Solution of Assignment 1

CMOS Analog IC Design

Prepared by Sachin Kumar(S23100)
Instructor : Prof. Hitesh Shrimali
School: SCEE (IIT MANDI)
September 20, 2025

Figure 1: Caption

1 Solution (Equivalent small-signal low frequency AC

model for BJT)
1.1 Small-Signal BJT Model in Z–parameters
Consider a BJT in common-emitter configuration (emitter as reference). Port 1 is between
base and emitter, Port 2 between collector and emitter.
Parameters:

rπ = input resistance, gm = transconductance, ro = output resistance.

Step 1: Hybrid-π model equations

With V1 = vbe and V2 = vce :
V1 V2
I1 = , I2 = −gm V1 − .
rπ ro

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Step 2: Express V1 and V2 in terms of I1 and I2
From the first equation:
V1 = rπ I1 .
From the second:
V2  
I2 = −gm V1 − ⇒ V2 = −ro I2 + gm V1 .
ro
Substitute V1 = rπ I1 :
V2 = −gm ro rπ I1 − ro I2 .
Thus " # " #" #
V1 rπ 0 I1
= .
V2 − gm ro rπ − ro I2

Step 3: Z–parameter matrix

" # " #
z11 z12 rπ 0
Z= = .
z21 z22 − gm ro rπ − ro

Step 4: Physical meaning of each term

• z11 = rπ : input impedance at the base with collector open.

• z12 = 0: no direct dependence of base voltage on collector current in this simple model.

• z21 = −gm ro rπ : collector voltage generated by base current due to transconductance

and output resistance.

• z22 = −ro : output impedance at collector with base open (negative sign from current
convention).

1.2 h-parameters

V1 = h11 I1 + h12 V2 (1)

I2 = h21 I1 + h22 V2 (2)

h-parameter for BJT:

VBE = h11 IB + h12 VCE (3)

IC = h21 IB + h22 VCE (4)

2
 I1
I2
B h11 C
+ +
h21 I1
V1 +
- h22
V2
h12 V2
- -
E
Figure 2: Small-signal model of BJT using h-parameter

BJT h-parameters (Common Emitter Configuration):

VBE
h11 = (Input impedance, VCE = 0) (5)
IB
VBE
h12 = (Reverse voltage gain, IB = 0) (6)
VCE
IC
h21 = (Current gain, VCE = 0) (7)
IB
IC
h22 = (Output admittance, IB = 0) (8)
VCE

B C
+ +
hie
VBE VCE
hfe ib
- -
E
Figure 3: small signal model of BJT using h-parameter

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 VBE
h12 = hre = =0 (9)
VCE
h12 = hre = 0 (neglected) (10)
IC
h22 = hoe = =0 (11)
IB
h22 = hoe = 0 (12)
1
= (very large = replace by open circuit ) (13)
hoe

1.3 Y-parameters

    
I1 A B V1
= (14)
I2 C D V2

1. Current at base:
VBE
IBE = (15)
rπ
2. Current at collector:
VCE
IC = gm VBE + (16)
ro
where:
VCE = VC − VE = V2 (17)

3. Voltage at input port:

V1 = VBE + V2 (18)
which gives:
VBE = V1 − V2 (19)

4. Current at output port:

I2 = −IC (20)
(The negative sign is due to the port direction convention.)

VBE V1 − V2
Base current: I1 = IB = =
rπ rπ
VCE
Collector current: IC = gm VBE +
ro
V2
= gm (V1 − V2 ) +
ro

4
 So,

I2 = −IC
V2
= −gm (V1 − V2 ) −
ro
V2
= −gm V1 + gm V2 −
ro
 
1
= −gm V1 + gm − V2
ro

Now we can write I1 and I2 in terms of V1 and V2 :

V1 V2
I1 = −
rπ rπ
 
1
I2 = −gm V1 + gm − V2
ro

Thus, in matrix form:

1 1
 
 
I1 −  
V1
 rπ rπ 
= 1  V2
I2 −g gm −
m
ro
This represents the Y-parameter form.

1.4 ABCD Parameter

I1 -I2
V1 + ABCD +V
2
- PARAMETRS -

Figure 4: Two-port network of ABCD parameter

Now, from the above Y-parameter equations, we can derive the ABCD parameters using I1
and I2 .
From I1 :
V1 − V2
I1 =
rπ
V1 = I1 rπ + V2

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 V2
I2 = −gm (V1 − V2 ) −
ro
V2
Substitute V1 : I2 = −gm (I1 rπ ) − + gm V 2
ro
 
1
I2 = −gm rπ I1 + gm − V2
ro
 
1
Solve for I1 : gm rπ I1 = −I2 + gm − V2
ro

−I2 gm − r1o
I1 = + V2
gm rπ gm rπ
Alternative approach: isolate VBE from I2 .

