- 2068 Chaitra · 7 marks
Positive/negative bias current effect and reduction
Short answer
Positive and negative bias currents produce opposite polarity output errors: Vo=IB−·Rf − IB+·Rcomp(1+Rf/R1). Choosing Rcomp=R1||Rf cancels the main error, leaving Rf·Ios. Additional reduction: FET-input op-amps, smaller resistors, offset-null, and ac coupling.
Explain the effect of positive and negative input bias current on output voltage of an op-amp and suggest methods of reduction.
Input bias currents are the dc base (or gate) currents of the input transistors. The current into the non-inverting input is called the positive input bias current IB⁺ (or IB2), and the current into the inverting input the negative input bias current IB⁻ (or IB1). (The currents flow into the op-amp for NPN input transistors and out of it for PNP inputs; only the sign of the error changes.)
Rf
+----/\/\/----+
| |
Vi --R1--+--> IB1 -(-) |
>--+---- Vo
+-----> IB2 ---(+)
|
Rcomp
|
GND
Effect of IB⁻ (inverting input)
With Vi = 0 and Rcomp = 0, the inverting input is a virtual ground, so no current flows in R1 and IB⁻ must flow from the output through Rf:
Vo1 = +IB⁻ · Rf
Effect of IB⁺ (non-inverting input)
IB⁺ flows through Rcomp and makes V+ = −IB⁺·Rcomp. The circuit amplifies this as a non-inverting amplifier:
Vo2 = -IB⁺ · Rcomp · (1 + Rf/R1)
Total effect
By superposition:
Vo = IB⁻·Rf - IB⁺·Rcomp(1 + Rf/R1)
The two currents produce outputs of opposite polarity, so they can be made to cancel.
Methods of reduction
- Compensating resistor: choose Rcomp(1 + Rf/R1) = Rf, i.e. Rcomp = R1‖Rf. Then
Vo = Rf(IB⁻ - IB⁺) = Rf·Ios
which is much smaller since Ios << IB. 2. Use FET-input or super-β op-amps with very small bias currents (pA). 3. Use smaller resistor values (or a T-feedback network for high gain), since the error is proportional to Rf. 4. Null the remaining error with the offset-null pot (741: 10 kΩ between pins 1 and 5). 5. In ac amplifiers, use coupling capacitors so the dc gain is unity.
✓ Answer checked by Claude Opus 5.5 (high effort)
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