Exercises for

Cascode Amplifiers

ECE 102, Fall 2012, F. Najmabadi

 Exercise 1: Compute R (assume ro1 = ro2 = ro3 and gm1 = gm2 = gm3)

Double Cascode ro 3 (1 + g m 3 g m ro2 ) + g m ro2

≈ g m2 ro3 + g m ro2 ≈ ( g m ro ) 2 ro

ro 2 (1 + g m 2 ro1 ) + ro1
≈ g m ro2 + ro ≈ ( g m ro )ro g m ro2

Every Cascode stage increases R by gmro

F. Najmabadi, ECE102, Fall 2012 (2/26)

 Exercise 2: Compute all indicated R’s, v’s, and i’s . Assume transistors are
identical. (S&S Problem 7.34)

F. Najmabadi, ECE102, Fall 2012 (3/26)

 R1 = ro

R2 = ro (1 + g m ro ) + ro
≈ g m ro2 + ro ≈ ( g m ro )ro

ro + R2 r + ( g m ro )ro
R3 = = o = ro
1 + g m ro 1 + g m ro
F. Najmabadi, ECE102, Fall 2012 (4/26)
 From previous slide:
ro + R2 r + ( g m ro )ro
R3 = ≈ o = ro
1 + g m ro 1 + g m ro

CS amplifier:

Av = − g m (ro || R3 ) = − g m (ro || ro )

v1 = −0.5 g m ro vi

v1 0.5 g m ro vi
i3 = − = = 0.5 g m vi
R3 ro
F. Najmabadi, ECE102, Fall 2012 (5/26)
 From previous slide:
R2 ≈ ( g m ro )ro

Cascode amplifier
(we can use cascode gain to find v2)

A simpler method: i4 = i3 = 0.5 g m vi

v2 = −i4 R2 = −0.5 g m vi ( g m ro2 ) = −0.5( g m ro ) 2 vi

F. Najmabadi, ECE102, Fall 2012 (6/26)

 From previous slide:
R1 = ro

Q3 is NOT an amp configuration

It is a part of the active load!

i5 = i4 = 0.5 g m vi

v3 = −i5 R1 = −0.5 g m vi ro = −0.5 g m ro vi

F. Najmabadi, ECE102, Fall 2012 (7/26)

 i3 = i4 = i5 Currents are the same!

R1 = ro Looking into drain of Q4

Looking into drain of Q3,

R2 ≈ ( g m ro )ro
R1 is increased by 1+gmro

R3 ≈ ro Looking into source of Q2,

R2 is decreased by 1+gmro

F. Najmabadi, ECE102, Fall 2012 (8/26)

 CS Amp with a load of ro Cascode Amp with a
Q3 increased R1 and Q2 practical load
decreased R2 by the same (Good circuit)
amount leading to R3 = ro
(Bad circuit)

v1 = −0.5 g m ro vi v2 = −0.5( g m ro ) 2 vi

F. Najmabadi, ECE102, Fall 2012 (9/26)

 Output is NOT taken across
the load (BAD circuit)

ro

Q3 is NOT an Amp
(input at the drain)

ro (1+gm ro )
R2 = ro (1 + g m ro ) + ro
v2
CG Amp
(input at the source
Output at the drain)

CS Amp
(input at the gate
Output at the drain)

v2 = −0.5( g m ro ) 2 vi
ro
v3 = v2 ≈ −0.5 g m ro vi
ro (1 + g m ro )
F. Najmabadi, ECE102, Fall 2012 (10/26)
 Exercise 3: Due to manufacturing error, the input terminal of a cascode
amplifier is mis-configured as is shown. Compute vo. Assume identical
transistors are identical.

F. Najmabadi, ECE102, Fall 2012 (11/26)

 From previous Problem:
R2 = ( g m ro )ro

RS = ro

Av = g m (ro || R2 ) ≈ g m ro
This is a CG configuration

F. Najmabadi, ECE102, Fall 2012 (12/26)

 R2 = ( g m ro )ro

RS = ro From previous
problem: R3 = r o

Q1 does not affect the gain, but it changes the input resistance to:
Ri = RS || R3 = ro || ro = 0.5ro

F. Najmabadi, ECE102, Fall 2012 (13/26) Exercise: Compute Ro (answer: Ro ≈ ro )

 Exercise 4: Due to manufacturing error, a parasitic resistance has appeared
between drain and source of Q1. Compute vo. Assume identical transistors are
identical.

