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 Differential Amplifiers The input stage of every op amp is a differential amplifier Immunity to temperature effects Ability to amplify dc signals Well-suited for IC fabrication because (a) they depend on matching of elements (b) they use more components Less sensitive to noise and interference Enable to bias amplifier and connect to other stage without the use of coupling capacitors ECE 442 Jose Schutt Aine 2
Differential Amplifiers Practical Considerations Both inputs to a differential amplifier may have different voltages applied to them In the ideal situation with perfectly symmetric stages, the common-mode input would lead to zero output Temperature drifts in each stage are often common-mode signals Power supply noise is a common-mode signal and has little effect on the output signal ECE 442 Jose Schutt Aine 3
MOS Differential Pair Assume current source is ideal Transistors should not enter triode region ECE 442 Jose Schutt Aine 4
Common-Mode Operation Input voltage v cm to both gates Difference in voltage between the two drains is zero ECE 442 Jose Schutt Aine 5
Differential Input Voltage Differential pair responds to differntial input signals by providing corresponding differential output signal between the two drains. ECE 442 Jose Schutt Aine 6
MOS Differential Pair Assume current source is ideal v ID =v gs1 -v gs2 Output is collected as v D2 -v D1 ECE 442 Jose Schutt Aine 7
MOS Differential Pair - If v ID is positive, v D2 -v D1 is positive v ID >0 v gs1 >v gs2 I D1 > I D2 v D1 lower voltage point than v D2 For proper operation, MOSFETS should not enter triode region ECE 442 Jose Schutt Aine 8
DC Analysis IRD VD 1 VDD 2 VD2 VDD I CoxW 2 I D ID VGS VT 2 2L IR 2 D V GS LI VT VSQ VT CoxW LI C W ox ECE 442 Jose Schutt Aine 9
Incremental Analysis 1 1 vg1 vcm v v id 2 v v 2 2 v g R ' o1 m D Neglecting the body effect v 2 in v g R ' o2 m D g cm id v 2 in R R r ' D D out A v v g R o2 o1 ' D m D vin ECE 442 Jose Schutt Aine 10
Frequency Response When driven by a low-impedance signal source, the upper corner frequency is determined by the output circuit f high 1 2C R out ' D ECE 442 Jose Schutt Aine 11
Common-Mode Rejection Ratio vo 1 vo2 RD v 1 icm vicm 2R g m SS Assume R SS >> 1/g m ECE 442 Jose Schutt Aine 12
Common-Mode Rejection Ratio (a) For single-ended output: v v R v v 2R o1 o2 D icm icm SS RD 1 A, A g R 2R 2 cm d m D SS CMRR A A d cm g R m SS ECE 442 Jose Schutt Aine 13
Common-Mode Rejection Ratio (b) For differential output: A cm v v o2 o1 0 icm v A v v g R o2 o1 d m D vid CMRR ECE 442 Jose Schutt Aine 14
BJT Differential Pair Assume perfect match between the devices and symmetry in the circuit ECE 442 Jose Schutt Aine 15
Rin 2 r Base currents: BJT Differential Pair i in i v 2 b1 in r v 2 R r R r R in ib 2 r ' C out1 c1 out2 c2 R R R vin 2 c1 c2 C r ECE 442 Jose Schutt Aine 16
BJT Differential Pair Incremental Model ' ' ' vr in gr m C vo 1 gmv1rc gmrc vin r 2 ' ' ' vr in gr m C vo2 gmv 2RC gmrc vin r 2 ' A g R Single-ended gain of first stage: S1 m C Double-ended differential gain (with v out =v o2 -v o1 ): A g R ' ' ' gmrc m C ' RC D gmrc 2 2 r ECE 442 Jose Schutt Aine 17
BJT Differential Pair General R R R B1 B2 B R R R C1 C2 C R 2 2 1 in RB RE 2r A D ' RC R 1 r R E B ECE 442 Jose Schutt Aine 18
Differential Amplifiers - Observations Observations The differential pair attenuates the input signal of each stage by a factor of one-half cutting the gain of each stage by one-half The double-ended output causes the two singleended gains to be additive Thus, the voltage gain of a perfectly matched differential stage is equal to that of a single stage ECE 442 Jose Schutt Aine 19
