ECE 255, Miller Effect, Etc.

26 April 2018

1 Introduction

In this lecture, we will focus on the study of the Miller effect and other high-frequency analysis tools for transistors.

Printed on April 26, 2018 at 13 : 52: W.C. Chew and S.K. Gupta.

1

2 High-Frequency Response of the CS and CEAmplifier with Miller Effect

Figure 1: Models for the high-frequency response of a CS amplifier. (a) Equiv-alent circuit. (b) A simplified circuit by consolidation: The source signal isreplaced by a Thevenin equivalence with effective source V ′

sig and Thevenin re-sistor R′

sig, and the load is replaced by an effective load R′L (Courtesy of Sedra

and Smith).

2

Figure 2: Continuation of the previous Figure 1. (c) Further simplificationby using Ceq. (d) A single-time-constant frequency response Bode plot for asingle-pole system (Courtesy of Sedra and Smith).

Figure 1 shows the small-signal equivalence of a CS amplifier. The overallvoltage gain is given by

AM =VoVsig

= − RG

RG +Rsig(gmR

′L) (2.1)

In order to simplify the circuit, it can be consolidated using Thevenin theoremso that the source is modeled by only two elements as shown in Figure 1(b).Also, R′

L consolidates the three resistances at the output end.One further simplification is to replace the capacitor Cgd with an equivalence

capacitor Ceq as shown in Figure 1(c). To this end, we shall calculate the loadcurrent, in accordance with Figure 1(b), which is given by (−gmVgs + Igd) viaKCL at node D. Then using generalized Ohm’s law, the output voltage is givenby

Vo = (−gmVgs + Igd)R′L ≈ −gmR′

LVgs (2.2)

3

One assumes that |gmVgs| � |Igd| in the last approximation. In the above,R′

L = ro ‖ RD ‖ RL. The current Igd can now be found using Vo from (2.2) as

Igd = sCgd(Vgs − Vo) ≈ sCgd [Vgs − (−gmR′LVgs)] = sCgd(1 + gmR

′L)Vgs (2.3)

Next, one can assume that this extra current Igd is due to an equivalent capacitorCeq connected in parallel to Cgs as shown in Figure 2(c). The current that flowsinto this equivalent capacitance Ceq is given by Igd, which is given by

sCeqVgs = sCgd(1 + gmR′L)Vgs (2.4)

The above results in that

Ceq = Cgd(1 + gmR′L) (2.5)

This equivalent capacitance Ceq is much larger than Cgd, and this effect is knownas the Miller effect, and the factor (1 + gmR

′L) is the Miller multiplier.

Hence, the larger the gain of the amplifier is, the larger this effect is.One can see that Ceq is much larger than Cgd because when a positive Vgs

is applied at the gate, the negative voltage Vo is generated at the output nodeaccording to (2.2). This negative Vo siphons more current from the gate to thedrain as result, increasing Igd in accordance with (2.3).

One has to be reminded that the above derivation is predicated on the as-sumption that |gmVgs| � |Igd|. However, it is seen that the Miller effect ampli-fies Igd as shown in (2.3). Hence, the |gmVgs| � |Igd| has to be checked whenone uses the above simplified formula for the equivalence capacitance Ceq.

2.1 Single Pole Approximation

The above approximation replaces a double-pole system with a single pole sys-tem or a single-time-constant (STC) circuit. Then the frequency dependentfunction with one pole and no zero can now be represented as

Vgs =

(RG

RG +RsigVsig

)1

1 + s/ω0(2.6)

The effective capacitance of this system is Cin = Cgs + Ceq. This capacitancerelaxes its charge via the effective resistor R′

sig. Hence, the pole frequency ofthis STC circuit is then

ω0 = 1/(CinR′sig) (2.7)

withCin = Cgs + Ceq = Cgs + Cgd(1 + gmR

′L) (2.8)

andR′

sig = Rsig ‖ RG (2.9)

A sanity check of (2.6) shows that this in fact is the correct formula: it reducesto the correct formula when s = ω = 0. The system in Figure 2(c) can only

4

have one pole with the corresponding relaxation frequency given by (2.7). Thezero of the system is at s = ∞, or unimportant. Therefore, it is the correctformula which can be confirmed by a longer derivation.

