EE105 – Fall 2014 Microelectronic Devices and Circuits

Prof. Ming C. Wu [email protected] 511 Sutardja Dai Hall (SDH)

Lecture23- Frequency Response 1

Common-Emitter Amplifier – ωH Open-Circuit Time Constant (OCTC) Method

At high frequencies, impedances of coupling and bypass capacitors are small enough to be considered short circuits. Open-circuit time constants associated with impedances of device capacitances are considered instead. 1 ωH ≅ m ∑RioCi i=1

where Rio is resistance at terminals of ith capacitor C with all other v ! R $ i R = x = r #1+ g R + L & capacitors open-circuited. µ 0 π 0 m L ix " rπ 0 % For a C-E amplifier, assuming C = 0 L 1 1 ωH ≅ = Rπ 0 = rπ 0 Rπ 0Cπ + Rµ 0Cµ rπ 0CT

Lecture23-Amplifier Frequency Response 2

1 Common-Emitter Amplifier High Frequency Response -

• First, find the simplified small -signal model of the C-E amp. • Replace coupling and bypass capacitors with short circuits • Insert the high frequency small -signal model for the

! # rπ 0 = rπ "rx +(RB RI )$

Lecture23-Amplifier Frequency Response 3

Common-Emitter Amplifier – ωH High Frequency Response - Miller Effect (cont.) v R r Input is found as A = b = in ⋅ π i v R R r r i I + in x + π R || R || (r + r ) r = 1 2 x π ⋅ π R + R || R || (r + r ) r + r I 1 2 x π x π

Terminal gain is vc Abc = = −gm (ro || RC || R3 ) ≅ −gm RL vb

Using the Miller effect, we find CeqB = Cµ (1− Abc )+Cπ (1− Abe ) the equivalent capacitance at the base as: = Cµ (1−[−gm RL ])+Cπ (1− 0)

= Cµ (1+ gm RL )+Cπ

Chap 17-4 Lecture23-Amplifier Frequency Response 4

2 Common-Emitter Amplifier – ωH High Frequency Response - Miller Effect (cont.)

! # • The total equivalent resistance ReqB = rπ 0 = rπ "rx +(RB RI )$ at the base is • The total capacitance and CeqC = Cµ +CL resistance at the collector are ReqC = ro RC R3 = RL

• Because of interaction through 1 ω = Cµ, the two RC time constants p1 ! # rπ 0 "Cπ +Cµ (1+ gm RL )$+ RL (Cµ +CL ) interact, giving rise to a dominant pole. 1 ω p1 = where rπ 0CT

RL CT = Cπ +Cµ (1+ gm RL ) + (Cµ +CL ) rπ 0 ! # rπ 0 = rπ "rx +(RB RI )$

Lecture23-Amplifier Frequency Response 5

Common-Source Amplifier – ωH Open-Circuit Time Constants

Analysis similar to the C-E case yields the following equations: Rth = RG RI

RL = RD R3 ro

RG vth = vi RI + RG

RL CT = CGS +CGD (1+ gm RL ) + (CGD +CL ) Rth

1 gm ωP1 = ωP2 = RthCT CGS +CL

gm ωZ = CGD

Lecture23-Amplifier Frequency Response 6

3 C-S Amplifier High Frequency Response Source Degeneration Resistance

First, find the simplified small -signal model of the C-S amp.

Recall that we can define an

effective gm to account for the unbypassed source resistance.

gm g'm = 1+ gm RS

Lecture23-Amplifier Frequency Response 7

C-S Amplifier High Frequency Response Source Degeneration Resistance (cont.)

v R R || R Input gain is found as A = g = G = 1 2 i v R + R R + R || R i i G i 1 2

vd ' −gm (RD || R3 ) Terminal gain is Agd = = −gm (RiD || RD || R3 ) ≅ vg 1+ gm RS

Again, we use the Miller effect CeqG = CGD (1− Agd )+CGS (1− Ags ) to find the equivalent " % capacitance at the gate as: (−gm RL ) gm RS = CGD $1− '+CGS (1− ) # 1+ gm RS & 1+ gm RS " % gm (RD || R3 ) CGS = CGD $1+ '+ # 1+ gm RS & 1+ gm RS

Lecture23-Amplifier Frequency Response 8

4 C-S Amplifier High Frequency Response Source Degeneration Resistance (cont.)

The total equivalent resistance ReqG = RG RI = Rth at the gate is

The total capacitance and C = C +C resistance at the drain are eqD GD L R = R R R ≅ R R = R Because of interaction through eqD iD D 3 D 3 L

CGD, the two RC time constants interact, giving rise to the dominant pole: 1 ω p1 = CGS gm RL RL Rth[ +CGD (1+ )+ (CGD +CL )] 1+ gm RS 1+ gm RS Rth And from previous analysis: ' ' gm gm +gm +gm ω p2 = = ωz = = (CGS +CL ) (1+ gm RS )(CGS +CL ) CGD (1+ gm RS )(CGD )