Remarks on Differential Amplifiers 1. In many applications, the differential amplifier is not fed in a complementary fashion 2. Rather, the input signal may be applied to one of the input terminals while the other terminal is grounded 3. In this case, the signal voltage at the emitters will not be zero and thus the resistor R EE will have an effect on the operation 4. However, if R EE is large (R EE >> r e ) as is usually the case, v id will still divide equally between the 2 junctions 5. The operation of the differential amplifier will still be almost identical to that of the symmetrical feed and the CE equivalence can still be employed ECE 442 Jose Schutt Aine 20
Common Mode R R R C2 C C R C1 R C Can show that v c1 v icm RC 2R EE r e v c2 v icm R C 2R EE R r e C ECE 442 Jose Schutt Aine 21
BJT Diff Pair - Common Mode v v v v o c1 c2 icm RC 2R r EE e A cm RC R 2R r 2R EE e EE C A cm R 2R C EE R R C C ECE 442 Jose Schutt Aine 22
Example - I =100 Collector resistance accurate within 1% Early voltage = 100V ECE 442 Jose Schutt Aine 23
Emitter current in both transistors is: 0.5 ma r e VT 25 mv 50 I 0.5 ma E R 2 1 r R 2101 50 150 40 k id e E Example I (cont ) vid Rid 40 v R R 5540 sig sig id 0.8 v v o id Total resistance in the collectors Total resistance in the emitters ECE 442 Jose Schutt Aine 24
Example - I (cont ) vo 2RC 210 v 2 r R 2 50 150 10 3 id e E 50 Overall differential gain: A d vo vid vo 0.850 40 v v v sig sig id A cm common-mode gain R 2R C EE R R C C Where R C is the worst case variation in collector resistance ECE 442 Jose Schutt Aine 25
Example I (cont ) A cm 10 0.02 510 2 200 4 Common-Mode Rejection ratio CMRR CMRR 20 log A A d cm CMRR 40 20 log 98 4 510 db ECE 442 Jose Schutt Aine 26
r o Example I (cont ) Input common-mode resistance: R icm VA 100 200 k I /2 0.5 r 1 o R 101200 100 icm REE k k 6.7 M 2 ECE 442 Jose Schutt Aine 27
Example - II In the circuit shown, the dc bias current is 4 ma. If = 0.993, R B1 = R B2 = R B3 = 1,000, R E = 30, R C = 1.6 k, V CC = 10 V, and V BE(on) = 0.7 V, (a) Calculate the dc collector currents (b) Calculate the dc or quiescent collector voltages (c) Calculate the maximum peak value of vout before serious distortion results (d) Calculate the incremental differential voltage gain of the circuit (e) If the base resistor of Q2 is changed to R B2 = 400, calculate the dc collector current through each device ECE 442 Jose Schutt Aine 28
Example - II ECE 442 Jose Schutt Aine 29
Example - II (a) Assuming perfect match between Q1 and Q2, DC bias current will split equally I E1 = I E2 = 2mA. I C =I E =1.986 ma (b) The quiescent collector voltages will equal V I R 10 1.9861.6 6.82V CC C C (c) Maximum collector voltage is 10 V (at cutoff) minimum is 0 V (at saturation).therefore, positive peak voltage is 10-6.82 = 3.18 V, and negative peak is 6.82 V p-p voltage = 6.36 V ECE 442 Jose Schutt Aine 30
Example - II (d) The incremental differential voltage gain of the circuit is defined as: Calculate r e and A D v v v v v out o2 o1 in in r e 26 26 13 I 2 E 0.993 142 1 0.007 ECE 442 Jose Schutt Aine 31
AD Example - II Applying the gain equation and assuming r out >> 1.6 k gives 1421600 143 13 30 1000 31.8 V / V (e) The voltage at the node above the dc current source can be found from V I R V 1 I R 1 B1 B1 BE( on) B1 E V I R V 1 I R 2 B2 B2 BE( on) B2 E ECE 442 Jose Schutt Aine 32
