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Effect of input bias current on a simple unity gain inverting amplifier

I'm trying to understand how input bias current affects the behaviour of this configuration in relation to a real op amp. 
I've read that the inputs should be fed from similar resistances so the effect of input bias current is minimised. 

Here is my circuit, I'm using the characteristics of a 741 as my example. R1 = R2 = 10M (yes I know thats large for a 741, just to illustrate my idea.)

and Ibias for a 741 is quoted as 0.08uA

We have resistors R1, R2 of 10M, which gives us a 10M input resistance because the - input is a virtual earth.

The thevenin equiv cct for the resistance seen by the inverting input is 0V / 10M //10M  = 0V 5M

The voltage we measure at the output will have an error because Ib1 is fed from a resistance of 5M  while Ib2 is fed from a zero resistance.

The error due to the input bias current of 0.08uA is Verror = 0.08uA * 5M = 0.4 volts!

However another view is as follows:

The problem with the simplification is that it ignored how the op-amp will respond when something changes.

The circuit can't be simplified to:
+3v
|
| 10M resistor
|
+----Op Amp -ve input
|
| 10M resistor
|
-3V
Because it will change when something else changes.

The action of the op-amp will be to try to maintain 0V at it's negative input when something else changes. To test what source impedance the -Ve input sees imagine drawing some current from that point. In the circuit above with no feedback the voltage will change according to Ohms law and you can calculate the source impedance from that.

With negative feedback, which is what you actually have, if you draw some current at the mid point then, as long as you remain inside the operating limits of the circuit (supply voltage in particular), the op-amp will adjust it's output to prevent the voltage at its -ve input from changing. If you have voltage source that does not change when you draw varying amounts of current from it then the source impedance of that voltage is zero ohms (or very low).  

Sorry, I know this is long but grateful for any clarification.



  • Hi John,

    yes, the source impedance must be taken into calculation. And if it changes for some reason, the input bias cancelation no longer works properly but becomes imbalanced.

    Does this answer your question?

    Kai

  • Former Member
    0 Former Member

    Hi John,

    You can think of the bias current as creating a DC bias error which can be added in superposition with other error sources, such as an input offset voltage.  So long as the amplifier is operating in its linear region, this error will be present unless it is canceled out, for example with the same error at the opposite input.

    Oftentimes the input bias current has a very negligible effect on the output voltage of the amplifier.  The exceptions to this case are when you have a very large resistance at one or both of the amplifier's inputs, a part with a large bias current, high temperatures, or a combination of these factors.  Depending on the type of amplifier, some have significantly better specifications at room temperature and some get worse at high temperature.

    There are two ways to calculate the corresponding error.  You can use superposition and set Vin = 0V (see the example at the end of this presentation for this method).  Alternatively, you can consider the equivalent impedance at the input, multiply that impedance by the bias current at that input which will give you the input voltage error due to the bias current, and then multiply by the non-inverting gain to give you the output voltage error due to the bias current (see the example worked in this presentation).  Both methods will give you the same output voltage error.  I find the second to be more intuitive.

    I highly recommend you look at this presentation on input bias current and offset voltage, which is the same as the first one linked above, and this great bit by Bruce Trump on input bias current resistors.

    Regards,
    Daniel