1

22-1

EE105 – Fall 2015 Microelectronic Devices and Circuits

Prof. Ming C. Wu

wu@eecs.berkeley.edu

511 Sutardja Dai Hall (SDH)

22-2

Typical Frequency Response of Discrete MOS or BJT Amplifiers

2

22-3

Frequency Response of Direct-Coupled (dc) Amplifier

Gain does not fall off at low frequencies, and the midband gain AM extends down to zero frequency

22-4

Cross Section of MOSFET Showing Internal Capacitances

MOSFET has several internal capacitances, which take time to charge/discharge, limitingthe transistor speed.

Gate/Source capacitance:

Cgs =23WLCox +Cov

where Cox = εox / t [F/cm2 ]W : transistor widthL : gate lengthCov =WLovCox

Lov : overlap bet. gate/sourceor gate/drain, ~ 0.05 to 0.1L

Gate/Drain capacitance:Cgd =Cov

3

22-5

(Simplified) High-Frequency Equivalent-Circuit Model for MOSFET

Hybrid-pi Model T Model

Capacitances between source/body, Csb, and between drain/body, Cdb, are neglected

22-6

Unity-Gain Frequency, fT

Vgs =Ii

sCgs + sCgd

KCL: Io +Vgs

1/ sCgd

= gmVgs

Io = gmVgs − sCgdVgs ≈ gmVgs

= gmIi

sCgs + sCgd

AI =IoIi=

gmsCgs + sCgd

s = jω

IoIi=

gmω(Cgs +Cgd )

ωT =gm

Cgs +Cgd

fT : defined as frequency at which short-circuit current gain = 1fT : a figure-of-merit for transistor speed

Drain is grounded (short-circuit load)

As gate length reduces in advanced technology node, Cgs reduces and fT increases

4

22-7

High Frequency Response of CS Amplifier

Frequency response can be solvedby analyzing the gain of the high-frequency equivalent circuit.However, it is tedious, and doesnot show intuitive interpretation.

Vo = (Igd − gmVgs )RL' ≈ −gmVgsRL

'

Igd << gmVgs (Cgd is small)

RL' = ro || RD || RL

Igd =Vgs −Vo1/ sCgd

=Vgs + gmRL

'Vgs1/ sCgd

Igd =Vgs (1+ gmRL

' )1 / sCgd

22-8

Miller Capacitance Igd =Vgs (1+ gmRL

' )1 / sCgd

=Vgs

1/ s(1+ gmRL' )Cgd

=Vgs

1/ sCeq

Ceq = (1+ gmRL' )Cgd

Vgs =RG

Rsig + RGVsig

!

"##

$

%&&

1/ sCin

Rsig' +1/ sCin

=RG

Rsig + RGVsig

!

"##

$

%&&

1sωH

+1

ωH =1/CinRsig'

Cin =Cgs + (1+ gmRL' )Cgd

AV (s) = −RG

Rsig + RGgmRL

'!

"##

$

%&&

1sωH

+1

Large input capacitance in CS amplifier due to Miller effect, which greatly reduced the bandwidth of CS amp

5

22-9

Miller Theorem Miller Equivalent Circuit

If V2 = KV1

Input side:

I = V1 −V2

Z=

(1−K )V1

Z=

V1

Z(1−K )

=V1

Z1

Z1 =Z

(1−K )

Ouput side:

I = V2 −V1

Z=

(1−1/K)V2

Z=

V2Z

1−1/K

=V2

Z2

Z2 =Z

1−1/K

22-10

Open-Circuit Time Constant (OCTC) Method for High Cut-off Frequency

1. Replace all capacitance by open circuit2. Replace signal source by short circuit3. Consider one capacitor at a time, find resistance Ri "seen" by the i-th capacitor, Ci

4.

ωH ≈1CiRi

i∑

6

22-11

Applying OCTC to CS Amplifier

Rgs = Rsig'

Ix = gmVgs +VdRL' = gmVgs +

Vx +VgsRL'

Vgs = −IxRsig'

Rgd =VxIx= Rsig

' (1+ gmRL' )+ RL

'

RCL= RL

'

22-12

Applying OCTC to CS Amplifier

τ H = Rsig' Cgs + Rsig

' (1+ gmRL' )+ RL

'( )Cgd + RL' CL

Rearranging :τ H = Rsig

' Cgs + Rsig' (1+ gmRL

' )Cgd + RL' Cgd + RL

' CL

≈ Rsig' Cgs + (1+ gmRL

' )Cgd( )+ RL' Cgd +CL( )

Time constant from input port of

Miller Equivalent Circuit

Time constant from Output port of

Miller Equivalent Circuit

7

22-13

High-Frequency Response of CG Amplifier

Rsig ||1gm

+RLgmro

!

"#

$

%&

Note: Cgd and CL can be lumped togetheer since they are in parallel.Rgd = RL || Ro ≈ RLRo = ro + (1+ gmro )Rsig

τ H = Rgd Cgd +CL( )+ RgsCgs

= RL Cgd +CL( )+ Rsig ||1gm

+RLgmro

!

"#

$

%&Cgs

fH =12π

1τ H

•  No Miller effect since both capacitance are grounded •  The dominant term is likely to be (1/gm)Cgs, which is small à High fH à Common-Gate is a broadband amplifier

22-14

Impedance Transformation of Common Gate Amplifier

Impedance transformation:

Look into Drain : Rs amplified by gmro( )Look into Source : RL reduced by gmro( )

From Lecture Slide: 17-5

8

22-15

More Detailed Analysis of Resistances in OCTC