Example - II Effects of non-balance ECE 442 Jose Schutt Aine 33
Example - II 1 I 1 I 4mA B1 B2 I B1 13.1 A I B2 14.8 A The corresponding emitter and collector currents are I E1 1.88 ma I E 2 2.12 ma IC1 1.86 ma IC 2 2.10 ma The two quiescent collector voltages are no longer equal, resulting in a nonzero quiescent output voltage VCQ 2 10 1.62.1 6.64V VCQ1 10 1.61.86 7.02V ECE 442 Jose Schutt Aine 34
Example - II V V 2 V 1 6.64 7.02 0.38V outq CQ CQ Nonzero quiescent voltage serious consequences when this stage is followed by additional gain stages, creating an output offset voltage when the inputs are shorted together ECE 442 Jose Schutt Aine 35
Nonideal Characteristics Input offset voltage of MOS differential pair Mismatch can result in a dc output voltage V o (output dc offset voltage) V os =V o /A d is input offset voltage ECE 442 Jose Schutt Aine 36
Nonideal Characteristics If V os is applied (differentially) at the input, a zero voltage difference should result at the output Factors contributing to dc offset voltage 1. Mismatch in load resistance 2. Mismatch in W/L 3. Mismatch in V T V os 2 V / ov RD V W L ov V 2 2 2 W / L 2 T 2 ECE 442 Jose Schutt Aine 37
Input Offset Voltage for BJT Diff Pair Offset results from 1. Mismatch in RC s 2. Mismatch in 3. Mismatch in junction area V os V T 2 2 R C I S R I C S ECE 442 Jose Schutt Aine 38
Offset Current for BJT Diff Amp In a perfectly symmetric differential pair, the 2 input terminals carry equal dc current to support bias I /2 IB 1 IB2 1 Mismatches (primarily from ) make the 2 input dc currents unequal I I I I os B1 B2 os IB ECE 442 Jose Schutt Aine 39
Differential-to-Single-Ended Conversion - Beyond first stage, signal can be converted from differential to single-ended - Simply ignore the drain current in Q 1 and eliminate its drain resistor ECE 442 Jose Schutt Aine 40
Differential-to-Single-Ended Conversion Limitations Factor of 2 (6 db) is lost in the gain if drain current of Q 1 is not used Much better approach consists of using drain current of Q 1 Active load approach allows to perform conversion without loss of gain by making use of drain current in Q 1 ECE 442 Jose Schutt Aine 41
MOS Differential Amp with Active Load Replacing drain resistances with current sources, results in much higher voltage gain and savings in chip area in diff amp ECE 442 Jose Schutt Aine 42
MOS Differential Amp - Equilibrium ECE 442 Jose Schutt Aine 43
MOS Differential Amp with Active Load Current mirror action makes it possible to convert the signal to single-ended form without loss of gain. The differential gain is: v A g r r o d m o2 o4 vid If r r r o2 o4 o A d 1 2 g r m o ECE 442 Jose Schutt Aine 44
MOS Differential Amp with Active Load The active-loaded MOS differential amplifier has a low common-mode gain high CMRR The common-mode gain is: A cm vo 1 ro4 v 2R 1 g r icm SS m3 o3 Usually, g r 1 and r r A cm m3 o3 o3 o4 2g 1 R m3 SS R SS is internal impedance of current source ECE 442 Jose Schutt Aine 45
MOS Differential Amp with Active Load Since R SS is large, A cm will be small A 2 d CMRR gm ro2 ro4 gm3rss A cm If r r r and g g o2 o4 o m3 m CMRR g mro gmrss ECE 442 Jose Schutt Aine 46
BJT Differential Amp with Active Load Current mirror & active load Differential stage ECE 442 Jose Schutt Aine 47
Active Loaded BJT Pair Incremental Model Virtual ground develops at common-emitter terminal ECE 442 Jose Schutt Aine 48
