and becomes poor at maximum output swings because of the nonlinearity of the voltage transfer curve for largesignal swing CMOS Analog Circuit Design Page 5.57 Source Follower Two types of source followers: v IN V SS i OUT vout v IN V GG V SS i OUT vout V SS Source follower with a MOS diode load. Large signal considerations: V SS Source Follower with a currentsink load. Fig (min) V SS and (max) = V T V ON V T But, we need to include the bulk effect on V T : V T = V T0 γ φ F v BS φ F V T0 γ v SB = V T0 γ (max) V SS (max) V SS V T0 γ (max) V SS or putting this equation in quadratic form, (max) V SS γ (max) V SS ( V T0 ) = 0 Solving the quadratic gives, (max) γ 4 V T0 γ γ 4( V SS V T0 ) Therefore if = V SS =.5V, then (max) =.46V
Chapter 5 Introduction (2/25/03) Page 5.0-1
Chapter 5 Introduction (/5/03) Page 5.0 CHAPTER 5 CMOS AMPLIFIERS Chapter Outline 5. Inverters 5. Differential Amplifiers 5.3 Cascode Amplifiers 5.4 Current Amplifiers 5.5 Output Amplifiers 5.6 HighGain
More informationLECTURE 19 DIFFERENTIAL AMPLIFIER
Lecture 19 Differential Amplifier (6/4/14) Page 191 LECTURE 19 DIFFERENTIAL AMPLIFIER LECTURE ORGANIZATION Outline Characterization of a differential amplifier Differential amplifier with a current mirror
More informationLecture 040 CE and CS Output Stages (1/11/04) Page ECE Analog Integrated Circuits and Systems II P.E. Allen
Lecture 040 CE and CS Output Stages (1/11/04) Page 0401 LECTURE 040 COMMON SOURCE AND EMITTER OUTPUT STAGES (READING: GHLM 8498, AH 181) Objective The objective of this presentation is: Show how to design
More informationLecture 240 Cascode Op Amps (3/28/10) Page 240-1
Lecture 240 Cascode Op Amps (3/28/10) Page 2401 LECTURE 240 CASCODE OP AMPS LECTURE ORGANIZATION Outline Lecture Organization Single Stage Cascode Op Amps Two Stage Cascode Op Amps Summary CMOS Analog
More informationLecture 030 ECE4430 Review III (1/9/04) Page 030-1
Lecture 030 ECE4430 Review III (1/9/04) Page 0301 LECTURE 030 ECE 4430 REVIEW III (READING: GHLM Chaps. 3 and 4) Objective The objective of this presentation is: 1.) Identify the prerequisite material
More informationLecture 350 Low Voltage Op Amps (3/26/02) Page 350-1
Lecture 350 Low Voltage Op Amps (3/26/02) Page 3501 LECTURE 350 LOW VOLTAGE OP AMPS (READING: AH 415432) Objective The objective of this presentation is: 1.) How to design standard circuit blocks with
More informationLecture 300 Low Voltage Op Amps (3/28/10) Page 300-1
Lecture 300 Low Voltage Op Amps (3/28/10) Page 300-1 LECTURE 300 LOW VOLTAGE OP AMPS LECTURE ORGANIZATION Outline Introduction Low voltage input stages Low voltage gain stages Low voltage bias circuits
More informationAdvanced Operational Amplifiers
IsLab Analog Integrated Circuit Design OPA2-47 Advanced Operational Amplifiers כ Kyungpook National University IsLab Analog Integrated Circuit Design OPA2-1 Advanced Current Mirrors and Opamps Two-stage
More informationLecture 200 Cascode Op Amps - II (2/18/02) Page 200-1
Lecture 200 Cascode Op Amps II (2/18/02) Page 2001 LECTURE 200 CASCODE OP AMPS II (READING: GHLM 443453, AH 293309) Objective The objective of this presentation is: 1.) Develop cascode op amp architectures
More informationLecture 330 Low Power Op Amps (3/27/02) Page 330-1
Lecture 33 Low Power Op Amps (3/27/2) Page 33 LECTURE 33 LOW POWER OP AMPS (READING: AH 39342) Objective The objective of this presentation is:.) Examine op amps that have minimum static power Minimize
More informationLecture 34: Designing amplifiers, biasing, frequency response. Context
