Download The Input Bias Current

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Example: The Input
Bias Current
Q: How do input bias currents I B 1 and IB 2 affect amplifier
operation?
A: Consider both inverting and non-inverting configurations.
Inverting Configuration
R2
i2
vin
R1
i1
vIB 1
v+
IB 2
vout
+
In this case, we apply KCL and we find:
i1  i2  IB 1
However, we still find v  v   0 (neglecting the input offset
voltage) by virtue of the virtual short.
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Therefore, from KVL and Ohm’s Law:
i1 
vin v  vin

R1
R1
and
i2 
v  vout vout

R2
R2
Combining these results:
vin vout

 IB 1
R1
R2
The output voltage is thus:
R 
vout    2 vin  R1 IB 1
 R1 
Note again that if IB 1  0 , the result reduces to the expected
inverting amplifier equation:
R 
vout    2 vin
 R1 
The second term in the above expression ( IB 1 R1 ) therefore
represents another output offset voltage!
It appears that we should keep the value of R1 small to minimize
the output offset voltage.
However, what will this do to the amplifier input resistance?
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Non-Inverting Configuration
R1
R2
i1
i2
IB 1
vIB 2
vin
vout
v+
+
Neglecting the input offset voltage, we can use the virtual
short to determine that:
v  vin
and KCL provides the same result as that of the inverting
amplifier:
i1  i2  IB 1
where from KVL and Ohm’s Law:
i1 
and likewise:
i2 
0 v  vin

R1
R1
v  vout vin vout

R2
R2
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Combining, we find:
vin vin vout

 IB 1
R1
R2
or rearranging:
 R 
vout   1  2 vin  IB 1R2
 R1 
Again, we find that this result is simply the ideal non-inverting
expression:
 R 
vout   1  2 vin
 R1 
with an added output offset voltage term:
IB 1 R2
In this case, we find that this offset voltage is minimized by
making feedback resistor R2 small.
In general, we find that the effects of the input bias currents
can be minimized by using small resistor values.
However, we will find that there is an additional strategy for
minimizing the effects of input bias currents!