BJT Differential Amp with Active Load Output resistance is parallel equivalent of the output resistance of the differential pair and the output resistance of the current mirror The differential gain is: v A g r r o d m o2 o4 vid If r r r A o2 o4 o d 1 2 g r m o The differential input impedance is: R id 2 r ECE 442 Jose Schutt Aine 49
BJT Differential Amp with Active Load The active-loaded BJT differential amplifier has a low common-mode gain high CMRR The common-mode gain is: A cm v v o icm r R 3 o4 EE It is assumed that g g, m3 m4 and r r and r r, r 4 3 o3 3 4 R EE is internal impedance of current source ECE 442 Jose Schutt Aine 50
BJT Differential Amp with Active Load A CMRR g r r A d 3 EE m o2 o4 cm R ro 4 If r r r o2 o4 o CMRR 1 3gmR 2 EE For large CMRR, bias current source should have large output resistance R EE ECE 442 Jose Schutt Aine 51
Frequency Response of MOS Diff Amp Resistively Loaded 1. Resistance R SS is between node S and ground 2. Capacitance C SS is between node S and ground 3. C SS includes C db, C gd, and C sb ECE 442 Jose Schutt Aine 52
Frequency Response Differential Half Gain function of differential half will be identical to that of common-source amplifier ECE 442 Jose Schutt Aine 53
Frequency Response Common-Mode C SS /2 will form dominant real axis zero at much lower frequency Zero dominates frequency dependence of A cm Common-mode gain is found by analyzing the effect of a mismatch R D in R D ECE 442 Jose Schutt Aine 54
Frequency Response Common-Mode R D R D A () s 1sC R 2RSS RD cm SS SS A cm picks up a zero on the negative real axis of the complex s-plane. The frequency is Z Z 1 C R SS SS f Z 1 2C R SS SS ECE 442 Jose Schutt Aine 55
Frequency Response Common-Mode Gain ECE 442 Jose Schutt Aine 56
Frequency Response Differential Gain ECE 442 Jose Schutt Aine 57
Frequency Response CMRR ECE 442 Jose Schutt Aine 58
Frequency Response Actively Loaded MOS C C C C C C m gd1 db1 db3 gs3 gs4 C C C C C C L gd2 db2 gd4 db4 load Capacitance at input node Capacitance at output node ECE 442 Jose Schutt Aine 59
Frequency Response Actively Loaded MOS Cm 1 s v 1 o 2g m3 Ad() s gmro vid 1 sc C LRo m 1 s g 1 m3 First pole: fp 1 2C R Second pole: Zero at: f P2 L o g g 2 C 2 2C m3 m3 f Z m 2g 2C m3 m f T gs3 ECE 442 Jose Schutt Aine 60 f T /2 Mirror pole and zero occur at very high frequencies
Actively Loaded MOS - Transconductance ECE 442 Jose Schutt Aine 61
CMOS OP Amp Example In the differential amplifier shown, Q 1 and Q 2 form the differential pair while the current source transistors Q 4 and Q 5 form the active loads for Q 1 and Q 2 respectively. The dc bias circuit that establishes an appropriate dc voltage at the drains of Q 1 and Q 2 is not shown. The following specifications are desired: differential gain A d = 80V/V, I REF = 100 A, the dc voltage at the gates of Q 6 and Q 3 is +1.5V; the dc voltage at the gates of Q 7, Q 4 and Q 5 is 1.5V. The technology available is specified as follows: n C ox =3 p C ox = 90A/V 2 ; V tn = V tp =0.7V, V An = V Ap = 20V. Specify the required value of R and the W/L ratios for all transistors. Also, specify I D and V GS at which each transistor is operating. For dc bias calculations, you may neglect channel-length modulation. Fill in the entries in the table provided to show your results. ECE 442 Jose Schutt Aine 62
CMOS OP Amp Example ECE 442 Jose Schutt Aine 63