V2
I2 = −gm VBE −
ro
V2
gm VBE = −I2 −
ro
V2
−I2 − ro
VBE =
gm
VBE
Now, using I1 = :
rπ
V2
!
1 −I2 − ro
I1 =
rπ gm

−I2 V2
I1 = −
gm rπ gm rπ ro
Let’s now directly write the expressions.
From V1 :

V1 = I1 rπ + V2
A = 1, B = rπ

From I1 :
We already have it.
From I2 :

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 Start again:
V2
I2 = −gm (V1 − V2 ) −
ro
V2
I2 = −gm (I1 rπ + V2 − V2 ) −
ro
V2
I2 = −gm rπ I1 −
ro
Rearrange:
V2
I2 + = −gm rπ I1
ro
 
−1 V2
I1 = I2 +
gm rπ ro
This form helps isolate I1 as needed.
Now, solving for C and D:

I1 = CV2 + DI2
Comparing:
−1 −1
I1 = V2 + I2
gm rπ ro gm rπ
Thus:
−1 −1
C= , D=
gm rπ ro gm rπ
The ABCD parameters for the BJT small-signal model are:
ro 1 ro
A=1+ , B = ro (1 + gm rπ ), C=− , D=− (21)
rπ rπ rπ

S–Parameter Derivation for the Small-Signal BJT Model

We start from the Z–parameter matrix of the BJT in common-emitter configuration:
" #
rπ 0
Z= ,
− gm ro rπ − ro
where rπ is the input resistance, gm the transconductance, and ro the output resistance.

General Formula
For a two-port with reference impedance Z0 :


−1
S = Z − Z0 I Z + Z0 I .
Let
A = Z + Z0 I, B = Z − Z0 I.
−1
Then S = BA .

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Step 1: Write A and B
" # " #
rπ + Z0 0 rπ − Z0 0
A= , B= .
− gm ro rπ −ro + Z0 − gm ro rπ −ro − Z0

Step 2: Invert A
Since A is lower-triangular,
 
1
rπ +Z0
0
A−1 =  gm ro rπ
.
1
(rπ +Z0 )(−ro +Z0 ) −ro +Z0

Step 3: Multiply to get S

" #
S11 S12
S = BA−1 = .
S21 S22
Compute each entry:
rπ − Z 0
S11 = , S12 = 0,
rπ + Z 0
2Z0 gm ro rπ ro + Z0
S21 = − , S22 = .
(rπ + Z0 )(−ro + Z0 ) ro − Z0

Step 4: Final S–parameter Matrix

 rπ − Z0 
" # 0
S11 S12  rπ + Z0 
= .
S21 S22  2Z0 gm ro rπ ro + Z0 
−
(rπ + Z0 )(−ro + Z0 ) ro − Z0

Step 5: Meaning of Each Term

• S11 : Reflection coefficient at the input (base).

• S12 = 0: Reverse isolation (no signal from output to input in this model).

• S21 : Forward gain from base to collector, including gm , ro and rπ .

• S22 : Reflection coefficient at the output (collector).

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2 Solution (Equivalent small-signal low frequency AC
model for the MOSFET)
2.1 y-parameters
The small-signal y-parameter equations for a MOSFET are expressed as:
1. General equation of y-parameter:
I1 = Y11 V1 + Y12 V2 (22)
I2 = Y21 V1 + Y22 V2 (23)

2. y-parameters for MOSFET:

IGS = y11 VGS + y12 VDS (24)
IDS = y21 VGS + y22 VDS (25)

3. Input Conductance:
IGS = 0 (26)
∂IGS
=0 (27)
∂VGS
∂IGS
=0 (28)
∂VDS
y11 = 0
(29)
y12 = 0

4. Output Conductance:
∂IDS λKn
y22 = = (VGS − VT )2 (30)
∂VDS 2

5. Transconductance:
∂IDS
gm = = kn (VGS − VT )(1 + λVDS ) (31)
∂VGS

9
 D G D
+ gmVgs
ro
G Vgs
M1
-
S
S

Figure 5: Small-signal model of MOSFET using Y-parameter

2.2 h-parameters
Similarly, the h-parameter representation is written as:

V1 = h11 I1 + h12 V2 (32)

