Lecture 21: Voltage/Current Buffer Freq Response

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1 Lecture 21: Voltage/Current Buffer Freq Response Prof. Niknejad

2 Lecture Outline Last Time: Frequency Response of Voltage Buffer Frequency Response of Current Buffer Current Mirrors Biasing Schemes Detailed Example

3 Common-Collector Amplifier Procedure: 1. Small-signal twoport model 2. Add device (and other) capacitors

4 Two-Port CC Model with Capacitors Gain ~ 1 Find Miller capacitor for C π -- note that the base-emitter capacitor is between the input and output

5 Voltage Gain A vcπ Across C π A R /( R + R ) 1 vc out out L π Note: this voltage gain is neither the two-port gain nor the loaded voltage gain C g R >> 1 m = 1 ) in L C + C = C + ( A C µ M µ vcπ π 1 C = C + C in 1 + g R µ π m L C C in µ R out = g 1 m

6 Bandwidth of CC Amplifier Input low-pass filter s 3 db frequency: ω ( R ) + + S Rin Cµ 1 gmrl 1 p = π Substitute favorable values of R S, R L : R 1/ S g m L R >>1/ g m C ω 1 p C 1+ BIG π ( 1/ g m ) Cµ + Cµ / gm Model not valid at these high frequencies ω g / C µ > ω p m T

7 CB Current Buffer Bandwidth Same procedure: start with two-port model and capacitors

8 Two-Port CB Model with Capacitors No Miller-transformed capacitor! Unity-gain frequency is on the order of ω T for small R L

9 Summ of Single-Stage Amp Freq Resp CE, CS: suffer from Miller-magnified capacitor for high-gain case CC, CD: Miller transformation nulled capacitor wideband stage CB, CG: no Millerized capacitor wideband stage (for low load resistance)

10 CMOS Diode Connected Transistor Short gate/drain of a transistor and pass current through it Since VGS = VDS, the device is in saturation since VDS > VGS-VT Since FET is a square-law (or weaker) device, the I-V curve is very soft compared to PN junction diode What s the input impedance of circuit?

11 Diode Equivalent Circuit R D di OUT = 0 1 OUT = = dv R OUT D I 1 g m v i t t Equivalent Circuit: R D i OUT V D v OUT -

12 The Integrated Current Mirror High Res Low Resis M 1 and M 2 have the same V GS If we neglect CLM (λ=0), then the drain currents are equal Since λ is small, the currents will nearly mirror one another even if V out is not equal to V GS1 We say that the current I REF is mirrored into i OUT Notice that the mirror works for small and large signals!

13 Current Mirror as Current Source The output current of M 2 is only weakly dependent on v OUT due to high output resistance of FET M2 acts like a current source to the rest of the circuit

14 Small-Signal Resistance of I-Source

15 Improved Current Sources Goal: increase r oc Approach: look at amplifier output resistance results to see topologies that boost resistance R out >> r o Looks like the output impedance of a commonsource amplifier with source degeneration

16 Effect of Source Degeneration R eq 1 g m v R = + g R r Equivalent resistance loading gate is dominated by the diode resistance assume this is a small impedance Output impedance is boosted by factor v = ( i g v ) r + v t t m gs o R S v v v gs R S R t S S = ir v = ( i + g R i ) r + ir ( 1 ) t o m S o it t t m S t o t S ( 1+ gmrs )

17 Cascode (or Stacked) Current Source Insight: V GS2 = constant AND V DS2 = constant Small-Signal Resistance r oc : ( 1 ) R + g R r o m S o ( 1 ) R + g r r o m o o R g r >> r 2 o m 0 o

18 Drawback of Cascode I-Source Minimum output voltage to keep both transistors in saturation: i OUT V = V + V OUT, MIN DS 4, MIN DS 2, MIN V > V V = V DS 2, MIN GS 2 T 0 DSAT 2 V > V + V = V + V V D4 DSAT2 GS4 GS2 GS4 T0 V = V + V V OUT, MIN GS 2 GS 4 T 0 v OUT

19 Current Sinks and Sources Sink: output current goes to ground Source: output current comes from voltage supply

20 Current Mirrors Idea: we only need one reference current to set up all the current sources and sinks needed for a multistage amplifier.

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