Lecture23-Amplifier Frequency Response 9

C-E Amplifier High Frequency Response Emitter Degeneration Resistance

Analysis similar to the C-S case yields the following equations: ! # Rπ 0 = Rth "rπ +(βo +1) RE $

Rth = RI RB = RI R1 R2 RL = RiC RC R3 ≅ RC R3 1 ωP1 = Rπ 0CT 1 ωP1 = Cπ gm RL RL Rπ 0[ +Cµ (1+ )+ (Cµ +CL )] 1+ gm RE 1+ gm RE Rπ 0 ' gm gm ωP2 = = (Cπ +CL ) (1+ gm RE )(Cπ +CL ) ' +gm +gm ωz = = Cµ (1+ gm RE )(Cµ )

Lecture23-Amplifier Frequency Response 10

5 Gain-Bandwidth Trade-offs Using Source/Emitter Degeneration Resistors

Adding source resistance to the C-S (or C-E) amp causes vd gm (RD || R3 ) gain to decrease and dominant Agd = = − pole frequency to increase. vg 1+ gm RS 1 ω = p1 C g R R R [ GS +C (1+ m L )+ L (C +C )] th 1+ g R GD 1+ g R R GD L m S m S th

However, decreasing the gain gm also decreased the frequency ω p2 = of the second pole. (1+ gm RS )(CGS +CL ) Increasing the gain of the C-E/C-S stage causes pole -splitting, or increase of the difference in frequency between the first and second poles.

Lecture23-Amplifier Frequency Response 11

High Frequency Poles Common-Base Amplifier

1 gm 1 Ai ≅ = (1 gm ) + RI 1+ gm RI

Aec = gm (RiC RL ) ≅ gm RL " $ RiC = ro #1+ gm (rπ RE RI )%

Since Cµ does not couple input and output, input and output poles can be found directly.

C C eqE = π CeqC = Cµ +CL 1 R = R R R = R R ReqC = RiC || RL ≅ RL eqE iE E I g E I m 1 1 −1 ω = ≅ (" 1 % + g p2 m (RiC || RL )(C +CL ) RL (C +CL ) ωP1 ≅ *$ RE RI 'Cπ - ≅ > ωT µ µ )# gm & , Cπ

Lecture23-Amplifier Frequency Response 12

6 High Frequency Poles Common-Gate Amplifier

Similar to the C-B, since CGD does not couple the input and output, input and output poles can be found directly.

C C eqS = GS CeqD = CGD +CL 1 R = r !1+ g R || R # ReqS = || R4 || RI iD o " m ( 4 I )$ gm ReqD = RiD || RL ≅ RL 1 gm ω p1 = ≅ 1 1 ! 1 $ C ω = ≅ || R || R C GS p2 # 4 I & GS (RiD || RL )(CGD +CL ) RL (CGD +CL ) " gm %

Lecture23-Amplifier Frequency Response 13

High Frequency Poles Common-Collector Amplifier

ve gm RL Abe = = vb 1+ gm RL " % v R gm RL b in C = C (1− A )+C (1− A ) = C (1− 0)+C 1− Ai = = eqB µ bc π be µ π $ ' v R R # 1+ gm RL & i i + in € Cπ CeqB = Cµ + 1+ gm RL

CeqE = Cπ +CL €

ReqB = (Rth + rx ) || RiB = [(RI || RB )+ rx ]||[rπ + (βo +1)RL ] = (Rth + rx ) ||[rπ + (β +1)RL ] " % 1 (Rth + rx ) ReqE = RiE || RL ≅ $ + ' || RL # gm βo +1 &

Lecture23-Amplifier Frequency Response 14

7 High Frequency Poles Common-Collector Amplifier (cont.)

The low impedance at the output makes the input and output time constants relatively well decoupled, leading to two poles. The feed-forward high-frequency path through Cπ leads to a zero 1 in the C-C response. Both the ω p1 = zero and the second pole are ! C $ π quite high frequency and are (Rth + rx ) || [rπ + (βo +1)RL ]#Cµ + & " 1+ gm RL % often neglected, although their 1 effect can be significant with ω p2 ≅ large load capacitances. (! $ + 1 Rth + rx *# + & || RL -(Cπ +CL ) g 1 )" m β + % , gm ωz ≅ Cπ

Lecture23-Amplifier Frequency Response 15

High Frequency Poles Common-Drain Amplifier

1 g Similar the the C-C m amplifier, the C-D high ω p1 = ωz ≅ CGS C R (C + ) GS frequency response is th GD 1+ g R dominated by the first pole m L 1 1 due to the low impedance at ω = ≅ the output of the C-C p2 " % (RiS || RL )(CGS +CL ) 1 amplifier. $ || RL '(CGS +CL ) # gm &

Lecture23-Amplifier Frequency Response 16

8 Frequency Response Amplifier

There are two important poles: the input pole for the C-E and the output pole for the C-B stage. The intermediate node pole can usually be neglected because of the low impedance at

the input of the C-B stage. RL1 is small, so the second term in the first pole can be neglected.

