3/9/2011
Reducing the Effect of Input Bias Current lecture
1/7
Reducing the Effect
of Input Bias Current
R2
i2
vin
R1
vIB 1
i1
v+
R3
IB 2
-
We found that the input bias current
will cause an offset in the output
voltage.
There is a solution to this problem—
place a resistor (R3) on the noninverting input!
vout
+
Q: Maria, why is
this resistor here?
I don’t see how it
can do any good.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
3/9/2011
Reducing the Effect of Input Bias Current lecture
2/7
The voltage v+ is non-zero!
A: Let’s analyze this circuit to determine how this new resistor helps.
First, notice that the voltage at the non-inverting terminal is now non-zero!
The bias current IB 2 means that, by virtue of KVL:
v + = 0 − R3 IB 2 = −R3 IB 2
R2
Now, because of the virtual short:
v − = v + = −R3 IB 2
And from KCL:
vin
i2
R1
vIB 1
i1
v+
i1 = i2 + IB 1
where from KCL and Ohm’s Law:
i1 =
R3
IB 2
-
vout
+
vin − v − vin + R3IB 2
=
R1
R1
Jim Stiles
The Univ. of Kansas
Dept. of EECS
3/9/2011
Reducing the Effect of Input Bias Current lecture
3/7
It seems like this just
made the offset even larger
R2
And also from KCL and Ohm’s Law:
i2 =
vin − vout − (R3IB 2 + vout )
=
R2
R2
i2
vin
Combining these results:
R1
vIB 1
i1
vin + R3IB 2 − (R3IB 2 + vout )
=
+ IB 1
R1
R2
v+
R3
IB 2
-
vout
+
Performing the usual algebraic gymnastics, we rearrange this result and find
that the output voltage is:
⎛ R2 ⎞
⎛
⎞
R2R3
−
+
−
v
R
I
I
R
I
⎟ in ⎜ 3 B 2
2 B1 ⎟
B2
R1
⎝ R1 ⎠
⎝
⎠
vout = ⎜ −
Jim Stiles
The Univ. of Kansas
Dept. of EECS
3/9/2011
Reducing the Effect of Input Bias Current lecture
4/7
Awl be baak
Again we find the output consists of two terms. The first term is the ideal
inverting amplifier result:
−
R2
vin
R1
and the second is an output D.C. offset:
−(R3IB 2 +
R2R3
IB 2 − R2IB 1 )
R1
Q: Resistor R3 was supposed to reduce the
D.C. offset, but it seems to have made
things even worse. Fix this or I shall be
forced to pummel you.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
3/9/2011
Reducing the Effect of Input Bias Current lecture
5/7
We must choose the proper value of R3…
A: Say we set the value of resistor R3 to equal R3 = R1 R2 , i.e.:
R1R2
R1 + R2
R3 =
In this case, the D.C. offset becomes:
⎛ R1R2
⎞
R22
IB 2 +
IB 2 − R2IB 1 ⎟
−⎜
R1 + R2
⎝ R1 + R2
⎠
⎛ (R + R2 )R2
⎞
= −⎜ 1
IB 2 − R2IB 1 ⎟
⎝ R1 + R2
⎠
= R2 (IB 1 − IB 2 )
= R2Ios
Typically, the bias currents IB 1 and IB 2 are approximately equal, so that offset
current IB 1 − IB 2 = Ios is very tiny.
Therefore, the resulting output offset voltage is likewise very tiny!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
3/9/2011
Reducing the Effect of Input Bias Current lecture
6/7
…and this is that proper value
Therefore, when designing an amplifier with real op-amps, always include a
resistor R3 equal to the value:
R3 = R1 R2 =
R1R2
R1 + R2
This is true regardless of whether we use the inverting or non-inverting
configurations!
R2
i2
vin
R1
i1
v-
R2
-
i2
IB 1
R3 = R1 R2
v+
IB 2
R1
+
i1
vin
vIB 1
R3 = R1 R2
v+
IB 2
Jim Stiles
The Univ. of Kansas
-
+
Dept. of EECS
3/9/2011
Reducing the Effect of Input Bias Current lecture
7/7
This is just the type of subtle point
that shows up on an exam
If the impedances are complex (i.e., Z 1 (ω ) and Z 2 (ω ) ), then set the resistor R3
based on the D.C. values of the impedances:
R3 = Z 1 (ω = 0 ) Z 2 (ω = 0 )
In other words, set the capacitors to open circuits and inductors to short
circuits.
Jim Stiles
The Univ. of Kansas
Dept. of EECS