3. 3
Boosted Output Impedances
S
O
S
m
out
E
O
E
m
out
R
r
R
g
R
r
R
r
r
R
g
R
1
||
||
1
2
1
4. 4
MOS Cascode Stage
2
1
1
2
1
2
1
1
O
O
m
out
O
O
O
m
out
r
r
g
R
r
r
r
g
R
2
1
2
1
1 ,
As, O
O
O
O
m r
or
r
r
r
g
5. 5
Another Interpretation of MOS Cascode
• Similar to its bipolar counterpart, MOS cascode can be
thought of as stacking a transistor on top of a current source.
• Unlike bipolar cascode, the output impedance is not limited
by .
7. 7
Example: Parasitic Resistance
• RP will lower the output impedance, since its parallel combination
with rO1 will always be lower than rO1. Consider original formula.
2
1
2
1 )
||
)(
1
( O
P
O
O
m
out r
R
r
r
g
R
8. 8
MOS Cascode Amplifier
2
2
1
1
2
1
2
2
1 )
1
(
O
m
O
m
v
O
O
O
m
m
v
out
m
v
r
g
r
g
A
r
r
r
g
g
A
R
G
A
• Cascode is a modified version of CS topology. Uses CS load to
increase gain.
Here, 𝐺𝑚 =
𝑖𝑜𝑢𝑡
𝑣𝑖𝑛
iout =iD1 and vin=vgs1
Gm=gm1
Finding Gm of the circuit.
9. 9
Improved MOS Cascode Amplifier
• Similar to its bipolar counterpart, the output impedance of a MOS
cascode amplifier can be improved by using a PMOS cascode current
source.
op
on
out
O
O
m
op
O
O
m
on
R
R
R
r
r
g
R
r
r
g
R
||
4
3
3
1
2
2
11. 11
Temperature and Supply Dependence
of Bias Current
• Since VT, IS, n, and VTH all depend on temperature, I1 for
both bipolar and MOS depends on temperature and supply.
2
2
1
2
1
1
2
1
2
2
1
)
ln(
)
(
TH
DD
ox
n
S
T
CC
V
V
R
R
R
L
W
C
I
I
I
V
R
R
V
R
12. Current Mirrors
• Supply dependent bias such as rechargeable battery looses voltages
overtime.
• VTh, μn is temperature dependent.
• An elegant method/ckt known as ‘Bandgap reference ckt’ solve the
dependence problem. But it has complexity. A bandgap reference can
provide a golden current. Current mirror is a technique of copying the
golden current without duplicating the complex circuitry of BG ref.
Fig. BG ref implemented with 5T OTA.
13. 13
Concept of Current Mirror
• The motivation behind a current mirror is to sense the
current from a “golden current source” and duplicate
this “golden current” to other locations.
14. 14
MOS Current Mirror
• The same concept of current mirror can be applied
to bipolar transistors as well.
𝑉𝑋 =
2𝐼𝑅𝐸𝐹
𝜇𝑛𝐶𝑂𝑋(
𝑊
𝐿
)1
+ 𝑉𝑇𝐻1
15. 15
Bad MOS Current Mirror Example
• This is not a current mirror since the relationship between
VX and IREF is not clearly defined.
• The only way to clearly define VX with IREF is to use a diode-
connected MOS since it provides square-law I-V
relationship.
16. 16
Example: Current Scaling
An integrated circuit employs the source follower and the common-source stage shown in Fig.
Design a current mirror that produces I1 and I2 from a 0.3-mA reference.
• MOS current mirrors can also scale IREF up or down (I1 = 0.2mA, I2 = 0.5mA).
17. 17
CMOS Current Mirror
• The idea of combining NMOS
and PMOS to produce CMOS
current mirror is shown above.
19. Problem – 9.34(b)
• Determine the voltage gain of each circuit in Fig. Assume gmrO >>1.
𝑖𝑜𝑢𝑡 = 𝑔𝑚1𝑉𝑖𝑛 +
𝑉𝑑1
𝑟𝑜1||𝑅𝑃
Again,
𝑉𝑑1 = −
𝑖𝑜𝑢𝑡𝑟𝑜2
1 + 𝑔𝑚2𝑟𝑜2
≈ −
𝑖𝑜𝑢𝑡
𝑔𝑚2
𝐺𝑚 = 𝑔𝑚1
𝑟𝑜1||𝑅𝑃
1
𝑔𝑚2
+ 𝑟𝑜1||𝑅𝑃
𝑅𝑜𝑢𝑡 = 𝑅𝑜𝑝||𝑅𝑜𝑛 =
𝐴𝑣 = −𝐺𝑚𝑅𝑜𝑢𝑡 =
20. Problem – 9.59
• Design the CMOS cascode amplifier of Fig. 9.76 for a voltage gain of 200 and a
power budget of 2 mW with VDD = 1.8 V. Assume (W/L)1 = ・・・ = (W/L)4 =
20/0.18 and λp = 2λn = 0.2 V−1. Determine the required dc levels of Vin and Vb3.
For simplicity, assume Vb1 = Vb2 = 0.9 V. Given μnCox =100 μA/V2, μpCox = 50
μA/V2, VTH,n = 0.4 V, and VTH,p =−0.5 V
𝑃 = 𝐼𝐷𝑉𝐷𝐷 ⇒ 𝐼𝐷 = 1.11𝑚𝐴
𝐴𝑣 = −𝑔𝑚1(𝑔𝑚2𝑟𝑜2𝑟𝑜1| 𝑔𝑚3𝑟𝑜3𝑟𝑜4 = 200
𝑉𝑖𝑛 = 𝑉𝐺𝑆1 = 1.07 𝑉
𝑉𝑏3 = 0.67 𝑉
21. Problem – 9.65
• Figure 9.80 shows an arrangement where M1 and M2 serve as
current sources for circuits 1 and 2. Design the circuit for a power
budget of 3 mW.
𝑃 = 𝐼𝑡𝑜𝑡𝑎𝑙𝑉𝐷𝐷 ⇒ 𝐼𝑡𝑜𝑡𝑎𝑙 = 1.667𝑚𝐴
𝐼𝑅𝐸𝐹 = 0.1667𝑚𝐴
(
𝑊
𝐿
)1
(
𝑊
𝐿
)𝑅𝐸𝐹
= 500: 167
(
𝑊
𝐿
)2
(
𝑊
𝐿
)𝑅𝐸𝐹
= 1000: 167
