MIT6_012S09_rec23

8/13/2019 MIT6_012S09_rec23

1/6

Recitation 23 Frequency Response of C-C & C-B Amps 6.012 Spring 2009

Recitation 23: Frequency Response of Common

Collector & Common-Base Amplier Yesterday, we used OCT technique for the frequency response of Common-Drain andCommon-Gate ampliers. Today we will look at C-C, C-B frequency response.

Common-Collector Amplier

One way to study the frequency response is to

First nd the small signal equivalent model for the circuit

Do KCL, KVL nodal analysis, to nd CO 3dB

Or use OCT + Miller Approximation to nd w3dB

However, the small signal model of this circuit is quite complicated (as the C-D Amp. wetalked about yesterday). What we can do is directly use the two-port model for the circuit,and add in the capacitances. So the methodology is as outlined below.

Methodology

1. Start with low frequency two port model, obtain Av, Ai, G m at low frequency

2. Identify the nodes (S/D/G/B for MOS; B/E/C for BJT) and add in capacitance inactive device

3. Use Miller Approximation in conjunction with OCT to estimate bandwidth ( w3dB ).

Advantage: can directly use the R in , R out from two-port model, only need Av,Ai orGmmuch easier.

1

8/13/2019 MIT6_012S09_rec23

2/6


So taking the C-C Amplier as an example, the two port model is:

R in = + o ( o || oc || RL )1 Rs

R out = +gm o

Avo = 1

Av ,LF = V out

= Rin

(1) RL

V s Rs + R in

RL + R out

Large gm , o will give desired resistances for voltage buffer. High Rin , low R out

= + o ( oc || o || RL ) (1)

R1

L

RL + + o ( oc || o || RL ) RL + gm + R s

Identify the B/E/C and add in capacitances

Note: the other end of C is to the right of Rout ! That is where E node is! C is in theinput/output feedback position.

2

8/13/2019 MIT6_012S09_rec23

3/6


Use Miller Approximation:

where C M = C (1 Av C ). Av C is the voltage gain across C (not across overall amplier).

What is the voltage gain across C ? V out

instead of V out

V in V sOr, it is

V outwhen Rs = 0.

V s

typically o || oc R L

V out V out

+ o RL RL=

= (1)

R s R s o

+ + o 1V in V s RLR s =0 RL + + gmRL

= Av C RL + 1 /g m

C M = C (1 Av C ) = C 1/g m

= C 1

RL + 1 /g m 1 + gm RL

1If g

1m

RL , then Av C 1 = C M 0 w3dB = (R s || R in ) (C + C M )In contrast to C-S or C-E amplier, the Miller effect reduces the capacitance in this case,which will give better frequency response: (or another way to look at it, effect of C is verysmall, since voltage gain across C is 1. We do not need a lot of charges to go in/out thecapacitor. And typically the movement of charges is the source to slow down the frequencyresponse).

Therefore like C-D, Miller effect reduces capacitor value, = expect good frequency response.

Use of C-C: for multistage ampliers, can enable high Rin , low Rout , wontdegrade frequency response

3

8/13/2019 MIT6_012S09_rec23

4/6


Common-Base Amplier

Current buffer:

1. Two port model (for current amplier)

Low frequency current gain

iout Rs Rout= ( 1)

vs Rs + R in Rout + RL1

R in =gm

R out = oc || [ o (1 + gm ( || R s ))]

4

8/13/2019 MIT6_012S09_rec23

5/6


2. Label B/E/C, add in capacitances

No capacitor in the feedback position = Do not need Miller Approximation. UseOCT

1 C : RTH C = R s || R in = R s || gm C : RTH C = Rout || RL = RL || ( oc || ( o + gm o ( || R s )))

Let us try to make some simplications (if conditions are met) for a on w3dB :

If R s not so small, since 1 is small ( 100),gm

1 1 C = =R s ||gm gm

T C gmAnd if R s ( 10 k)

R out = oc || ( o + gm o ( || R s ))gm o ( || R s )) = gm o = o o

R out oc || o o can be quite large

= RTH C RL || ( oc || o o ) RL

w3dB + 1C RL can be approaching wT = C g+m C - a good current bufferC

5

8/13/2019 MIT6_012S09_rec23

6/6

Documents

MIT6_012S09_rec23