Then using Vgs from (2.6) in (2.2), one arrives at

VoVsig

= −(

RG

RG +Rsig

)(gmR

′L)

1

1 + s/ω0(2.10)

which can be simplified asVoVsig

=AM

1 + s/ωH(2.11)

where AM is the midband gain given by (2.1), and ωH is the upper 3-dB fre-quency point, or

ωH = ω0 =1

CinR′sig

, fH =ωH

2π=

1

2πCinR′sig

(2.12)

2.1.1 Validity of Single-Pole Approximation

The single-pole approximation is valid when the second pole is far away fromthe first pole. It can be shown that with exact analysis, when the inequality|gmVgs| � |Igd| is satisfied, the second pole is indeed far away. Then one has tocheck if the zeros of the transfer function are also far away.

The zeros of the amplifier can be obtained by looking at Figure 1(a). Thezero occurs as s = ∞ as Cgs becomes short circuit. This is the unimportantzero. Also, another possible zero occurs when Vo = 0. In this case, the voltageacross the capacitor Cgd is Vgs and the current through it is sCgdVgs. Then ifgmVgs and sCgdVgs are equal to each other, no current flows to the output nodegiving rise to Vo = 0. This is only possible if

gm = sCgd (2.13)

or when s = gm/Cgd which is the location of this zero. But Cgd is usually small,making this zero far away compared to ω0 in (2.10).

5

2.2 Miller Effect of the Common-Emitter Amplifier

Figure 3: Models for the high-frequency response of a CE amplifier. (a) Equiva-lent circuit. (b) A simplified circuit by consolidation. (c) Further simplificationby using Ceq (Courtesy of Sedra and Smith).

6

The analysis of the CE amplifier is very similar to that of the CS amplifier asshown in Figure 3.

2.3 Some Observations

1. The upper 3-dB frequency is determined by the interaction of R′sig with

Cin = Cgs +Cgd(1 + gmR′L). Further, Rsig ‖ RG ≈ Rsig when RG � Rsig.

Hence, a large Rsig will cause fH to be lowered, decreasing the bandwidthof the amplifier.

2. The total capacitance Cin is increased by the Miller effect which magnifiesCgd by the factor 1 + gmR

′L, which lowers fH .

3. To improve the high-frequency response of MOSFET, one has to reducethe Miller effect.

4. The STC is an approximation because we are replacing a double-pole sys-tem with a single-pole system. A system with two capacitors has twopoles, but replacing it with one capacitor or one single pole is only ap-proximately correct.

5. The dominant high-frequency pole of the system is given by by fP ≈ fH .

2.4 Miller’s Theorem

Figure 4: The Miller equivalent circuit (Courtesy of Sedra and Smith).

The Miller’s theorem allows the replacement of a bridging capacitance by twoequivalent capacitances as shown in Figure 4. This theorem relies on that

V2 = KV1 (2.14)

in Figure 4. In this case, it can be shown that

Z1 = Z/(1−K), Z2 = Z/(

1− 1

K

)(2.15)

7

The proof is given in the textbook and will not be reproduced here.1

3 Useful Tools for High-Frequency Response ofAmplifiers

Next, some more sophisticated and useful tools for high-frequency analysis oftransistor amplifiers will be given. When the simple analysis previously dis-cussed fails, one may resort to these more sophisticated tools. This happens forinstance, when the poles and zeros are not far apart.

3.1 High-Frequency Gain Function

The frequency gain as a function of frequency can be expressed as

A(s) = AMFH(s) (3.1)

where in general,

FH(s) =(1 + s/ωZ1)(1 + s/ωZ2) · · · (1 + s/ωZn)

(1 + s/ωP1)(1 + s/ωP2) · · · (1 + s/ωPm)(3.2)

Notice that the above function FH(s)→ 1 when s→ 0, which is what is desired.

3.2 Determining the 3-dB Frequency fH

3.2.1 Single-Pole Case

As in the low-frequency case, when the above function is dominated by a singlepole, then one has

FH(s) ≈ 1

1 + s/ωP1(3.3)

The 3-dB point is easily shown to be

ωH ≈ ωP1 (3.4)

by first letting s = jω and then ω = ωP1. The above single pole approximationis good when the next pole or zero is two octaves (4 times) further away fromthe dominant pole.

3.2.2 Double-Pole Case

If a dominant pole approximation is not possible, the aggregate effects of thepoles and zeros need to be considered in finding ωH . For simplicity, one considersfirst a simple two-pole and two-zero system. Then

FH(s) =(1 + s/ωZ1)(1 + s/ωZ2)

(1 + s/ωP1)(1 + s/ωP2)(3.5)

1An elegant proof is also given by Davidovic in 1999 in IEEE Trans. Education.