Lecture 34: Designing amplifiers, biasing, frequency response Prof J. S. Smith Context We will figure out more of the design parameters for the amplifier we looked at in the last lecture, and then we will
More informationECEN 474/704 Lab 6: Differential Pairs
ECEN 474/704 Lab 6: Differential Pairs Objective Design, simulate and layout various differential pairs used in different types of differential amplifiers such as operational transconductance amplifiers
More informationChapter 4. CMOS Cascode Amplifiers. 4.1 Introduction. 4.2 CMOS Cascode Amplifiers
Chapter 4 CMOS Cascode Amplifiers 4.1 Introduction A single stage CMOS amplifier cannot give desired dc voltage gain, output resistance and transconductance. The voltage gain can be made to attain higher
More informationChapter 5. Operational Amplifiers and Source Followers. 5.1 Operational Amplifier
Chapter 5 Operational Amplifiers and Source Followers 5.1 Operational Amplifier In single ended operation the output is measured with respect to a fixed potential, usually ground, whereas in double-ended
More informationIOWA STATE UNIVERSITY. EE501 Project. Fully Differential Multi-Stage Op-Amp Design. Ryan Boesch 11/12/2008
IOWA STATE UNIVERSITY EE501 Project Fully Differential Multi-Stage Op-Amp Design Ryan Boesch 11/12/2008 This report documents the design, simulation, layout, and post-layout simulation of a fully differential
More informationChapter 12 Opertational Amplifier Circuits
1 Chapter 12 Opertational Amplifier Circuits Learning Objectives 1) The design and analysis of the two basic CMOS op-amp architectures: the two-stage circuit and the single-stage, folded cascode circuit.
More informationRadivoje Đurić, 2015, Analogna Integrisana Kola 1
Low power OTA 1 Two-Stage, Miller Op Amp Operating in Weak Inversion Low frequency response: gm1 gm6 Av 0 g g g g A v 0 ds2 ds4 ds6 ds7 I D m, ds D nvt g g I n GB and SR: GB 1 1 n 1 2 4 6 6 7 g 2 2 m1
More informationDesign Analysis and Performance Comparison of Low Power High Gain 2nd Stage Differential Amplifier Along with 1st Stage
Design Analysis and Performance Comparison of Low Power High Gain 2nd Stage Differential Amplifier Along with 1st Stage Sadeque Reza Khan Department of Electronic and Communication Engineering, National
More informationDesign and Simulation of Low Voltage Operational Amplifier
Design and Simulation of Low Voltage Operational Amplifier Zach Nelson Department of Electrical Engineering, University of Nevada, Las Vegas 4505 S Maryland Pkwy, Las Vegas, NV 89154 United States of America
More informationOperational Amplifiers
CHAPTER 9 Operational Amplifiers Analog IC Analysis and Design 9- Chih-Cheng Hsieh Outline. General Consideration. One-Stage Op Amps / Two-Stage Op Amps 3. Gain Boosting 4. Common-Mode Feedback 5. Input
More informationECE 442 Solid State Devices & Circuits. 15. Differential Amplifiers
ECE 442 Solid State Devices & Circuits 15. Differential Amplifiers Jose E. Schutt-Aine Electrical & Computer Engineering University of Illinois jschutt@emlab.uiuc.edu ECE 442 Jose Schutt Aine 1 Background
More informationSolid State Devices & Circuits. 18. Advanced Techniques
ECE 442 Solid State Devices & Circuits 18. Advanced Techniques Jose E. Schutt-Aine Electrical l&c Computer Engineering i University of Illinois jschutt@emlab.uiuc.edu 1 Darlington Configuration - Popular
More informationAnalog Integrated Circuit Design Exercise 1
Analog Integrated Circuit Design Exercise 1 Integrated Electronic Systems Lab Prof. Dr.-Ing. Klaus Hofmann M.Sc. Katrin Hirmer, M.Sc. Sreekesh Lakshminarayanan Status: 21.10.2015 Pre-Assignments The lecture
More informationTWO AND ONE STAGES OTA
