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BJT Amplifier Gain and
the Active Region
Consider this simple MOSFET circuit:
5.0V
Q: Oh, goody—you’re
going to waste my time
with another of these
pointless academic
problems. Why can’t you
discuss a circuit that
actually does something?
RD=1K
vO
vI
K=0.75 mA/V2
Vt= 1.0 V
A: Actually, this circuit is a fundamental
electronic device! To see what this circuit does, we
need to determine its transfer function vO  f vI  .
Q: Transfer function! How can
we determine the transfer
function of a MOSFET circuit!?
A: Same as with junction diodes—we determine the output vO
for each device mode, and then determine when (i.e., for what
values of vI) the device is in that mode!
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5.0V
First, note that regardless of the
MOSFET mode:
iD
RD=1K
vGS  vI  0.0  vI
and:
K=0.75 mA/V2
Vt= 1.0 V
vDS  vO  0.0  vO
vI
From KVL, we can likewise conclude that:
+
+
vDS  vO  5.0  iDRD
vDS
-
vGS
-
Now let’s ASSUME that the MOSFET is in cutoff, thus
ENFORCING iD=0.
Therefore:
vO  5.0  iDRD
 5.0  0 1 
 5.0 V
Now, we know that MOSFET is in cutoff when:
vGS  vI Vt  1.0
Thus, we conclude that:
vO  5.0 V
vO
when vI  1.0 V
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Now, let’s ASSUME that the MOSFET is in saturation, thus
ENFORCE:
2
iD  K vGS Vt 
 K vI Vt 
2
 0.75 vI  1.0 
2
And thus the output voltage is:
vO  5.0  iD RD
 5.0  0.75 v I  1.0  1 
2
 5.0  0.75 v I  1.0 
2
Now, we know that MOSFET is in saturation when:
vGS  vI Vt  1.0
and when:
vDS  vO  vGS Vt  vI  1.0
This second inequality means:
vO  vI  1.0
5.0  0.75 vI  1.0   v I  1.0
2
0  0.75 vI  1.0   v I  1.0   5.0
2
Solving this quadratic, we find that the only consistent solution
is:
vI  1.0  2.0
vI  3.0
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Meaning that the MOSFET is in saturation when vI > 1.0 and vI >
3.0. Logically, this is same thing as saying the MOSFET is in
saturation when vI > 3.0.
Thus we conclude:
vO  5.0  0.75 vI  1.0 
2
when vI  3.0 V
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Therefore the output voltage is
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VO
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BJT in cutoff
VCC
BJT in active mode
BJT in saturation
VI
VCC
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Note that:
VI
VO
Mode
0
VCC
Cutoff
VCC
0
Saturation
Why, this device is not
useless at all! It is clearly a:
____________________
Digital devices made with BJTs typically work in either the
cutoff of saturation regions.
So, what
good is the
BJT Active
Mode ??
Sir, it appears to
me that the active
region is just a
useless BJT mode
between cutoff
and saturation.
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Actually, we will find that the active mode is extremely useful!
To see why, take the derivative of the above circuit’s transfer
function (i.e., d VO d VI ):
VO
VCC
d VO
d VI
VI
VCC
We note that in cutoff and saturation:
d VO
0
d VI
while in the active mode:
d VO
 1
d VI
Q: I’ve got better things to do than
listen to some Professor Egghead
mumble about derivatives. How are
these results even remotely important?
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A: Since in cutoff and saturation d VO d VI  0 , a small change
in input voltage VI will result in almost no change in output
voltage VO .
Contrast this with the active region, where d VO d VI  1 .
This means that a small change in input voltage VI results in a
large change in the output voltageVO !
I see. A small voltage
change results in a big
voltage change—it’s
voltage gain!
The active mode turns
out to be—excellent.
Whereas the important BJT regions for digital devices are
saturation and cutoff, bipolar junction transistors in linear (i.e.,
analog) devices are typically biased to the active region.
This is especially true for BJT amplifier. Almost all of the
transistors in EECS 412 will be in the active region—this is
where we get amplifier gain !