Also note the RL1 is equal to 1/gm2.

1 1 ω = = pB1 g R R rπ 0CT m1 L1 L1 rπ 01([Cµ1(1+ )+Cπ1]+ [Cµ1 +CL1]) 1+ gm1RE1 rπ 0 1 1 ω = ≅ pB1 g 1/ g m1 m2 rπ 01(2Cµ1 +Cπ1) rπ 01([Cµ1(1+ )+Cπ1]+ [Cµ1 +Cπ 2 ]) gm2 rπ 01 1 ω pC2 ≅ RL (Cµ 2 +CL )

Lecture23-Amplifier Frequency Response 17

Frequency Response of Multistage Amplifier

• Problem: Use open-circuit and short-circuit time constant methods to estimate upper and lower cutoff frequencies and bandwidth. • Approach: Coupling and bypass capacitors determine the low -frequency response; device capacitances affect the high -frequency response

• At high frequencies, ac model for the multi-stage amplifier is as shown. Lecture23-Amplifier Frequency Response 18

9 Frequency Response Multistage Amplifier Parameters (example)

Parameters and operation point information for the example multistage amplifier.

Lecture23-Amplifier Frequency Response 19

Frequency Response Multistage Amplifier: High-Frequency Poles

High-frequency pole at the gate of M1: Using our equation for the C-S input pole:

1 1 f = P1 ! $ 2π RL1 Rth1 #CGD1 (1+ gm RL1) +CGS1 + (CGD1 +CL1)& " Rth1 %

RL1 = RI12 (rx2 + rπ 2 ) ro1 = 598Ω (250Ω+ 2.39kΩ) 12.2kΩ = 469 Ω

CL1 = Cπ 2 +Cµ 2 (1+ gm2 RL2 ) R R R r R r r 1 R R r 3.33 k L2 = I 23 in3 o2 = I 23 ( x3 + π 3 +(βo3 + )( E3 L )) o2 = Ω ! $ CL1 = Cπ 2 +Cµ 2 (1+ gm2 RL2 ) = 39pF +1pF"1+ 67.8mS(3.33kΩ)% = 266 pF 1 1 f = = 689 kHz P1 2π ! 469Ω $ 9.9kΩ 1pF"!1+ 0.01S(3.33kΩ)%$+ 5pF + (1pF + 266pF) "# 9.9kΩ %&

Lecture23-Amplifier Frequency Response 20

10 Frequency Response Multistage Amplifier: High-Freq. Poles (cont.)

High-frequency pole at the base of Q2: From the detailed analysis of the C-S amp, we find the following expression for the pole at the

output of the M1 C-S stage:

1 CGS1gL1 +CGD1(gm1 + gth1 + gL1)+CL1gth1 fp2 = 2π [CGS1(CGD1 +CL1)+CGD1CL1]

For this particular case, CL1 (Q2 input capacitance) is much larger than the other capacitances, so fp2 simplifies to:

1 CL1gth1 1 1 fp2 ≅ ≅ 2π [CGS1CL1 +CGD1CL1] 2π Rth1(CGS1 +CGD1) 1 f = = 2.68 MHz p2 2π (9.9kΩ)(5pF +1pF)

Lecture23-Amplifier Frequency Response 21

Frequency Response Multistage Amplifier: High-Freq. Poles (cont.)

High-frequency pole at the base of Q3: Again, due to the pole-splitting behavior of the C-E second stage, we expect that the pole at the base

of Q3 will be set by equation 17.96:

gm2 fp3 ≅ CL2 2π[Cπ 2 (1+ )+CL2 ] Cµ 2

The load capacitance of Q2 is the input capacitance of the C-C stage.

Cπ 3 50pF CL2 = Cµ3 + =1pF + = 3.55 pF 1+ gm3 (RE3 || RL ) 1+ 79.6mS(3.3kΩ || 250Ω) 67.8mS[1kΩ (1kΩ+ 250Ω)] f ≅ = 47.7 MHz p3 3.55pF 2π[39pF(1+ )+ 3.55pF] 1pF

Lecture23-Amplifier Frequency Response 22

11 Frequency Response

Multistage Amplifier: fH Estimate

There is an additional pole at the output of Q3, but it is expected to be at a very high frequency due to the low of the C-C stage. We can estimate fH from eq. 16.23 using the calculated pole frequencies. 1 f = = 667 kHz H 1 1 1 2 + 2 + 2 fp1 fp2 fp3

The SPICE simulation of the circuit on the next slide shows an fH of 667 kHz and an fL of 530 Hz. The phase and gain characteristics of our calculated€ high frequency response are quite close to that of the SPICE simulation. It was quite important to take into account the pole-splitting behavior of the C-S and C-E stages. Not doing so would

have resulted in a calculated fH of less than 550 kHz.

Lecture23-Amplifier Frequency Response 23

Frequency Response Multistage Amplifier: SPICE Simulation

Lecture23-Amplifier Frequency Response 24

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