TWO AND ONE STAGES OTA F. Maloberti Department of Electronics Integrated Microsystem Group University of Pavia, 7100 Pavia, Italy franco@ele.unipv.it tel. +39-38-50505; fax. +39-038-505677 474 EE Department
More informationChapter 15 Goals. ac-coupled Amplifiers Example of a Three-Stage Amplifier
Chapter 15 Goals ac-coupled multistage amplifiers including voltage gain, input and output resistances, and small-signal limitations. dc-coupled multistage amplifiers. Darlington configuration and cascode
More informationLecture 16: MOS Transistor models: Linear models, SPICE models. Context. In the last lecture, we discussed the MOS transistor, and
Lecture 16: MOS Transistor models: Linear models, SPICE models Context In the last lecture, we discussed the MOS transistor, and added a correction due to the changing depletion region, called the body
More informationChapter 11. Differential Amplifier Circuits
Chapter 11 Differential Amplifier Circuits 11.0 ntroduction Differential amplifier or diff-amp is a multi-transistor amplifier. t is the fundamental building block of analog circuit. t is virtually formed
More informationECEN 474/704 Lab 5: Frequency Response of Inverting Amplifiers
ECEN 474/704 Lab 5: Frequency Response of Inverting Amplifiers Objective Design, simulate and layout various inverting amplifiers. Introduction Inverting amplifiers are fundamental building blocks of electronic
More informationAnalysis and Design of Analog Integrated Circuits Lecture 18. Key Opamp Specifications
Analysis and Design of Analog Integrated Circuits Lecture 8 Key Opamp Specifications Michael H. Perrott April 8, 0 Copyright 0 by Michael H. Perrott All rights reserved. Recall: Key Specifications of Opamps
More informationCHAPTER 7 - HIGH-PERFORMANCE CMOS OPERATIONAL AMPLIFIERS SECTION BUFFERED OP AMPS
CMOS Analog Circuit Design Page 7.01 CHAPTER 7 HIGHPERFORMANCE CMOS OPERATIONAL AMPLIFIERS Chapter Outline 7.1 Buffered Op Amps 7.2 HighSpeed/Frequency Op Amps 7.3 Differential Output Op Amps 7.4 Micropower
More informationMicroelectronics Part 2: Basic analog CMOS circuits
GBM830 Dispositifs Médicaux Intelligents Microelectronics Part : Basic analog CMOS circuits Mohamad Sawan et al. Laboratoire de neurotechnologies Polystim!! http://www.cours.polymtl.ca/gbm830/! mohamad.sawan@polymtl.ca!
More informationHomework Assignment 07
Homework Assignment 07 Question 1 (Short Takes). 2 points each unless otherwise noted. 1. A single-pole op-amp has an open-loop low-frequency gain of A = 10 5 and an open loop, 3-dB frequency of 4 Hz.
More informationMicroelectronic Circuits II. Ch 10 : Operational-Amplifier Circuits
Microelectronic Circuits II Ch 0 : Operational-Amplifier Circuits 0. The Two-stage CMOS Op Amp 0.2 The Folded-Cascode CMOS Op Amp CNU EE 0.- Operational-Amplifier Introduction - Analog ICs : operational
More informationMicroelectronics Exercises of Topic 5 ICT Systems Engineering EPSEM - UPC
Microelectronics Exercises of Topic 5 ICT Systems Engineering EPSEM - UPC F. Xavier Moncunill Autumn 2018 5 Analog integrated circuits Exercise 5.1 This problem aims to follow the steps in the design of
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science Microelectronic Devices and Circuits Fall 2009
1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.012 Microelectronic Devices and Circuits Fall 2009 SPECIAL PROBLEM ON CIRCUIT DESIGN 12/1/09 edition
More informationEE 140 / EE 240A ANALOG INTEGRATED CIRCUITS FALL 2015 C. Nguyen PROBLEM SET #7
Issued: Friday, Oct. 16, 2015 PROBLEM SET #7 Due (at 8 a.m.): Monday, Oct. 26, 2015, in the EE 140/240A HW box near 125 Cory. 1. A design error has resulted in a mismatch in the circuit of Fig. PS7-1.