8

Letting s = jω and taking the magnitude square of the above,2 one gets

|FH(jω)|2 =(1 + ω2/ω2

Z1)(1 + ω2/ω2Z2)

(1 + ω2/ω2P1)(1 + ω2/ω2

P2)(3.6)

By definition, at ω = ωH , the half-power point, |FH(jωH)|2 = 12 , and

1

2=

(1 + ω2H/ω

2Z1)(1 + ω2

H/ω2Z2)

(1 + ω2H/ω

2P1)(1 + ω2

H/ω2P2)≈

1 + ω2H

(1

ω2Z1

+ 1ω2

Z2

)+ · · ·

1 + ω2H

(1

ω2P1

+ 1ω2

P2

)+ · · ·

(3.7)

where we have kept only the quadratic terms in both the numerator and de-nominator. The remaining terms are proportional to ω4

H , which are negligiblewhen ωH is small, compared to the terms retained. The above equation can besolved approximately to yield3

ωH ≈ 1/√( 1

ω2P1

+1

ω2P2

)− 2

(1

ω2Z1

+1

ω2Z2

)(3.8)

Another statement that the above formula is making is that the half-powerfrequency point ωH exists only if

2

(1

ω2Z1

+1

ω2Z2

)>

(1

ω2P1

+1

ω2P2

)(3.9)

Otherwise, this frequency point is pure imaginary. This implies that the polelocations have to be small enough compared to the zero locations so that theabove inequality can be satisfied.

3.2.3 Multi-Pole Case

The above expression can be generalized to a multi-pole and multi-zero systemgiving

ωH ≈ 1/√( 1

ω2P1

+1

ω2P2

+ · · ·)− 2

(1

ω2Z1

+1

ω2Z2

+ · · ·)

(3.10)

Again, the half-power frequency point, ωH exists only if(1

ω2P1

+1

ω2P2

+ · · ·)> 2

(1

ω2Z1

+1

ω2Z2

+ · · ·)

(3.11)

2One has used that |A/B| = |A|/|B|, |AB| = |A||B|, and |x + jy|2 = x2 + y2 to arrive atthis.

3Without this approximation, the quartic equation is otherwise not easily solvable.

9

3.3 Low-Frequency Gain Function

A similar low-frequency gain function can be defined such that the frequencygain as a function of frequency can be expressed as

A(s) = AMFL(s) (3.12)

where

FL(s) =(1 + ωZ1/s)(1 + ωZ2/s) · · · (1 + ωZn/s)

(1 + ωP1/s)(1 + ωP2/s) · · · (1 + ωPm/s)(3.13)

Notice that the above function FL(s)→ 1 when s→∞, which is what is desired.A similar analysis shows that the half-power frequency point ωL is4

ωL ≈√

(ω2P1 + ω2

P2 + · · · )− 2 (ω2Z1 + ω2

Z2 + · · · ) (3.14)

Again, this half-power frequency point ωL exists only if(ω2P1 + ω2

P2 + · · ·)> 2

(ω2Z1 + ω2

Z2 + · · ·)

(3.15)

One can obtain the above result by comparing (3.2) and (3.13). One noticesthat the roles of ω’s and s’s are switched in the formulas. Hence, by symmetry,one arrives at the above formula by letting s ω.

3.4 The Method of Open-Circuit Time Constants

In finding fL, when a single pole dominates and they are far from each other, oneuses the short-circuit time-constant method to decouple the capacitors and findtheir respective time constants. The rationale is that at the highest frequencypole that decides fL, the frequency is high enough such that the other capacitorscan be considered short-circuited, and hence, the determination of the highestfrequency pole is a reasonable approximation (see Figure 5).

4This formula is given without derivation in the textbook in equation (10.21). Here’s thederivation for it.

10

Figure 5: Reminder figure for the Bode plot of the low-frequency response of atransistor amplifier for defining ωL (fL) (Courtesy of Sedra and Smith).

By the same token, when one finds fH , it is the lowest frequency pole thatdominates fH (see Figure 6). Again, this frequency is low enough that theother capacitors can be considered open-circuited. Then the time-constant forthe lowest frequency pole is fairly accurate. With this in mind, then

ωH = 2πfH ≈1∑

i CiRi(3.16)

In the above sum, it will automatically be dominated by the largest RC timeconstant term, since it is an ordinary mean.

11

Figure 6: Reminder figure for the Bode plot of the high-frequency response ofa transistor amplifier for defining ωH (fH) (Courtesy of Sedra and Smith).

In contrast, in the short-circuit time-constant method,

ωL = 2πfL ≈∑i

1

RiCi(3.17)

which is proportional to the harmonic mean of the RC time constants. Theterm with the shortest time constant automatically dominates this sum.

12