ECE35 / ECE55 Lecture 8 Date: 05.09.06 CS Amplifier with Constant Current Source Current Steering Circuits CS Stage Followed by CG Stage Cascode as Current Source Cascode as Amplifier
ECE35 / ECE55 CS Amplifier with Constant Current Source The important thing to realize when analyzing this circuit is that the gate-tosource voltage for transistors Q, Q 3, and Q 4 are DC values!?? Transistors Q, Q 3, and Q 4 form the current mirror that acts as the current source. Note that transistor Q 4 is an enhancement load it acts as the resistor in the current mirror circuit. source resistance r o of this current source requires the small-signal analysis there are four (count em ) transistors in this circuit, determining the small-signal circuit must take forever! the answer is actually a NO. A: In other words, the small signal voltages v v gs for each transistor are equal to zero: gs vgs3 vgs4 0
ECE35 / ECE55 CS Amplifier with Constant Current Source (contd.) Q: But doesn t the small-signal source v i (t) create small-signal voltages and currents throughout the amplifier? A: For some of the circuit yes, but for most of the circuit no! Note that for transistor Q there will be small-signal voltages v gs (t) and v ds (t), along with i d (t). Likewise for transistor Q, a small-signal voltage v ds (t) and current i d t will occur. But, for the remainder of the voltages and currents in this circuit (e.g., V DS4, V GS, I D3 ), the small-signal component is zero!
ECE35 / ECE55 CS Amplifier with Constant Current Source (contd.) Q: But wait! How can there be a smallsignal drain current i d t through transistor Q, without a corresponding small-signal v gs (t) gate-to-source voltage? A: Transistor Q is the important device in this analysis. Note its gate-to-source voltage is a DC value (no small-signal component, v gs t = 0 ), yet there must be (by KCL) a small-signal drain current! This is a case where we must consider the MOSFET output resistance r o. The small-signal drain current for a PMOS device is: vds Since v gs = 0, this equation simplifies to: id r o i g v d m gs v ds r o
ECE35 / ECE55 CS Amplifier with Constant Current Source (contd.) Equivalently, the small-signal PMOS model is: Thus for v gs = 0, the small-signal model becomes: Thus, the small-signal model of the entire current mirror is simply the output resistance of the MOSFET Q! Or, simplifying further:
ECE35 / ECE55 CS Amplifier with Constant Current Source (contd.) It is evident that the output resistance of the current mirror is simply equal to the output resistance of MOSFET Q!!!! ro ro is equivalent to this circuit The resulting small-signal circuit of this amp is: the small signal voltage gain is: A g r r K I r r v m o o ref o o
ECE35 / ECE55 CS Amplifier with Constant Current Source (contd.) Note this result is far different (i.e., larger) than the K result when using the enhancement load for R D : A v K However, we find that the output and input resistances of this amplifier are the same as R R r r i o o o with the enhancement load: A current mirror may consist of many MOSFET current sources! This circuit is particularly useful in integrated circuit design, where one resistor R is used to make multiple current sources. Q: What if we want to make the sources have different current values? Do we need to make additional current mirrors?
ECE35 / ECE55 Current Steering Circuits A: NO!! Recall that the current mirror simply ensures that the gate to source voltages of each transistor is equal to the gate to source voltage of the reference: Therefore, if each transistor is identical (i.e., K ref = K =, and V ref T = V T = V T = ) then: ref V V V V GS GS GS GS3 ref I K V V ref ref ref GS T In other words, if each transistor Q n is identical to Q ref, then each current I Dn will equal reference current I ref. I K ref n I Kref Kref K V V I n GSn Tn Dn But, consider what happens if the MOSFETS are not identical. Specifically, consider the case where K n K ref (but V Tn = V Ref T ). The drain ref ref In such a scenario I K V V K V V Dn n GSn Tn n GS T current is a the drain current scaled value of I Dn will now be: Kn ref I ref!
