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The MOSFET
Current Mirror
Consider the following MOSFET circuit:
VDD
Note VD  VG , therefore:
ID
VDS  VGS
R
and thus:
VDS  VGS Vt
+
VGS
So, if VGS  Vt , then the
-
MOSFET is in saturation!
We know that for a MOSFET in saturation, the drain current is
equal to:
ID  K VGS Vt 
2
Say we want this current ID to be a specific value—call it I ref .
Since Vs  0 , we find that from the above equation, the drain
voltage must be:
VD 
Iref
K
Vt
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Likewise, from KVL we find that:
Iref 
VDD VD
R
And thus the resistor value to achieve the desired drain current
I ref is:
V V
R  DD D
Iref
where:
VD 
Iref
K
Vt
Q: Why are we doing this?
A: Say we now add another component to the circuit, with a
second MOSFET that is identical to the first :
VDD
VDD
Iref
IL
R
+
VGS1
-
+
VGS2
-
RL
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Q: So what is current IL ?
A: Note that the gate voltage of each MOSFET is the same
(i.e., VGS 1  VGS 2 ), and if the MOSFETS are the same (i.e.,
K1  K2, Vt 1  Vt 2 ), and if the second MOSFET is likewise in
saturation, its drain current I L is:
I L  K2 VGS 2 Vt 2 
2
 K1 VGS 1 Vt 1   Iref
2
Therefore, the drain current of the second MOSFET is equal
to the current of the first!
Iref  I L
Q: Wait a minute! You mean to say that the current through
the resistor RL is independent of the value of resistor RL ?
A: Absolutely! As long as the second MOSFET is in
saturation, the current through RL is equal to Iref—period.
The current through RL is independent on the value of RL
(provided that the MOSFET remains in saturation). Think about
what this means—this device is a current source !
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VDD
VDD
VDD
R
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RL
Iref
RL
Iref
Current Source
Remember, the second MOSFET must be in saturation for the
current through RL to be a constant value I ref . As a result, we
find that:
VDS 2  VGS 2 Vt 2
or for this example, since Vs  0 :
VD 2  VG 2 Vt 2
Since VD 2  VDD  RLIref , we find that in order for the MOSFET
to be in saturation:
VDD  RLIref  VG 2 Vt 2  VG 1 Vt 1
Or, sated another way, we find that the load resistor RL can be
no larger than:
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VDD VG 1 Vt 1
Iref
Where we know that:
VG 1  VDD  R Iref
and thus we can alternatively write the above equation as:
RL  R 
Vt 1
Iref
If the load resistor becomes larger than R Vt 1 Iref , the
voltage VDS 2 will drop below the excess gate voltage VGS 2 Vt 2 ,
and thus the second MOSFET will enter the triode region. As a
result, the drain current will not equal I ref —the current source
will stop working!
Although the circuit presented here is
sometimes referred to as a current
sink, understand that the circuit is
clearly a way of designing a current
source.
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We can also use PMOS devices to construct a current mirror!
VSS
+
+
VGS1
-
VSS
VGS2
This better be
in saturation!
-
I L  I ref ,
Iref
R
RL
IL
regardless of the
value of RL!!!