I2 = h21 I1 + h22 V2 (33)
h-parameter for MOSFET:
VGS = h11 IGS + h12 VDS (34)
IDS = h21 IGS + h22 VDS (35)

IGS
h11 IDS
G D
+ +
h21 I1
+
VGS - h22 VDS
h12 V2
- -
S
Figure 6: Small-signal model of MOSFET using h-parameter

MOSFET h-parameters (Common Source Configuration):

h11 ≈ ∞ (Input impedance, gate draws no current) (36)

h12 ≈ 0 (Reverse voltage gain negligible) (37)
h21 ∼ gm (Forward transfer gain) (38)
1
h22 = (Output admittance from channel-length modulation) (39)
ro

2.3 Z-parameter
We want the Z-matrix defined by
    
v1 Z11 Z12 i1
= .
v2 Z21 Z22 i2

10
Step 1: Gate relation
v1
From i1 = we get
Rg
v1 = Rg i1 ⇒ Z11 = Rg .

Step 2: Why Z12 = 0

By definition,
v1
Z12 = .
i2 i1 =0

In the intrinsic low-frequency MOSFET model (ignoring Cgd , Cgs and body effect), the gate
node is electrically isolated from the drain. Thus, any current i2 injected into the drain
cannot affect the gate voltage. With i1 = 0, the gate current is zero, and therefore v1 = 0.
Hence,
Z12 = 0.

Step 3: Drain relation

From the drain equation
v2
i2 = gm v1 + ,
ro
we rearrange to obtain
v2 = ro (i2 − gm v1 ) .
Substituting v1 = Rg i1 gives
v2 = − gm ro Rg i1 + ro i2 ,
so
Z21 = −gm ro Rg , Z22 = ro .

Final Z-matrix
Thus, the low-frequency Z-matrix is
   
Z11 Z12 Rg 0
Z= = .
Z21 Z22 − gm ro Rg ro

2.4 ABCD parameter

ABCD Parameters of MOSFET Small-Signal Model
The transmission (ABCD) parameters are defined as
    
v1 A B v2
= .
i1 C D −i2

11
 From the Z-matrix of the MOSFET:
" #
Rg 0
Z= ,
− gm ro Rg ro

we use the standard Z-to-ABCD conversion:

 
  Z11 det(Z)
A B  Z21 Z21 
= .
C D  1 Z22 
Z21 Z21
Substituting,

det(Z) = Rg ro , Z11 = Rg , Z21 = −gm ro Rg , Z22 = ro ,

we obtain
1 1 1 1
A=− , B=− , C=− , D=− .
gm ro gm gm ro Rg gm Rg

1 1
 
 − gm ro −
" #
A B gm 
=
 
C D 1 1 
− −
gm ro Rg gm Rg

2.5 S-parameter
Assume a small–signal MOSFET with source grounded, ideal gate (no gate current), transcon-
ductance gm and output resistance ro . Port impedance is Z0 at both ports.

Step 1: Y–parameters
The current–voltage relationship is:
" #
I1
"
Y11 Y12
#" #
V1 Y11 = 0, Y12 = 0,
= with 1
I2 Y21 Y22 V2 Y21 = gm , Y22 = .
ro
Hence " #
0 0
Y= 1
.
gm ro

Step 2: Conversion to S–parameters

For equal Z0 at both ports,

S = (I + Z0 Y)−1 (I − Z0 Y)

12
 Compute " #
0 0
Z0 Y = Z0 .
Z0 gm ro

Then " # " #

1 0 1 0
I + Z0 Y = Z0 , I − Z0 Y = Z0 .
Z0 gm 1 + ro
− Z0 gm 1 − ro

Determinant of (I + Z0 Y):
Z0
∆=1+ .
ro
Inverse:  
1 0
(I + Z0 Y)−1 =  Z0 gm 1 .
 
−
1 + Zro0 1 + Zro0
Multiply by (I − Z0 Y) to obtain the S–parameters:

S11 = 1,
S12 = 0,
2Z0 gm
S21 = − ,
1 + Zro0
Z0
1− ro
S22 = Z0
.
1+ ro

Thus the complete S–matrix is

 
1 0
Z0
S =  2Z0 gm 1− .
 
ro
−
1 + Zro0 1+ Z0
ro

Problem Statement
Q3. Determine the impedance at various nodes (Gate, Drain, Source, and Bulk) of a MOS-
FET using Thevenin’s or Norton’s theorem.

Given/Assumption
For Thevenin/Norton analysis:

• All independent voltage sources are shorted.

