2.4 Measurement of input bias current
As mentioned eralier, input bias current is very small in magnitude - so, measuring it directly is not
a good idea. However, it can be measured cleverly using the following circuit.
Ra
(a)
R1
+Vcc
−
Ra
(b)
+Vcc
−
+
Rb
−Vcc
Vout
R2
+
Rb
−Vcc
Vout
Figure 3: Circuits to measure input bias currents Ib1 and Ib2, respectively.
Fig. 3(a) is just the circuit for an inverting amplifier, with the input grounded. So, the voltage at
the inverting input terminal should be ideally zero. But from the circuit above, one can see that
the voltage at the inverting input has two contributions - one, Vout reduced by the potential divider
made
out
of
Ra
and
Rb,
i.e.,
V Rb
Ra+Rb out
-
two,
the
voltage
drop
over
the
R1
if
there
is
a
non-zero
input bias current flowing. Thus, we can write
Vi
=
Ra
Rb
+
Rb
Vout
−
Ib1R1
=
0
(3)
or
Ib1
=
1
R1
Ra
Rb
+
Rb
Vout.
(4)
If Ra = 10 kΩ, Rb = 780Ω and R1 = 1 MΩ, we get
Ib1
=
1
1
× 106
780
10000 +
780
Vout
=
10−6Vout
13.82
A
=
Vout
13.82
µA.
(5)
(In the original manual, the value of Rb is wrongly assumed to be 330 Ω, which leads to the wrong
conclusion,
Ib1
=
10−6
330
10000+330
Vout
≈
Vout
31.3
µA)
Input bias current Ib2 can be similarly measured using the circuit in Fig. 3(b), which represents
a non-inverting amplifier, with the input grounded through the resistor R2. The voltage at the
non-inverting terminal would be Ib2R2, which gets amplified to give Vout. Using the relation for
non-inverting gain, one can write
Vout
=
Vi
Rb
+
Rb
Ra
=
Ib2
R2
Rb
+
Rb
Ra
.
(6)
So, the inverting input bias current is given by
Ib2
=
1
R2
Ra
Rb
+
Rb
Vout.
(7)
For R2 = 1 MΩ, one gets
Ib2
=
1
1
× 106
780
10000 +
780
Vout
=
10−6Vout
13.82
A
=
Vout
13.82
µA.
(8)
4