CMOS OP Amp Example 1.5 ( 1.5) 3V I REF 100 A R 30k R 0.1mA Drain currents are determined by symmetry and inspection V GS values are also determined by inspection for all transistors except Q 1 and Q 2. To determine V GS for Q 1 and Q 2, we do the following: the equivalent load resistance will consist of r o1 in parallel with r o4 for Q 1 and r o2 in parallel with r o5 for Q 5. Since the r o s are equal, this corresponds to r o /2. We have: ro 2Ad 280 gm Ad gm 0.4 ma / V 2 r 400k o ECE 442 Jose Schutt Aine 64
CMOS OP Amp Example g m 2ID 2ID 2 0.05 Vov V g 0.4 ov Take polarity into account for PMOS m 0.25 V GS 1,2 0.25 VT 0.95 To find W/L ratios, use W 2 W 2I D ID Cox ( VGS VT) 2 L L C ( V V ) ox GS T taking into account PMOS and NMOS devices separately 2 ECE 442 Jose Schutt Aine 65
CMOS OP-AMP DESIGN TABLE Q 1 Q 2 Q 3 Q 4 Q 5 Q 6 Q 7 Units C ox 30 30 30 90 90 30 90 A/V 2 I D 50 50 100 50 50 100 100 A V GS -.95 -.95-1 +1 +1-1 +1 V W/L 57.3 57.3 74 1. 12.3 12.3 73.1 24.7 ECE 442 Jose Schutt Aine 66
2-Stage CMOS Op Amp ECE 442 Jose Schutt Aine 67
2-Stage CMOS Op Amp Two-stage configuration with two power supplies which can range from +/- 2.5 V for 0.5 m technology to +/- 0.9 V for 0.18 m technology. I REF is generated either externally or using on-chip CKT. Current mirror formed by Q 5 -Q 8 supplies differential pair Q 1 -Q 2 with bias current. The W/L of Q 5 is selected to control I. The diff pair is actively loaded by current mirror Q 3 -Q 4 ECE 442 Jose Schutt Aine 68
2-Stage CMOS Op Amp Second stage is Q6 which is a CS amplifier for which Q7 is the current source. A capacitor Cc is included for negative feedback to enhance the Miller effect through Q6 compensation. This op amp does not have a low output impedance and is thus not suited for driving a lowimpedance load. The W/L ratios are given and listed below: Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 W/L 20/0.8 20/0.8 5/0.8 5/0.8 40/0.8 10/0.8 40/0.8 40/0.8 Let I 90 A, V 0.7 V, V 0.8V REF tn tp C 160 A/ V, C 40 A/ V 2 2 n ox p ox ECE 442 Jose Schutt Aine 69
2-Stage CMOS Op Amp V for all devices 10 V, V V 2.5V A DD SS Voltage Gain A g r r First stage: 1 m1 o2 o4 Since Q 8 and Q 5 are matched, I = I REF, Q 1, Q 2, Q 3 and Q 4 will have I/2 = 45 A. I Q7 =I REF = 90 A = I Q6 Let V GS -V T = V ov (overdrive voltage) ECE 442 Jose Schutt Aine 70
2-Stage CMOS Op Amp 1 2 From I C W / L V 2 D ox ov We find V ov for each transistor. Transconductance is: g m 2I D V ov r o V I A D ECE 442 Jose Schutt Aine 71
2-Stage CMOS Op Amp Voltage Gain Gain for first stage: A A g r r 1 m1 o2 o4 1 0.3 222 222 33.3 / V V A g r r Gain for second stage: A 2 0.6 111 111 33.3 / 2 m6 o6 o7 V V Overall dc open loop gain is (-33.3)(-33.3) = 1109 V/V 20 log1109 = 61 db ECE 442 Jose Schutt Aine 72
2-Stage Op Amp Design Table Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 W/L 20/0.8 20/0.8 5/0.8 5/0.8 40/0.8 10/0.8 40/0.8 40/0.8 I D (A) 45 45 45 45 90 90 90 90 V ov (v) 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 V GS (v) 1.1 1.1 1.0 1.0 1.1 1.0 1.1 1.1 g m (ma/v) 0.3 0.3 0.3 0.4 0.6 0.6 0.6 0.6 r o (k) 222 222 222 222 111 111 111 111 ECE 442 Jose Schutt Aine 73
2-Stage Op Amp Frequency Response Incremental Circuit G g g m1 m1 m2 R r r, C C C C C C 1 o2 o4 1 gd4 db4 gd2 db2 gs6 ECE 442 Jose Schutt Aine 74
2-Stage Op Amp Frequency Response G g m2 m6 R r r, C C C C C 2 o6 o7 2 db6 db7 gd7 L C is the load capacitance (usually large) C C L 2 1 V G G sc RR o V 1sAs B id m1 m2 C 1 2 2 ECE 442 Jose Schutt Aine 75
2-Stage Op Amp Frequency Response A CR CR C G RR R R 1 1 2 2 C m2 1 2 1 2 B CC C C C RR 1 2 C 1 2 1 2 Transmission zero at s = s Z with Z G C m2 C Two poles that are the root of the denominator 1 Gm p1 RC 1 CGm2R 2 p2 2 C 2 ECE 442 Jose Schutt Aine 76