EECS 105 Fall 1998Lecture 28

Frequency Response of Transistor Amplifiers

■ Simplest case: CE short-circuit current gain Ai(jω) as a function of frequency

Kirchhoff’s current law at the output node:

Kirchhoff’s current law at the input node:

■ Solving for Vπ at the input node:

Cπ

Cµ

gmVπrπ Vπ

+

−

Iin

Io

Io gmVπ Vπ jωCµ–=

Is

VπZπ------ VπjωCµ+= where Zπ rπ

1jωCπ-------------

=

VπIs

1 Zπ⁄( ) jωCµ+--------------------------------------=

28

EECS 105 Fall 1998Lecture 28

Short-Circuit Gain Frequency Response

■ Substituting Vπ into the output node equation--

■ Substituting for Zπ and simplifying --

Current gain has one pole:

and one zero

Io

Is----

gmZπ 1jωCµ

gm-------------–

1 jωCµZπ+------------------------------------------=

Io

Is----

gmrπ 1jωCµ

gm-------------–

1 jωrπ Cπ Cµ+( )+-----------------------------------------------

βo 1jωCµ

gm-------------–

1 jωrπ Cπ Cµ+( )+----------------------------------------------- βo

1 jωωz------–

1 jωωp------+

------------------= = =

ωp rπ Cπ Cµ+( )( ) 1–=

ωz gm1–Cµ( )

1–ωp»=

EECS 105 Fall 1998Lecture 28

Bode Plot of Short-Circuit Current Gain

■ Note low frequency magnitude of gain is βo

■ Frequency at which current gain is reduced to 0 dB is defined as the transition frequency ωT. Neglecting the zero,

1

ω

IoIin

1

0

−45

−90

−135

−180

IoIin

∠

ω

(a)

log scale

log scale

(b)

Cµ Cπ + Cµ

rπ(Cπ + Cµ)

1rπ(Cπ + Cµ)

gmCπ + Cµ

gm Cµ

gm Cµ

βo = gmrπ

ωT

gm

Cπ Cµ+( )-------------------------=

EECS 105 Fall 1998Lecture 28

Transition Frequency of the Bipolar Transistor

■ Dependence of transition time τT = ωT-1 on the bias collector current IC:

■ If the collector current is increased enough to make the second term negligible, then the minimum τT is the base transit time, τF.. In practice, the ωT decreases at very high values of IC due to other effects and the minimum τT may not be achieved.

■ Numerical values of fT = (1/2π)ωT range from 10 MHz for lateral pnp’s to 10 GHz for oxide-isolated npn’s

Note that the small-signal model is not valid above fT (due to distributed effects in the base) and the zero in the current gain is not observed

τT1

ωT-------

Cπ Cµ+

gm--------------------

gmτF CjE Cµ+ +

gm------------------------------------------= = =

τT τF

CjE Cµ+

gm----------------------

+ τF

Vth

IC-------- CjE Cµ+( )+= =

EECS 105 Fall 1998Lecture 28

Common-Source Current Gain

■ CS amplifier has a non-infinite input impedance for ω > 0 and we can measure its small-signal current gain.

■ Analysis is similar to CE case; result is

■ Transition frequency for the MOSFET is

+

−

IingmVgs

Cgd

Cgs

Io

Vgs

Io

I in------

gm 1jωCgd

gm----------------–

jω Cgs Cgd+( )--------------------------------------gm

ω Cgs Cgd+( )----------------------------------≈=

ωT

gm

Cgs Cgd+-------------------------≈

EECS 105 Fall 1998Lecture 28

Transition Frequency of the MOSFET

■ Substitution of gate-source capacitance and transconductance:

and

■ The transition time is the inverse of ωT and can be written as the average time for electrons to drift from source to drain

velocity saturation causes τT to decrease linearly with L; however,submicron MOSFETs have transition frequencies that are approaching those for oxide-isolated BJTs