More informationBasic Circuits. Current Mirror, Gain stage, Source Follower, Cascode, Differential Pair,
Basic Circuits Current Mirror, Gain stage, Source Follower, Cascode, Differential Pair, CCS - Basic Circuits P. Fischer, ZITI, Uni Heidelberg, Seite 1 Reminder: Effect of Transistor Sizes Very crude classification:
More informationChapter 8 Differential and Multistage Amplifiers
1 Chapter 8 Differential and Multistage Amplifiers Operational Amplifier Circuit Components 2 1. Ch 7: Current Mirrors and Biasing 2. Ch 9: Frequency Response 3. Ch 8: Active-Loaded Differential Pair 4.
More informationHomework Assignment 12
Homework Assignment 12 Question 1 Shown the is Bode plot of the magnitude of the gain transfer function of a constant GBP amplifier. By how much will the amplifier delay a sine wave with the following
More informationPreliminary Exam, Fall 2013 Department of Electrical and Computer Engineering University of California, Irvine EECS 170B
Preliminary Exam, Fall 2013 Department of Electrical and Computer Engineering University of California, Irvine EECS 170B Problem 1. Consider the following circuit, where a saw-tooth voltage is applied
More informationDesign of Miller Compensated Two-Stage Operational Amplifier for Data Converter Applications
Design of Miller Compensated Two-Stage Operational Amplifier for Data Converter Applications Prema Kumar. G Shravan Kudikala Casest, School Of Physics Casest, School Of Physics University Of Hyderabad
More informationCommon mode rejection ratio
Common mode rejection ratio Definition: Common mode rejection ratio represents the ratio of the differential voltage gaina d tothecommonmodevoltagegain,a cm : Common mode rejection ratio Definition: Common
More informationEECE488: Analog CMOS Integrated Circuit Design Set 7 Opamp Design
EECE488: Analog CMOS Integrated Circuit Design Set 7 Opamp Design References: Analog Integrated Circuit Design by D. Johns and K. Martin and Design of Analog CMOS Integrated Circuits by B. Razavi All figures
More informationImproving Amplifier Voltage Gain
15.1 Multistage ac-coupled Amplifiers 1077 TABLE 15.3 Three-Stage Amplifier Summary HAND ANALYSIS SPICE RESULTS Voltage gain 998 1010 Input signal range 92.7 V Input resistance 1 M 1M Output resistance
More informationMOS Field Effect Transistors
MOS Field Effect Transistors A gate contact gate interconnect n polysilicon gate source contacts W active area (thin oxide area) polysilicon gate contact metal interconnect drain contacts A bulk contact
More informationA 16Ω Audio Amplifier with 93.8 mw Peak loadpower and 1.43 quiscent power consumption
A 16Ω Audio Amplifier with 93.8 mw Peak loadpower and 1.43 quiscent power consumption IEEE Transactions on circuits and systems- Vol 59 No:3 March 2012 Abstract A class AB audio amplifier is used to drive
More informationA CMOS Low-Voltage, High-Gain Op-Amp
A CMOS Low-Voltage, High-Gain Op-Amp G N Lu and G Sou LEAM, Université Pierre et Marie Curie Case 203, 4 place Jussieu, 75252 Paris Cedex 05, France Telephone: (33 1) 44 27 75 11 Fax: (33 1) 44 27 48 37
More informationCMOS Operational Amplifier
The George Washington University Department of Electrical and Computer Engineering Course: ECE218 Instructor: Mona E. Zaghloul Students: Shunping Wang Yiping (Neil) Tsai Data: 05/14/07 Introduction In
More informationHomework Assignment 07
Homework Assignment 07 Question 1 (Short Takes). 2 points each unless otherwise noted. 1. A single-pole op-amp has an open-loop low-frequency gain of A = 10 5 and an open loop, 3-dB frequency of 4 Hz.