F. Najmabadi, ECE102, Fall 2012 (14/26)

 From previous
Problem:
R2 ≈ ( g m ro )ro

R3 ≈ ro

Q2: CG configuration: Q1: CS configuration:

Av 2 = g m (ro || R2 ) ≈ g m ro Av1 = − g m (ro || R3 || R p ) = − g m (ro || ro || R p )

Av = Av1 Av 2 = − g m2 ro (ro || ro || R p )
F. Najmabadi, ECE102, Fall 2012 (15/26)
 When a resistor is placed between Drain and Source of a transistor,
there is a simpler and more elegant way to solve the signal circuit

ro′ = ro || R

F. Najmabadi, ECE102, Fall 2012 (16/26)

 ro′ = ro || R

Av = Av1 Av 2 ≈ − g m2 ro (ro || ro || R p )

F. Najmabadi, ECE102, Fall 2012 (17/26)

 Exercise 5: Due to manufacturing error, a parasitic resistance has appeared on
the circuit. Compute vo. Assume identical transistors are identical.

F. Najmabadi, ECE102, Fall 2012 (18/26)

 R2 = ro (1 + g m ro ) + ro ≈ ( g m ro )ro

ro′ = ro || R p

ro′ + R2 1 ( g m ro )ro r
R3 = ≈ + ≈ ro × o
1 + g m ro′ g m g m ro′ ro′

Q2: CG configuration: Q1: CS configuration:

Av 2 ≈ g m (ro′ || R2 ) = g m (ro || R p || R2 ) ≈ g m (ro || R p ) Av1 = − g m (ro || R3 )

Av = Av1 Av 2 = − g m2 (ro || R p || R2 )(ro || R3 )

F. Najmabadi, ECE102, Fall 2012 (19/26)
 Exercise 6: Find the open-loop gain, overall gain and output resistance of the
circuit if VOV1 = 0.3 V and RL = 100 k (µnCox = 200 µA/V2, µpCox = 50
µA/V2, Vtp = − 0.6 V, λn = 0.1 /V, λp = 0.2/ V and (W/L) = 20/0.18 for all
transistors).

W  2 20
I D1 = 0.5µ nCox   VOV 1 = 0 . 5 × 200 × 10 −6
× × (0.3) 2
 L 1 0.18
11 final –P5
I D 4 = I D 3 = I D 2 = I D1 = 1 mA

I D 2 = I D1 → VOV 2 = VOV 1 = 0.3 V

W  2 20
I D 3 = 10 −3 = 0.5µ p Cox   VOV 3 = 0 . 5 × 50 × 10 −6
× × VOV
2
3
 L 3 0.18

VOV 3 = 0.6 V

I D 4 = I D 3 → VOV 4 = VOV 3 = 0.6 V

F. Najmabadi, ECE102, Fall 2012 (20/26)

 Previous page: I D 4 = I D 3 = I D 2 = I D1 = 1 mA VOV 2 = VOV 1 = 0.3 V & VOV 4 = VOV 3 = 0.6 V

2 I D1 2 ×10 −3 1 1
gm2 = g m1 = = = 6.67 ×10 −3 A/V ro 2 = ro1 = = = 10 k
VOV 1 0.3 λn I D1 0.1×10 −3

2 I D 3 2 ×10 −3 1 1
g m 4 = g m3 = = = 3.33 ×10 −3 A/V ro 4 = ro 3 = = =5k
VOV 3 0.6 λ p I D1 0.2 ×10 −3

From Elementary R forms:

R1 = ro 3 (1 + g m 3 ro 4 ) + ro 4
R1 = 5 ×103 (1 + 3.33 ×10-3 × 5 ×103 ) + 5 ×103 = 93.3 k

R2 = ro 2 (1 + g m 2 ro1 ) + ro1 = 687 k

 We need only to compute two of Avo, Av, and Ro set

as they are related: RL
Av = Avo ⋅
o We calculate all three below to RL + Ro
show the method.
o Note: Cascode Amp formula gives Avo

F. Najmabadi, ECE102, Fall 2012 (21/26)

 From Previous page: R1 = 93.3 k, R2 = 687 k, RL = 100 k
g m 2 = g m1 = 6.67 × 10 −3 A/V, ro 2 = ro1 = 10 k, g m 4 = g m 3 = 3.33 × 10 −3 A/V, ro 4 = ro 3 = 5 k