Cgs23---WLCox= Cgd» gm

WL-----µnCox VGS VTn–( )=

ωT

gm

Cgs---------

WL-----µnCox VGS VTn–( )

23---WLCox

---------------------------------------------------- 32---µn

VGS VTn–( )L

------------------------------ L= =≈

τTL

µn23---

VGS VTn–( )L

------------------------------

---------------------------------------------- L

vdr

----------= =

EECS 105 Fall 1998Lecture 28

Frequency Response of Voltage Amplifiers

■ Common-emitter amplifier:

Procedure: substitute small-signal model and perform phasor analysis

+

−

V +

V−

RS

RL

iSUP

iOUT

Vs

VBIAS

vOUT+−

+−

EECS 105 Fall 1998Lecture 28

Brute Force Phasor Analysis

■ “Exact” analysis: transform into Norton form at input to facilitate nodal analysis

Note that Ccs is omitted, along with rb

Details: see Section 10.4

Factor (approximately!) into one high-frequency zero and two poles ω1 << ω2

Vs rπ

RSCµ

CπVπ

+

−

VoutRL

+

−

gmVπ ro roc+−

Is R'in

R'in = RS rπ R'out = ro roc RL

R'out

Cµ

CπVπ

+

−

Vout

+

−

gmVπ

(a)

(b)

ω11–

rπ RS( ) Cπ 1 gmro roc RL+( )Cµ+( ) ro roc RL( )Cµ+=

ω21–

ro roc RL( ) rπ RS( )CµCπrπ RS( ) Cπ 1 gmro roc RL+( )Cµ+( ) ro roc RL( )Cµ+

-------------------------------------------------------------------------------------------------------------------------------------------------=

EECS 105 Fall 1998Lecture 28

The Miller Approximation

■ The “exact” analysis is not particularly helpful for gaining insight into the frequency response ... consider the effect of Cµ on the input only

neglect the feedforward current Iµ in comparison with gm Vπ ... a good approximation

It = (Vt - Vo) / Zµ

Vo = - gmVt RL / (RL + Rout’) = AvCµ Vt

where AvCµ is the low frequency voltage gain across Cµ

It = Vt (1- Av) / Zµ

Zeff = Vt / It = Zµ / (1 - Av)

is the Miller capacitor

Cµ

gmVt R'out = ro roc RL Vout

+

−

Vt

It

+−

Zeff1

jωCµ------------- 1

1 AvCµ–

--------------------- 1

jω Cµ 1 AvCµ–( )( )-------------------------------------------- 1

jωCM---------------= = =

CM 1 AvCµ–( )Cµ=

EECS 105 Fall 1998Lecture 28

Generalized Miller Approximation

■ An impedance Z connected across an amplifier with voltage gain AvZ canbe replaced by an impedance to ground ... multiplied by (1-AvZ)

■ Common-emitter and common-source:

AvZ = large and negative for Cµ or Cgd --> capacitance at the input is magnified

■ Common-collector and common-drain:

--> capacitance at the input due to Cπ or Cgs is greatly reduced

Vi

+

−Vo

+

−

Vi

+

−Vo

+

−

AvoAvo

Z

Zeff

Zeff = Z /(1 − Avo)

AvZ 1≈

EECS 105 Fall 1998Lecture 28

Voltage Gain vs. Frequency for CE AmplifierUsing the Miller Approximation

■ The Miller capacitance is lumped together with Cπ, which results in a single-pole low-pass RC filter at the input

Transfer function has one pole and no zero after Miller approximation:

from the exact analysis (final term is missing)

Vs rπ

RS

CMCπ

CM = Cµ(1 + gmR'out)

R'outVπ

+

−

Vout

+

−

gmVπ+−

ω3dB1–

rπ RS( ) Cπ CM+( )=

ω3dB1–

rπ RS( ) Cπ 1 gmro roc RL+( )Cµ+[ ]=

ω3dB1– ω1

1–≈ Rout′Cµ

EECS 105 Fall 1998Lecture 28