More informationCMOS Cascode Transconductance Amplifier
CMOS Cascode Transconductance Amplifier Basic topology. 5 V I SUP v s V G2 M 2 iout C L v OUT Device Data V Tn = 1 V V Tp = 1 V µ n C ox = 50 µa/v 2 µ p C ox = 25 µa/v 2 λ n = 0.05 V 1 λ p = 0.02 V 1 @
More informationBJT Amplifier. Superposition principle (linear amplifier)
BJT Amplifier Two types analysis DC analysis Applied DC voltage source AC analysis Time varying signal source Superposition principle (linear amplifier) The response of a linear amplifier circuit excited
More informationHomework Assignment 06
Homework Assignment 06 Question 1 (Short Takes) One point each unless otherwise indicated. 1. Consider the current mirror below, and neglect base currents. What is? Answer: 2. In the current mirrors below,
More informationDigital Electronics. Assign 1 and 0 to a range of voltage (or current), with a separation that minimizes a transition region. Positive Logic.
Digital Electronics Assign 1 and 0 to a range of voltage (or current), with a separation that minimizes a transition region Positive Logic Logic 1 Negative Logic Logic 0 Voltage Transition Region Transition
More information6.976 High Speed Communication Circuits and Systems Lecture 5 High Speed, Broadband Amplifiers
6.976 High Speed Communication Circuits and Systems Lecture 5 High Speed, Broadband Amplifiers Michael Perrott Massachusetts Institute of Technology Copyright 2003 by Michael H. Perrott Broadband Communication
More informationUnit 3: Integrated-circuit amplifiers (contd.)
Unit 3: Integrated-circuit amplifiers (contd.) COMMON-SOURCE AND COMMON-EMITTER AMPLIFIERS The Common-Source Circuit The most basic IC MOS amplifier is shown in fig.(1). The source of MOS transistor is
More informationYou will be asked to make the following statement and provide your signature on the top of your solutions.
1 EE 435 Name Exam 1 Spring 2018 Instructions: The points allocated to each problem are as indicated. Note that the first and last problem are weighted more heavily than the rest of the problems. On those
More informationNonlinear Macromodeling of Amplifiers and Applications to Filter Design.
ECEN 622(ESS) Nonlinear Macromodeling of Amplifiers and Applications to Filter Design. By Edgar Sanchez-Sinencio Thanks to Heng Zhang for part of the material OP AMP MACROMODELS Systems containing a significant
More informationRail-To-Rail Output Op-Amp Design with Negative Miller Capacitance Compensation
Rail-To-Rail Op-Amp Design with Negative Miller Capacitance Compensation Muhaned Zaidi, Ian Grout, Abu Khari bin A ain Abstract In this paper, a two-stage op-amp design is considered using both Miller
More informationLecture 110 Intro. and Characterization of the Op Amp (1/28/02) Page 110-1
Lecture 110 Intro. and Characterization of the Op Amp (1/28/02) Page 1101 LECTURE 110 INTRODUCTION AND CHARACTERIZATION OF THE OP AMP (READING: GHLM 404424, AH 243249) Objective The objective of this presentation
More informationNonlinear Macromodeling of Amplifiers and Applications to Filter Design.
