I have been working with integrated design. Now I am working on a discrete circuit design project. I want to understand how the bias at the input will change the operating point at the output of an op-amp.

Is there any way to calculate it? Since in integrated circuit design we have operating point decided by bias current, but in discrete design I am unable to find the bias current. How will the output operating be set up?

I am yet to design an op-amp on a breadboard, but trying to simulate in Cadence, so wondering how to keep the operating point at output.

I am updating my question here:

I am going to use single supply as +5 V. I will obtain input for the instrumentation amplifier from a transducer whose output might swing differentially between 20-120 mV, over common mode of +2.5 V (VDD/2). instrumentation amplifier

As seen in above image, if I consider R5 to be twice of RG, what will be the output common mode of the two opamps connected as non inverting amplifier (in black colour). Even though I am calculating the differential gain which is obtained to be 5, which means my output common mode voltage most likely to be saturated? I am asking this because the next differential amplifier stage needs to provide some more differential gain.

The calculation of gain is not ideal but approximate derivation.

  • \$\begingroup\$ The bias at the input (provided there's negative feedback) will change the output in the same way your simulator computes it. There's nothing special about it. Bring an example schematic if you'd like to be shown more. \$\endgroup\$
    – Designalog
    Feb 28 at 6:40
  • \$\begingroup\$ @ErnestoG I have edited my question, please help me to weather reduce the common mode operating voltage level . \$\endgroup\$ Feb 28 at 10:02
  • 5
    \$\begingroup\$ If you're using a single supply (and for other reasons), the LM741 is not remotely an appropriate op-amp. TI recommends +/-10V minimum. \$\endgroup\$ Feb 28 at 10:05
  • \$\begingroup\$ @SpehroPefhany understandable. I have mostly done projects on single supply low voltage in integrated circuit design. So, I was thinking in same way . \$\endgroup\$ Feb 28 at 10:07
  • \$\begingroup\$ I agree that a 54 years old 741 opamp will not work from a supply that is only 5V. Its old datasheet shows all circuits powered from +15V and -15V. \$\endgroup\$
    – Audioguru
    Feb 28 at 15:11

2 Answers 2


This is the rule for linear operation with negative feedback.

If both inputs are equal in the Vcm common mode range and the output is not saturated meaning in the linear range then the Vin- will match or follow the Vin+ voltage.. So the choice of Vin+ might be considered first if using a single or dual supply.


An op-amp, even a 741, has such a high gain (200,000 typically for the 741) that it's not particularly important what you assume the output voltage bias point to be.

If you have symmetrical +/- supplies it may be convenient to assume that the output voltage will be 0V for 0V in (0V in plus or minus an "input offset voltage" that may be several mV- as much as 5 or 6mV).

Since +/-1mV * 200,000 is +/-200V you can see that whatever assumption you make will be okay. At most it will change the measured offset voltage slightly.

The circuit you show (a classic 3-amplifier instrumentation amplifier) will produce about 0V out for 0V differential input and the CM 2.5V will be subtracted provided none of the three outputs saturate and provided all 6 inputs are within the common-mode range for that particular op-amp.

Too much gain in the first amplifier can cause problems, with outputs saturating or with the next stage inputs. Allowable input common mode voltage and output voltage is a diamond shape. Diagram below is from this article, which I think is worth a read.

enter image description here

  • \$\begingroup\$ So the offset might itself saturate the output if opamp used in open loop as comparator. I have edited my question, please have a look on it \$\endgroup\$ Feb 28 at 10:04

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