Lect. 22: MOSFET Current Mirror and CS Amplifier
Various bias techniques for MOSFET circuits
How do we make
a constant current source
with MOSFETs?
Electronic Circuits 1 (06/2)
Prof. Woo-Young Choi
Lect. 22: MOSFET Current Mirror and CS Amplifier
Constant current source:
I D1
1 ' ⎛W ⎞
2
= kn ⎜ ⎟ (VGS − Vtn )
2 ⎝ L ⎠1
I D1 = I REF
VDD − VGS
=
R
Assuming Q1, Q2 have same properties,
IO = I D 2 =
1 ' ⎛W ⎞
2
kn ⎜ ⎟ (VGS − Vtn )
2 ⎝ L ⎠2
IO
(R can be outside the IC chip)
Î Current mirror
Electronic Circuits 1 (06/2)
I REF
Limitation on Vo?
(W / L ) 2
=
(W / L )1
VO ≥ VGS − Vt n
Prof. Woo-Young Choi
Lect. 22: MOSFET Current Mirror and CS Amplifier
Mismatches between IREF and IO
Channel-length modulation
For two identical Q1 and Q2
IO = IREF when V0=VGS
As VO increased, IO increases from IREF
I O = I REF +
Since r0 I O = I REF +
Electronic Circuits 1 (06/2)
VO − VGS
r0
VA
I REF
⎛ V − VGS ⎞
VO − VGS
⋅ I REF = I REF ⎜ 1 + O
⎟
VA
V
⎝
⎠
A
Prof. Woo-Young Choi
Lect. 22: MOSFET Current Mirror and CS Amplifier
Example 6.5
VDD=3V, Q1 and Q2 are identical with
L= 1μm, W=100μm, Vt=0.7V, kn’=200μA/V2,
VA’ (Early voltage per L) =20V/μm
1. Determine R for IO=100μA.
2. What is the lowest value for VO?
3. How much IO changes when VO changes 1V?
I D1 = I REF =
100 =
1 ' ⎛W ⎞
kn ⎜ ⎟ (VGS − Vt )2
2 ⎝ L ⎠1
1
× 200 × 10 × (VGS − Vt )2 ,
2
(VGS − Vt ) = 0.316, VGS = 1.016
R=
VDD − VGS 3 − 1.016
=
= 19.84kΩ
I REF
0.1mA
VO min = VGS − Vt = VOV = 0.316 V
VA = 20 × 1 = 20 V
ro 2 =
20V
= 0.2MΩ
100 μ A
ΔI O =
Electronic Circuits 1 (06/2)
ΔVO
1V
=
= 5μ A
ro 2
0.2MΩ
Prof. Woo-Young Choi
Lect. 22: MOSFET Current Mirror and CS Amplifier
(W / L ) 2
I 2 = I REF
( W / L )1
(W / L ) 3
I 3 = I REF
( W / L )1
I3 = I4
(W / L ) 5
I5 = I4
(W / L ) 4
Current-steering circuits, current source (Q5), current sink (Q2)
Electronic Circuits 1 (06/2)
Prof. Woo-Young Choi
Lect. 22: MOSFET Current Mirror and CS Amplifier
CS amplifier
Where is RD ?
Current mirror as a resistor Î Active load
(Remember Q2 has rO)
Electronic Circuits 1 (06/2)
Prof. Woo-Young Choi
Lect. 22: MOSFET Current Mirror and CS Amplifier
Large change in vO with vI change!
Î Large amplifier gain
Load-line analysis
Q1
Q2
v = VDD-vo
Electronic Circuits 1 (06/2)
Prof. Woo-Young Choi
Lect. 22: MOSFET Current Mirror and CS Amplifier
Gain for CS amplifier with RD:
-gm (rO || RD || RL )
Gain for CS amplifier with PMOS current mirror
-gm (rO1 || rO2||RL)
PMOS current mirror provides
large “Drain” resistance (Active Load)
as well as bias current!
Î Good for IC!
Electronic Circuits 1 (06/2)
Prof. Woo-Young Choi
Lect. 22: MOSFET Current Mirror and CS Amplifier
Example 6.8
VDD=3V, Vtn=|Vtp|=0.6V, kn’=200μA/V2, kp’=65μA/V2,
L=0.4μm, W=4μm, VAn=20V, |VAp|=10V, IREF=100μA.
1. What is the small-signal voltage gain, vO/vI?
2. What is the maximum vO for which the above is valid?
1. Aυ = − gm 1 ( ro1 & ro 2 )
4
⎛W ⎞
gm 1 = 2kn' ⎜ ⎟ I REF = 2 × 200 ×
× 100 = 0.63mA/V
0.4
⎝ L ⎠1
V
20V
ro1 = An =
= 200kΩ
I D1 0.1mA
ro 2 =
| VAp |
I D2
=
10V
= 100kΩ
0.1mA
∴ Aυ = −0.63(mA/V) ⋅ (200 & 100)(kΩ ) = −42
2. For Q 3 , I REF
100 =
∴VSG
(
1 ⎛W ⎞
= k 'p ⎜ ⎟ VSG − Vtp
2 ⎝ L ⎠3
)
2
⎞
⎛ V
⎜ 1 + SD ,3 ⎟
⎜
VAp ⎟⎠
⎝
V ⎞
1
2⎛
⎛ 4 ⎞
× 65 ⎜
VSG − 0.6 ) ⎜ 1 + SG ⎟
(
⎟
2
10 ⎠
⎝ 0.4 ⎠
⎝
= 1.13V
For vO ,max , VSD 2,min = VSG − | Vtp |= 1.13 − 0.6 = 0.53V
∴ vO ,max = VDD − VSD 2,min = 2.47V
Electronic Circuits 1 (06/2)
Prof. Woo-Young Choi
Lect. 22: MOSFET Current Mirror and CS Amplifier
Homework: Do your design project: Part 1 and 2 (No need to hand it in)
ITOTAL ≤ 150 μ A
Electronic Circuits 1 (06/2)
Prof. Woo-Young Choi