ECEN 622 Nonlinear Macromodeling of Amplifiers and Applications to Filter Design. By Edgar Sanchez-Sinencio Thanks to Heng Zhang for part of the material OP AMP MACROMODELS Systems containing a significant
More informationWeek 9a OUTLINE. MOSFET I D vs. V GS characteristic Circuit models for the MOSFET. Reading. resistive switch model small-signal model
Week 9a OUTLINE MOSFET I vs. V GS characteristic Circuit models for the MOSFET resistive switch model small-signal model Reading Rabaey et al.: Chapter 3.3.2 Hambley: Chapter 12 (through 12.5); Section
More informationSKEL 4283 Analog CMOS IC Design Current Mirrors
SKEL 4283 Analog CMOS IC Design Current Mirrors Dr. Nasir Shaikh Husin Faculty of Electrical Engineering Universiti Teknologi Malaysia Current Mirrors 1 Objectives Introduce and characterize the current
More informationUNISONIC TECHNOLOGIES CO., LTD LM321
UNISONIC TECHNOLOGIES CO., LTD LM321 LOW POWER SINGLE OP AMP DESCRIPTION The UTC LM321 s quiescent current is only 430µA (5V). The UTC LM321 brings performance and economy to low power systems, With a
More informationAmplifiers Frequency Response Examples
ECE 5/45 Analog IC Design We will use the following MOSFET parameters for hand-calculations and the µm CMOS models for corresponding simulations. Table : Long-channel MOSFET parameters. Parameter NMOS
More informationECE 546 Lecture 12 Integrated Circuits
ECE 546 Lecture 12 Integrated Circuits Spring 2018 Jose E. Schutt-Aine Electrical & Computer Engineering University of Illinois jesa@illinois.edu ECE 546 Jose Schutt Aine 1 Integrated Circuits IC Requirements
More informationClass-AB Low-Voltage CMOS Unity-Gain Buffers
Class-AB Low-Voltage CMOS Unity-Gain Buffers Mariano Jimenez, Antonio Torralba, Ramón G. Carvajal and J. Ramírez-Angulo Abstract Class-AB circuits, which are able to deal with currents several orders of
More informationYou will be asked to make the following statement and provide your signature on the top of your solutions.
1 EE 435 Name Exam 1 Spring 216 Instructions: The points allocated to each problem are as indicated. Note that the first and last problem are weighted more heavily than the rest of the problems. On those
More informationCourse Outline. 4. Chapter 5: MOS Field Effect Transistors (MOSFET) 5. Chapter 6: Bipolar Junction Transistors (BJT)
Course Outline 1. Chapter 1: Signals and Amplifiers 1 2. Chapter 3: Semiconductors 3. Chapter 4: Diodes 4. Chapter 5: MOS Field Effect Transistors (MOSFET) 5. Chapter 6: Bipolar Junction Transistors (BJT)
More informationSAMPLE FINAL EXAMINATION FALL TERM
ENGINEERING SCIENCES 154 ELECTRONIC DEVICES AND CIRCUITS SAMPLE FINAL EXAMINATION FALL TERM 2001-2002 NAME Some Possible Solutions a. Please answer all of the questions in the spaces provided. If you need
More informationCMOS Operational-Amplifier
CMOS Operational-Amplifier 1 What will we learn in this course How to design a good OP Amp. Basic building blocks Biasing and Loading Swings and Bandwidth CH2(8) Operational Amplifier as A Black Box Copyright
More informationLow Quiescent Power CMOS Op-Amp in 0.5µm Technology
Kevin Fronczak - Low Power CMOS Op-Amp - Rochester Institute of Technology EE610 1 Low Quiescent Power CMOS Op-Amp in 0.5µm Technology Kevin C. Fronczak Abstract This paper analyzes a low quiescent power
More informationCurrent Mirrors. Prof. Tai-Haur Kuo, EE, NCKU, Tainan City, Taiwan 4-1
Current Mirrors Prof. Tai-Haur Kuo, EE, NCKU, Tainan City, Taiwan 4- 郭泰豪, Analog C Design, 08 { Prof. Tai-Haur Kuo, EE, NCKU, Tainan City, Taiwan 4- 郭泰豪, Analog C Design, 08 { Current Source and Sink Symbol
More informationApplied Electronics II
Applied Electronics II Chapter 2: Differential Amplifier School of Electrical and Computer Engineering Addis Ababa Institute of Technology Addis Ababa University Daniel D./Abel G. April 4, 2016 Chapter
More information6.012 Microelectronic Devices and Circuits
Page 1 of 13 YOUR NAME Department of Electrical Engineering and Computer Science Massachusetts Institute of Technology 6.012 Microelectronic Devices and Circuits Final Eam Closed Book: Formula sheet provided;
More informationLaboratory 1 Single-Stage MOSFET Amplifier Analysis and Design Due Date: Week of February 20, 2014, at the beginning of your lab section
Laboratory 1 Single-Stage MOSFET Amplifier Analysis and Design Due Date: Week of February 20, 2014, at the beginning of your lab section Objective To analyze and design single-stage common source amplifiers.
