Download Depletion Loads

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Depletion Loads
We can also make a transistor “load” by using a depletion
MOSFET!
Note vG  vs , therefore vGS  0 !
+
v
i
Q: Doesn’t that mean that the
MOSFET is in cutoff!?
A: No! Note we are using a
depletion MOSFET. Recall that for
these devices, a conducting channel
is implanted—we do not need to
induce a channel!
Instead, the channel is cutoff only if it is fully depleted, where
depletion is accomplished by making the gate-to-source voltage
negative (for NMOS). Thus, the threshold voltage for a
depletion NMOS transistor is negative (Vt  0 ).
For the depletion load shown above, VGS  0 Vt , so that the
MOSFET can never be in cutoff—the channel is always
conducting!
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The question then is whether the MOSFET is in triode or
saturation. Recall an n-channel MOSFET is in triode if:
vDS  vGS Vt
and since v  vDS and vGS  0 , we find that the depletion load
MOSFET is in triode if:
v  Vt
Note that since the threshold voltage for a depletion NMOS
device is negative, the value Vt is a positive number!
Recall that a MOSFET in triode has drain current:
2

iD  K 2 vGS Vt vDS  vDS
And since for the depletion load, i  iD , v  vDS and vGS  0 :
i  K  2Vt v  v 2 
 K 2Vt  v v
Since v  Vt , this current value is actually positive (i  0) if
voltage v is positive.
Now, if v  Vt (i.e., vDS  vGS Vt ), we find that the depletion
MOSFET in the load will be in saturation, and thus
2
iD  K vGS Vt  . Since for the depletion load i  iD , v  vDS and
vGS  0 , we find that in saturation:
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i  K Vt 2
A constant!
If we account for channel-length modulation effects (i.e., the
MOSFET output resistance), we modify the above equation to
be:
v
i  K Vt 2 
ro
Summarizing, we find that the i-v relationship for a depletion
load is:

K (2V  v )v for v  V
t
t

i 

v
K Vt 2 
for v  Vt
r

o
Thus the i-v curve is:
i  K Vt
v  Vt
i
2
resistor
i  K Vt 2
i  K Vt 2 
i  K (2Vt  v )v
(triode)
(saturation)
v
Vt
v
ro
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Therefore, we can build a common source amplifier with either
a resistor, or in the case of an integrated circuit, a depletion
load.
VDD
V
DD
Q2
vO
vO
vI
vI
Q1
The “load curve” thus looks like:
iD
Depletion Load (Q2)
NMOS (Q1)
ID
Resistor Load
vDS
VDS
And the circuit transfer function is:
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vO
Both Q1 and Q2
in saturation
dvO
dvI
1
vI
Vt
We can likewise determine the small signal circuit for this load.
Step1 – DC Analysis
In saturation:
+
I  KVt 2
V
-
and:
I
VGS  0
Step 2 – Determine Small-signal Parameters
gm  2KVt

ro 
a positive number!
1
1

 I  K Vt 2
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Steps 3 & 4 – Small-Signal Analysis
Inserting the MOSFET small-signal model, we get:
i  id
D
G
+
vgs = 0
-
gm v gs
ro

vds  v

S
Note that the dependent current source is zero !!! Redrawing
this circuit, we get:
i  id
D
G
ro

vds  v

S
Further simplifying, we get this simple small-signal equivalent
circuit:
i
D

ro
S
v

The Depletion Load
Small-Signal Model
G
WARNING!: We have ignored body effects!