ECE35 / ECE55 Current Steering Circuits (contd.) For example, if K is twice that of K ref (i.e., K = K ref ), then I D will be twice as large as I ref (i.e., I D = I ref ). K K n ref W L W L n ref From the standpoint of integrated circuit design, we can change the value of K by modifying the MOSFET channel width-tolength ratio (W/L) for each transistor.
Example- ECE35 / ECE55 Determine all the currents in the following and find the value of R C p C n ox ox VTP 0 A / V 50 A / V V I D I REF W / L W / L 0 / 5 A* 50A 5 / VTN V D3 ID4 I 5A D5 D4 W / L 3 W / L I I 5 A REF REF W / L 5 W / L I I 5 A 4
ECE35 / ECE55 Example- (contd.) R R 5 ( VGSP ) VGSN ( 5) I REF 0 VGSP VGSN I REF V V GSN GSP V V TN TP ID4 C ( W / L) p ox ID C ( W / L) n ox 4 The resistor R can be replaced by an active load such as the PFET shown here what will be its W/L?
Example- An integrated circuit employs the following CD and CS stages. Design an NMOS type current mirror that produces I and I from a 0.8mA reference. ECE35 / ECE55 Required current for CD stage is 0. ma, hence mirror equation gives: I W / L I W / L DMI MI REF R EF 0. W / L 0.8 W / L MI R EF They are in the ratio of to 8 Similarly, for CS stage the ratio is 5 to 8 8 W L 0.8mA W L = 0. ma = 0.5 ma 5 W L
Common Gate Stage ECE35 / ECE55 RD ro RD Rin ( g g ) r ( g g ) r ( g g ) m mb o m mb o m mb Case-I: RD 0 VX ro Rin I ( g g ) r X m mb o Case-II: R D is an ideal current source ie, R D the input impedance of common-gate stage is low only if the load impedance connected to the drain is low Rin VX R ( g g ) r R r R I out m mb o S o D X
ECE35 / ECE55 CS-stage followed by a CG-stage CG-stage CS-stage Thevenin Equivalent of M Req ro gmb gm o out m mb o mb Av RD in r ( ) o o gm gmb r r o ro R D g gmb gm mb gm V in, eq r ro gmb g g o m mb Use the CG stage expression to obtain the small signal voltage gain as: r V ( g g ) r g V significantly larger gain when compared to CS stage V in
Example 3 ECE35 / ECE55 The CS-stage in both the following circuits senses V at node X and delivers a proportional current to a 50Ω transmission line. (a) Calculate small signal gain at low frequencies. (b) What condition is necessary to minimize wave reflections at node X X
ECE35 / ECE55 Cascode Stage Terminology: It stems from erstwhile vacuum tube in which cathodes were cascaded. In practice, the output of first tube (anode) feeds the input of second tube (cathode) Configuration: CG-stage in cascade with CS-stage actually CS-stage is called the main device whereas CG is called the cascode device Basic Idea: combines high input impedance and large transconductance of CS with the current buffering property and the superior high frequency response of CG stage Cascode Provides: wider bandwidth, increased small-signal gain, high input impedance, customized output impedance Applications: Current Source, Small-Signal Amplifier
Cascode as a current source ECE35 / ECE55 Current Source: requires very high output impedance Cascode Transistor (CG Stage): always in saturation and is the main device that provides a constant current source Degeneration Transistor (CS Stage): in saturation and acts as a degeneration resistor for a fixed bias point provides an impedance of r o KVL in loop- V I r X o KVL in loop- V I V I r ( I g V g V ) r X X o X m mb o X X r r r ( g r g r ) R o o o m o mb o out
Cascode as a current source ECE35 / ECE55 R r r r ( g r g r ) r r r ( g g ) out o o o m o mb o o o o m mb However: ro ( gm gmb ) R ( g g ) r r out m mb o o Very High Output Impedance r ( g r g r ) r o m o mb o o M already in saturation In this case, M boosts the output resistance of M by a factor of (g m +g mb )r o R r r r ( g g ) out o o o m mb Appropriate candidate for constant current source
