Download The Common Source Amp with Enhancement Load

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The Common Source Amp
with Enhancement Load
VDD
Consider this NMOS amplifier
using an enhancement load.
* Note no resistors or
Q2
capacitors are present!
vO(t)
* This is a common source
Q1
amplifier.
* ID stability could be a
problem
vi(t)
+
_
VG
Q: What is the small-signal open-circuit voltage gain, input
resistance, and output resistance of this amplifier?
A: The values that we will determine when we follow precisely
the same steps as before!!
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Step 1 – DC Analysis
Note that:
The DC circuit of this amplifier is:
ID 1  ID 2
VDD
ID
and that:
ID1
VGS 1 VG  0 VG
Q2
and also that:
VO
VDS 2 VGS 2
Q1
VG
and finally that:
ID2
VDS 1 VDD VDS 2
Let’s of course ASSUME that both Q1 and Q2 are in
saturation. Therefore we ENFORCE:
ID 1  K1 VGS 1 Vt 1 
2
 K1 VG Vt 1 
2
Note that there are no unknowns in the previous equation. The
drain current is explicitly determined from K1 , VG , and Vt 1 !
Continuing with the ANALYSIS, we can find the drain current
through the enhancement load (ID2), since it is equal to the
current through Q1:
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ID 2  ID 1  K1 VG Vt 1 
2
Yet we also know that VGS2 must be related to this drain
current as:
ID 2  K2 VGS 2 Vt 2 
2
and therefore combining the above equations:
ID 1  ID 2
K1 VG Vt 1   K2 VGS 2 Vt 2 
2
2
Note this last equation has only one unknown (VGS 2 ) !
Rearranging, we find that:
VGS 2 
K1
VG Vt 1  Vt 2
K2
Since VDS 2 VGS 2 and VDS 1 VDD VDS 2 , we can likewise state
that:
VDS 2 
K1
VG Vt 1  Vt 2
K2
and:
VDS 1 VDD Vt 2 
K1
VG Vt 1 
K2
Now, we must CHECK to see if our assumption is correct.
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The saturation assumption will be correct if:
VDS 1  VGS 1 Vt 1
 VG Vt 1
and:
VGS 1 Vt 1
 if VG Vt 1
Step 2 – Calculate small-signal parameters
We require the small-signal parameters for each of the
transistors Q1 and Q2. Therefore:
gm 1  2K1 VG Vt 1 
and
1
1 ID
and
and:
ro 1 
gm 2  2K1 VGS 2 Vt 2 
ro 2 
1
2 ID
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Step 3 – Determine the small-signal circuit
First, let’s turn off the DC sources:
Q2
vO(t)
Q1
vi(t)
+
_
We now replace MOSFET Q1 with its equivalent small-signal
model, and replace the enhancement load with its equivalent
small-signal model.
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G2
+
ro 2
vgs2
gm 2 v gs 2
S2
id
G1
D1
+
vi
vgs1
-
+
-
vo
ro 1
gm v gs 1
S1
Note S1 and G2 are at small-signal ground!
Therefore, we can rewrite this circuit as:
id
G1
+
vi
+
-
vgs1
-
ro 1
gm v gs 1
S1
D1
S2
gm 2 v gs 2
G2
vgs2
+
vo
ro 2
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Simplifying further, we find:
-
+
vi
+
-
vo
vgs1
-
g
m1
Therefore, we find that:
and that:
v gs 1  gm 2v gs 2

vgs2
+
v gs 1  vi
v gs 2  vo
as well as that:
vo  (gm 1 v gs 1  gm 2 v gs 2 ) (ro 1 ro 2 )
 (gm 1 vi  gm 2 vo ) (ro 1 ro 2 )
Rearranging, we find:
Avo 
But recall that:
(ro 1 ro 2 )gm 1
g
vo

 m1
vi 1  (ro 1 ro 2 )gm 2
gm 2
gm  2K VGS Vt 
 2 K ID
where we have used the fact that ID  K VGS Vt  to
2
determine that VGS Vt   ID K .
ro 1 ro 2
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Therefore:
Avo 
 gm 1
gm 2

2 K1 ID
2 K2
K
 1 
K2
ID
W L 
W L 
1
2
In other words, we adjust the MOSFET channel geometry to
set the small-signal gain of this amplifier!
Now let’s determine the small-signal input and output
resistances of this amplifier!
ii
-
+
vi
+
_
vgs1
-
g
m1
v gs 1  gm 2v gs 2

vgs2
vooc
ro 1 ro 2
+
It is evident that since ii  i g  0 :
Ri 
vi

ii
(Great!!!)
Now for the output resistance, we know that the open-circuit
output voltage is:
vooc  (gm 1 v gs 1  gm 2 v gs 2 ) (ro 1 ro 2 )
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iosc
ii
vi
+
-
-
+
vgs
-
g
m1
v gs 1  gm 2v gs 2

vgs2
ro 1 ro 2
+
Likewise, the short-circuit output current iosc is:
ios  (gm 1 v gs 1  gm 2 v gs 2 )
Thus, the small-signal output resistance of this amplifier is
equal to:
Ro 
vooc
iosc

(gm 1 v gs 1  gm 2 v gs 2 ) (ro 1 ro 2 )
(gm 1 v gs 1  gm 2 v gs 2 )
 (ro 1 ro 2 )
The input resistance and open-circuit voltage gain
of this common source amplifier are good, but the
output resistance stinks!!
Smells like a common emitter amplifier!
(Doh!!!)