Вы находитесь на странице: 1из 10

Microelectronic Circuits -10EC63

Prof. Dr. Noorullah Shariff C.

Single Stage IC Amplifiers (Unit-3)
S No Question and Answer Mar mm-
k yy
1 Q. Briefly explain common source amplifier with active load.
Fig shows CS Amplifier with active load ( with constant
current source I between drain and VDD)

The current source is implemented using a PMOS.

Q1 is biased in such a way that it operates in

saturation region.

Small signal equivalent circuit is shown in fig b. From

this fig it is clear that

Ri = , Ro = r o &
Avo = (vo/vgs) = (-gm vgs ro)/vgs = (-gm ro )

Also intrinsic gain Av = Avo = (gm ro )

2 Q. Find Ao for an NMOS transistor fabricated in a 0.4 m
CMOS process for which kn = 200A/V2 and VA =20V/m. L=
0.4 m & overdrive voltage is 0.25V. Find W when I D = 100
A. Also find gm & ro .
gm = 2 ID/VOV = 2 x 100 x 10-6/0.25 = 80A/V
ro =VA/ ID = VA L/ ID =20 x 0.4/100 x 10-6 =80K
Av = (gm ro ) = 64
ID =(1/2) kn W/L V2OV
W= (2 L ID )/ (kn V2OV) = 6.4 m
3 Q. Explain CMOS implementation of a CS amplifier & also
draw its characteristic curve & transfer characteristic.


Here load current source I is implemented using Q 2.

Q2 is the output transistor of the current mirror (Q 2
Q3) which is fed by bias current I REF. Q2 & Q3 are
matched transistors.

i-v characteristic of Q2 (PMOS) which is iD vs vSD

characteristic is as shown in the fig b. Q2 is in
saturation so that its finite output resistance ro2 is
given by
ro2 = VA2/ IREF , where VA2 is the early voltage
i-v characteristic of Q1 (NMOS) which is iD vs vDS
characteristic is as shown in the fig c. This is
superimposed with the load curve which is the flipped
& shifted version of fig b (i-v characteristic of Q 2 ).
The intersection of each curve with the load line gives
the corresponding value of vDS, which is equal to vo.
Using this vO- vI characteristics is found point by point
which is sketched in fig d. This has four regions I,II,III &
Segment III is the region is of interest which is linear.
Here both Q2 & Q1 are in saturation. End points of
segment III are A & B.
A is Pinch-off point of Q2 & vO = VDD VOV2
B is Pinch-off point of Q1 & vO = vI - Vtn .
Small signal voltage gain is obtained by replacing Q!
by its small signal model and Q2 by its output
resistance ro2.

where ro = ro1 ro2

vo = - gm1 vgs1 (ro1 ro2)

Av = vo/ vgs1= - gm1 (ro1 ro2)

In case when ro1 = ro2 the value of small signal gain is
given by
Av = - gm1ro1/2

Q. A CMOS common source amplifier shown in
fig is fabricated with W/L= 100m/1.6m for all
transistors., with kn = 90A/V2, kp =30A/V2,
IREF = 100A, VAn= 8V/m & VAp = 12V/m.
Determine gm1, ro1,ro2, Avo
ID1 = IREF = (1/2) kn (W/L) V2ov1 = ID2
100 = (1/2) 90 (100/1.6) V2ov
Vov1 =0.188 V
gm1 = 2 ID1/ Vov1 = 2 x 100/0.188 = 1060A/V
ro1 = VAn/ ID1 = VAn x L/ ID1 =8x 1.6/100 = 0.144M
ro2 = VAp/ ID2 = VAp x L/ ID2 =12 x 1.6/100
Avo =- gm1 (ro1 ro2) = 1060 (0.144 x 0.192)

5 Q. Draw the high frequency equivalent circuit of MOS CS

amplifier & explain the significance of each element.


Input Signal source is represented by Vsig and Rsig,

which are the signal source voltage & internal
resistance, taking bias resistance etc. into account,
RL represents actual load RL, output resistance ro of
MOSFET & input resistance of succeeding amplifier
Cgs & Cgd are gate to source & gate to drain
capacitances respectively which consist of gate oxide
capacitance and depletion region capacitance
CL is the output capacitance which consists of drain to
body capacitance & input capacitance of succeeding
amplifier stage.
gm is the trans-conductance.

High frequency analysis for CS amplifier

i)Using Millers Effect

AM = Vo/Vsig = -gm Vgs RL/ Vgs = -gm


ii) Using open circuit time constants

The open-circuit time constant of Cgs is given by

gs = CgsRgs

The resistance Rgd as seen by Cgd is found by setting Cgs

= 0 and short-circuiting Vsig. The result is the circuit in
Fig. to which we apply a test current Ix. Writing a node
equation at G gives
i.e., Vgs = - Ix Rsig.(1)

A node equation at D Provides

Ix = gm Vgs [(Vgs + Vx)/RL]

Substituting for Vgs from Eq. (1) and

Ix = - gm Ix Rsig [(- Ix Rsig + Vx)/RL]

Ix RL + gm Ix Rsig RL+ Ix Rsig = Vx

Rgd = Vx / Ix = RL+ Rsig + gm Rsig RL

Rgd = Vx / Ix = RL+ Rsig (1+ gm RL)

The open-circuit time constant of Cgd is
gd = Cgd Rgd

The upper 3-dB frequency H can now be determined from

H = 1/ ( gs + gd )

6 Q. In CG amplifier with active load obtain 3bd

frequency fH using open circuit time constants. Draw
the circuit required for determining Rgs and Rgd .

Using Millers theorem

From the above fig it is clear that there are two

poles: one at the input side with a frequency fP1

and the other at the output side with a frequency fP2.

fP2 is the dominant pole, therefore fH = fP2

Using open circuit time constants

Fig above shows circuits for determining Rgs and Rgd

Rgs = Rs Rin and

Rgd = RL Rout

Input time constant is T1 = Cgs Rgs

Output time constant is T2 = (CL + Cgd) Rgd

i.e., fH = 1/(T! +T2)

7 Q. Draw the small signal model of a common base amplifier

and hence derive the expression for Rin, Rout & Avo

To Find Rin
From fig b,
& ii = vi/re + iro
i.e., ii = vi/re + (vi vo)/ro
i.e., ii = vi/re + vi/ro (ioRL)/ ro
Using (i)
i.e., ii = vi/re + vi/ro (ii vi/r)RL/ ro
i.e., ii + ii RL/ ro = vi(1/re + 1i/ro + RL /r ro)
i.e., ii (1+ RL/ ro )= vi(1/re + 1i/ro+RL /r ro)
i.e., ii (ro + RL)/ ro = vi/ ro (1 + ro /re+ RL /r)
Rin = vi/ii

where r =(1+ ) re
To Find Avo
From (c)
Vo = gm Vi ro + Vi
i.e., Vo = Vi (gm ro + 1 )

i.e., Av =Vo /Vi =(1 + gm ro )

To Find Rout