ECE35 / ECE55 Cascode Amplifier Cascode device (in CG) in saturation routes the current generated by main device to R D main device (in CS) in saturation converts and amplifies an input voltage signal into output current Bias Conditions of Cascode: For M to be in saturation: VX Vin VT V V V V b GS in T V V V V b in GS T For M to be in saturation: Vout VX VGS VT V V V V V out in T GS T V X V OV V OV Minimum output voltage equals overdrive voltage of M and M addition of M reduces the output voltage swing by V GS -V T
ECE35 / ECE55 Cascode Amplifier (contd.) It is a CS stage and therefore generates: V g r V X m o in R D Contribution from M V X V X Small Signal Model of Cascode: Thevenin Equivalent KVL in loop-: V V r g r V out o m o in 0 RD V V r g r V out o m o in RD V m r o g V gmbv gm ro V in r o KVL in loop-: V r V g V g V V r g r V out out out o m mb o m o in RD RD
Cascode Amplifier (contd.) Simplification gives: V R out D ECE35 / ECE55 ( ) ( ) R r r g g r r g g g r r g r V D o o m mb o o m m mb o o m o in V g ( g g ) r r g r A V R r r g g r r out m m mb o o m o v in D o o ( m mb) o o R D g ( g g ) r r g r Now: m m mb o o m o A v gm ( gm gmb) ro ro R r r ( g g ) r r D o o m mb o o R D Output Impedance: ability to synthesize desired output impedance R g g r r out m mb o o
Cascode Amplifier (contd.) ECE35 / ECE55 the gain of Cascode stage: A v gm ( gm gmb) ro ro R r r ( g g ) r r D o o m mb o o R D A constant current source possesses very high output impedance (R D ), therefore the gain equation changes to: A g ( g g ) r r ( g r ).( g g ) r v m m mb o o m o m mb o It is apparent that the maximum small-signal voltage gain is the multiplication of gains from CS and CG stages definitely a big plus! R g g r r out m mb o o
ECE35 / ECE55 Cascode Amplifier (contd.) Discussion with respect to alteration in dimension What happens if Length (L) of the main device is quadrupled while the Width (W) remains same? W Av gmro ncox I D L ID Av A A v v Av W g gm m ncox I D gm gm gm L I V Cascode more suited for noise applications TH W ncox L Quadruppling of Length (L) while keeping the Width (W) results in doubling of overdrive voltage D V GS For identical devices, cascode also exhibits doubling of overdrive voltage
Shielding Property of Cascode ECE35 / ECE55 Cascode amplifier and Cascode current source could be used in applications where the output varies drastically due to any reason! This variation doesn t affect the subsequent sections greatly shielding property of Cascode Example: Two identical NFETs are used to generate constant current sources. However, due to internal circuitry of the system, V X is higher than V Y by V. Q: determine the resulting difference between I D and I D if λ 0 I C V V V D n ox b T X I C V V V D n ox b T Y I D I D ncox Vb VT V V DS V DS
ECE35 / ECE55 Shielding Property of Cascode (contd.) Q: Add Cascode devices to M and M and then check the difference between I D and I D if λ 0 Cascode devices I I C V V V D D n ox b T PQ o VPQ V m3 mb3 o3 o o3 r V g g r g g r r r m3 mb3 o3 V ID ID ncox Vb VT gm3 gmb3 r o3 This is a large value and thus the Cascode structure gives smaller variation perfect example of Shielding property!!!
Triple Cascode ECE35 / ECE55 Cascoding can be extended to three or more devices to achieve higher output impedance It limits the voltage swing Here the minimum output voltage equals the sum of three overdrive voltages Folded Cascode Self Study