• All independent current sources are opened.

• Small-signal model of NMOS transistor is used with:

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 – gm : transconductance
– ro : drain-source output resistance
– RG : gate resistor
– RD : drain resistor (load)
– RS : source resistor (degeneration)

Solution

VD

ID
RG RG
IG ID
+
+ + VGS
− −
VG VG - ro
IS
IS
RS RS
+ +
−
VS −
VS

Figure 7: Caption

1. Small-Signal Model
The MOSFET small-signal current is:
vds
id = gm vgs + , vgs = vg − vs , vds = vd − vs
ro

2. Gate Impedance (ZG )

At low frequency (ignoring capacitances):
(
RG , if gate bias resistor present
ZG =
∞, if gate is ideal (no DC path)

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3. Drain Impedance (ZD )
Attach a test voltage Vtest at drain:

ZD = ro ∥ RD
With source degeneration RS , feedback modifies the impedance:

ZD ≈ ro ∥ (RD + (1 + gm RS )RS ) (approximate)

4. Source Impedance (ZS )

Attach a test voltage at the source:
(
1
, if source grounded and gate at AC ground
ZS = gm 1
RS ∥ gm , if source resistor RS present

5. Bulk Impedance (ZB )


0,
 if bulk tied to AC ground
ZB = ZS , if bulk tied to source

rj , if considering junction dynamic resistance


6. High-Frequency Considerations
Including gate-source Cgs and gate-drain Cgd capacitances:

1
ZG (jω) ≈ RG ∥
jω (Cgs + Cgd (1 − Av ))
1
ZD (jω) ≈ ro ∥ RD ∥
jωCgd
1 1
ZS (jω) ≈ RS ∥ ∥
gm jωCgs
ZB (jω) ≈ junction capacitance Csb in parallel with other paths

Conclusion
Using Thevenin’s theorem, the small-signal impedances of MOSFET nodes are summarized:

ZG ≈ RG (or infinite for ideal gate)

ZD ≈ ro ∥ RD
1
ZS ≈ (or RS ∥ 1/gm )
gm
ZB ≈ 0 (if bulk grounded) or ZS if bulk tied to source

15
 Vc

ic
Rb Vc
Rb Vb
ib
ic
+ +
− Ve
− gmVbe
Vb ro
Vb ie

Re Re

+ +
− −
Ve Ve

Figure 8: bjt and small signal model for impedence calculation

Problem Statement
Q4. Determine the impedance at various nodes (Collector, Emitter, and Base)
of a BJT using Thevenin’s or Norton’s theorem.

Given/Assumptions
For Thevenin/Norton analysis:

• All independent voltage sources are shorted.

• All independent current sources are opened.

• Small-signal model of NPN BJT is used with:

– rπ : base-emitter resistance
– gm : transconductance (gm = IC /VT )
– ro : collector-emitter output resistance
– RC : collector resistor
– RE : emitter resistor (degeneration)
– RB : base resistor (biasing)

16
Solution
1. Small-Signal Model
The small-signal current equation is:
vce
ic = gm vbe + , vbe = vb − ve , vce = vc − ve
ro

2. Base Impedance (ZB )

Attach a test voltage Vtest at the base and compute resulting current Itest . For low frequency:

ZB ≈ RB ∥ rπ ∥ (β + 1)(RE ∥ ro )

3. Collector Impedance (ZC )

Attach a test voltage at the collector:

ZC ≈ RC ∥ ro (if emitter is AC grounded)

With emitter degeneration RE :

ZC ≈ RC ∥ [ro + (1 + gm RE )RE ] (approximate)

4. Emitter Impedance (ZE )

Attach a test voltage at the emitter:
rπ
ZE ≈ RE ∥ (if base is AC grounded)
β+1
If collector has a finite load, include it in parallel with ro as seen through gm .

5. High-Frequency Considerations
Including base-emitter capacitance Cπ and base-collector capacitance Cµ :

1
ZB (jω) ≈ RB ∥
jω(Cπ + Cµ (1 − Av ))
1
ZC (jω) ≈ RC ∥ ro ∥
jωCµ
rπ 1
ZE (jω) ≈ RE ∥ ∥
β + 1 jω(Cπ + Cµ )

17
Conclusion
Using Thevenin’s theorem, the small-signal impedances of the BJT nodes are summarized
as:

ZB ≈ RB ∥ rπ ∥ (β + 1)(RE ∥ ro )
ZC ≈ RC ∥ ro
rπ
ZE ≈ RE ∥
β+1

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