More informationINF3410 Fall Book Chapter 3: Basic Current Mirrors and Single-Stage Amplifiers
INF3410 Fall 2013 Amplifiers content Simple Current Mirror Common-Source Amplifier Interrupt: A word on output resistance Common-Drain Amplifier with active load / Source Follower Common-Gate Amplifier
More informationCMOS Operational-Amplifier
CMOS Operational-Amplifier 1 What will we learn in this course How to design a good OP Amp. Basic building blocks Biasing and Loading Swings and Bandwidth CH2(8) Operational Amplifier as A Black Box Copyright
More informationDesign of a low voltage,low drop-out (LDO) voltage cmos regulator
Design of a low,low drop-out (LDO) cmos regulator Chaithra T S Ashwini Abstract- In this paper a low, low drop-out (LDO) regulator design procedure is proposed and implemented using 0.25 micron CMOS process.
More informationRadivoje Đurić, 2015, Analogna Integrisana Kola 1
OTA-output buffer 1 According to the types of loads, the driving capability of the output stages differs. For switched capacitor circuits which have high impedance capacitive loads, class A output stage
More informationCHAPTER 8 DIFFERENTIAL AND MULTISTAGE AMPLIFIERS
CHAPTER 8 DIFFERENTIAL AND MULTISTAGE AMPLIFIERS Chapter Outline 8.1 The CMOS Differential Pair 8. Small-Signal Operations of the MOS Differential Pair 8.3 The BJT Differential Pair 8.4 Other Non-ideal
More informationINF3410 Fall Book Chapter 6: Basic Opamp Design and Compensation
INF3410 Fall 2015 Book Chapter 6: Basic Opamp Design and Compensation content Introduction Two Stage Opamps Compensation Slew Rate Systematic Offset Advanced Current Mirrors Operational Transconductance
More informationA High-Gain, Low-Power CMOS Operational Amplifier Using Composite Cascode Stage in the Subthreshold Region
Brigham Young University BYU ScholarsArchive All Theses and Dissertations 2011-03-15 A High-Gain, Low-Power CMOS Operational Amplifier Using Composite Cascode Stage in the Subthreshold Region Rishi Pratap
More informationECE315 / ECE515 Lecture 7 Date:
Lecture 7 ate: 01.09.2016 CG Amplifier Examples Biasing in MOS Amplifier Circuits Common Gate (CG) Amplifier CG Amplifier- nput is applied at the Source and the output is sensed at the rain. The Gate terminal
More informationECE4902 C Lab 5 MOSFET Common Source Amplifier with Active Load Bandwidth of MOSFET Common Source Amplifier: Resistive Load / Active Load
ECE4902 C2012 - Lab 5 MOSFET Common Source Amplifier with Active Load Bandwidth of MOSFET Common Source Amplifier: Resistive Load / Active Load PURPOSE: The primary purpose of this lab is to measure the
More informationExperiment 1: Amplifier Characterization Spring 2019
Experiment 1: Amplifier Characterization Spring 2019 Objective: The objective of this experiment is to develop methods for characterizing key properties of operational amplifiers Note: We will be using
More informationBuilding Blocks of Integrated-Circuit Amplifiers
Building Blocks of ntegrated-circuit Amplifiers 1 The Basic Gain Cell CS and CE Amplifiers with Current Source Loads Current-source- or active-loaded CS amplifier Rin A o R A o g r r o g r 0 m o m o Current-source-
More informationECEN 5008: Analog IC Design. Final Exam
ECEN 5008 Initials: 1/10 ECEN 5008: Analog IC Design Final Exam Spring 2004 Instructions: 1. Exam Policy: Time-limited, 150-minute exam. When the time is called, all work must stop. Put your initials on
More information55:041 Electronic Circuits
55:041 Electronic Circuits MOSFETs Sections of Chapter 3 &4 A. Kruger MOSFETs, Page-1 Basic Structure of MOS Capacitor Sect. 3.1 Width = 1 10-6 m or less Thickness = 50 10-9 m or less ` MOS Metal-Oxide-Semiconductor