Ro = R1 || R2 = 93.3 k || 687 k = 82.2 k

Av Avo (RL→ ∞ )
Q2 (CG):
RL′ = RL || R1 = 100k || 93.3k = 48.4k RL′ = R1 = 93.3k
ro 2 || RL′ = 10k || 48.4k = 8.29k ro 2 || RL′ = 10k || 93.3k = 9.03k
Av 2 ≈ g m 2 (ro 2 || RL′ ) = 55.3 Av 2 ≈ g m 2 (ro 2 || RL′ ) = 60.2

R’L1 = Ri2:

R = R1 || RL = 93.3 k || 100 k = 48.3 k R = R1 = 93.3 k

ro 2 + R ro 2 + R Resistance in the
Ri 2 = = 0.862 k Ri 2 = = 1.53 k
1 + g m 2 ro 2 1 + g m 2 ro 2 drain of Q2
Q1 (CS):
Av1 = − g m1 (ro1 || Ri 2 ) = −5.29 Av1 = − g m1 (ro1 || Ri 2 ) = −8.83
Av = Av1 × Av 2 = −293 Avo = Av1 × Av 2 = −532 Exercise: Show
RL
Av = Avo ⋅
F. Najmabadi, ECE102, Fall 2012 (22/26) RL + Ro
 Exercise 7: In the folded cascode circuit below, all transistors have the same
µCox (W/L), the same λ and ID2 = 2ID1 . Find the gain and the output
resistance of the amplifier (in terms of gm1 and ro1 only).

F. Najmabadi, ECE102, Fall 2012 (23/26)

 Problem: ID2 = 2ID1 , the same µCox (W/L), and the same λ

KCL : I D 2 = 2 I D1 = I D1 + I D 3 → I D 3 = I D1 Bias
I D 5 = I D 4 = I D 3 = I D1 & I D 2 = 2 I D1

W  2
I D = 0.5µCox  VOV →
L
VOV 5 = VOV 4 = VOV 3 = VOV 1 & VOV 2 = 2 VOV 1

2I D g m 5 = g m 4 = g m 3 = g m1
gm = →
VOV 2 I D 2 2 × 2 I D1 2I
gm2 = = = 2 × D1 = 2 g m1
VOV 2 2 VOV 1 VOV 1

1
ro = → ro 5 = ro 4 = ro 3 = ro1
λI D
1 1 1
ro 2 = = = 0.5 × = 0.5ro1
λI D 2 λ × 2 I D1 λI D1

F. Najmabadi, ECE102, Fall 2012 (24/26)

 g m 5 = g m 4 = g m 3 = g m1 , g m 2 = 2 g m1
ro 5 = ro 4 = ro 3 = ro1 , ro 2 = 0.5ro1

CG Amp

Signal

CS Amp
R2 = ro 4 (1 + g m 4 ro 5 ) + ro 5
R2 ≈ ( g m1ro1 )ro1

Q3(CG): Av 3 = vo / v x ≈ g m 3 (ro 3 || R2 ) = g m1[ro1 || ( g m1ro1 )ro1 ] ≈ g m1ro1

ro 3 + R2 r + ( g m1ro1 )ro1
Ri 3 = = o1 = ro1
1 + g m 3 ro 3 1 + g m1ro1

Q1 (CS): RL′ 1 = ro 2 || Ri 3 = (0.5ro1 ) || ro1 = 0.33ro1

Av1 = v x / vi = − g m1 (ro1 || RL′ 1 ) = − g m1[ro1 || (0.33ro1 )] = −0.25 g m1ro1

Av = vo / vi = Av1 × Av 3 = −0.25 g m1ro1 × ( g m1ro1 ) = −0.25( g m1ro1 ) 2

F. Najmabadi, ECE102, Fall 2012 (25/26)
 g m 5 = g m 4 = g m 3 = g m1 , g m 2 = 2 g m1
ro 5 = ro 4 = ro 3 = ro1 , ro 2 = 0.5ro1

Ro = R1 || R2

R2 = ro 4 (1 + g m 4 ro 5 ) + ro 5 ≈ g m1ro21

R1 :
Ro1 = ro 2 || R4 = (0.5ro1 ) || ro1 = 0.33ro1
R1 = ro 3 (1 + g m 3 Ro1 ) + Ro1
= ro1 (1 + 0.33 g m1ro1 ) + 0.33 g m1ro1 ≈ 0.33 g m1ro21

(Assumed 0.33 gm1 ro1 >> 1)

R2 = ro 4 (1 + g m 4 ro 5 ) + ro 5
R2 ≈ ( g m1ro1 )ro1
Ro = R1 || R2
Ro = (0.33 g m1ro21 ) || g m1ro21 = 0.25 g m1ro21

F. Najmabadi, ECE102, Fall 2012 (26/26)