DIFFERENTIAL AMPLIFIERS USING
TRANSISTOR LOADS.
Nagendra Krishnapura (nagendra@iitm.ac.in)
28/2/06
DIFFERENTIAL AMPLIFIER USING NMOS LOAD.
☛ The advantage over resistive load is that gain is a ratio of like components, so the variations are less.
☛ The gain is
gm1
gm3 +gmbs3 .
Vdd
M3
M4
Vo
vin1
M1
vin2
M2
Io
Figure 1: DIFFERENTIAL AMPLIFIER USING NMOS TRANSISTOR LOAD.
1
Vdd
M4
M3
Vo
Vbias +vin
2
Vbias-vin
M1
2
M2
Io
Figure 2: DIFFERENTIAL AMPLIFIER USING PMOS TRANSISTOR LOAD.
☛ To avoid the body effect we can use PMOS transistors.The gain is then
gm1
gm3
2
Vdd
Vbias
M3
M4
Vbias
Vo
vin1
M1
vin2
M2
Io
Figure 3: DIFFERENTIAL AMPLIFIER USING PMOS TRANSISTOR AND A PMOS
CURRENT SOURCE AS LOAD.
☛ We can increase gain by reducing gm3 .This can be done by reducing the
current through M3,so we connect a current source in parallel.
3
-gm vi
2
gm vi
2
vi
2
vi
2
Figure 4: NEGATIVE RESISTANCE
☛ Another option to increase gain is by connecting negative resistance in
parallel across the load.
Negative small signal resistance =
4
−2
gm
Vdd
M5
M6
M3
M4
Vo
Vbias + vin
2
M1 Vbias-vin
2
M2
Io
Figure 5: DIFFERENTIAL AMPLIFIER WITH NEGATIVE RESISTANCE IN PARALLEL AS LOAD
5
Vdd
RL
RL
Vo
Vbias + vin
2
vin
M1 Vbias- 2
M2
Io
Mo
Figure 6: DIFFERENTIAL AMPLIFIER WITH RESISTANCE AS LOAD
☛ For a differential amplifier using RL as load,
Vbias > VT +
v
u
u
u
t
v
u
u
2 I2o
2Io
u
t
+
W
µn Cox L
µn Cox W
L
(1)
Also,
Vbias +
vin
Io RL gm vinRL
− vth < VDD −
−
2
2
2
I o RL
vin
Vbias < VDD −
− (1 + gm RL )
+ vth
2
2
(2)
(3)
✧ Vbias should be such that the transistors M0 and M1 are kept in
saturation.
✧ To maximise signal swing Vbias must be at the lowest possible value.
✧ With Vbias at the minimum value
I o RL
vin
+ (gm RL + 1)
2
2
☛ For a differential pair using current sources as load the lower constraint
is the same but the upper constraint is given by the sum of overdrive of
M3 and vth .
v
u
u
2 I2o
u
Vbias < VDD − (t
+ vth3) + vth1
(4)
µn Cox W
L
VDD > VDSAT 1 + VDSAT 0 +
6