More informationISSN:
468 Modeling and Design of a CMOS Low Drop-out (LDO) Voltage Regulator PRIYADARSHINI JAINAPUR 1, CHIRAG SHARMA 2 1 Department of E&CE, Nitte Meenakshi Institute of Technology, Yelahanka, Bangalore-560064,
More informationSession 2 MOS Transistor for RF Circuits
Session 2 MOS Transistor for RF Circuits Session Speaker Chandramohan P. Session Contents MOS transistor basics MOS equivalent circuit Single stage amplifiers Opamp design Session objectives To understand
More informationLow voltage, low power, bulk-driven amplifier
University of Arkansas, Fayetteville ScholarWorks@UARK Electrical Engineering Undergraduate Honors Theses Electrical Engineering 5-2009 Low voltage, low power, bulk-driven amplifier Shama Huda University
More informationINTRODUCTION TO ELECTRONICS EHB 222E
INTRODUCTION TO ELECTRONICS EHB 222E MOS Field Effect Transistors (MOSFETS II) MOSFETS 1/ INTRODUCTION TO ELECTRONICS 1 MOSFETS Amplifiers Cut off when v GS < V t v DS decreases starting point A, once
More informationLECTURE 09 LARGE SIGNAL MOSFET MODEL
Lecture 9 Large Signal MOSFET Model (5/14/18) Page 9-1 LECTURE 9 LARGE SIGNAL MOSFET MODEL LECTURE ORGANIZATION Outline Introduction to modeling Operation of the MOS transistor Simple large signal model
More information55:041 Electronic Circuits The University of Iowa Fall Exam 3. Question 1 Unless stated otherwise, each question below is 1 point.
Exam 3 Name: Score /65 Question 1 Unless stated otherwise, each question below is 1 point. 1. An engineer designs a class-ab amplifier to deliver 2 W (sinusoidal) signal power to an resistive load. Ignoring
More informationHomework Assignment EE 435 Homework 4 Spring 2014 Due Wednesday Feb 26
Homework Assignment EE 435 Homework 4 Spring 2014 Due Wednesday Feb 26 In the following problems, if reference to a semiconductor process is needed, assume processes with the following characteristics:
More informationDifferential Amplifier Design
Fall - 2009 EE114 - Design Project Differential Amplifier Design Submitted by Piyush Keshri (0559 4497) Jeffrey Tu (0554 4565) On November 20th, 2009 EE114 - Design Project Stanford University Page No.
More informationFinal Exam. 1. An engineer measures the (step response) rise time of an amplifier as t r = 0.1 μs. Estimate the 3 db bandwidth of the amplifier.
Final Exam Name: Score /100 Question 1 Short Takes 1 point each unless noted otherwise. 1. An engineer measures the (step response) rise time of an amplifier as t r = 0.1 μs. Estimate the 3 db bandwidth
More informationHigh Voltage and Temperature Auto Zero Op-Amp Cell Features Applications Process Technology Introduction Parameter Unit Rating
Analogue Integration AISC11 High Voltage and Temperature Auto Zero Op-Amp Cell Rev.1 12-1-5 Features High Voltage Operation: 4.5-3 V Precision, Auto-Zeroed Input Vos High Temperature Operation Low Quiescent
More informationAmplifier Frequency Response, Feedback, Oscillations; Op-Amp Block Diagram and Gain-Bandwidth Product
Amplifier Frequency Response, Feedback, Oscillations; Op-Amp Block Diagram and Gain-Bandwidth Product Physics116A,12/4/06 Draft Rev. 1, 12/12/06 D. Pellett 2 Negative Feedback and Voltage Amplifier AB
More informationMicroelectronic Devices and Circuits Lecture 22 - Diff-Amp Anal. III: Cascode, µa Outline Announcements DP:
6.012 Microelectronic Devices and Circuits Lecture 22 DiffAmp Anal. III: Cascode, µa741 Outline Announcements DP: Discussion of Q13, Q13' impact. Gain expressions. Review Output Stages DC Offset of an
More information