In Floyd book it says that 2 input bias currents for 2 inputs ideally equal. I understand if they are equal in a common mode, but how they can be equal in a differential mode if in this case, one input is connected to the ground, or voltages are in opposite phase which at least gives opposite directions of the currents? enter image description here

  • \$\begingroup\$ In fact, ideally they're not even there! \$\endgroup\$ – Scott Seidman Dec 26 '13 at 19:06
  • \$\begingroup\$ That's what I thought too. \$\endgroup\$ – flup Dec 26 '13 at 19:13
  • \$\begingroup\$ The authors should know better than to use "ideal" in the context of op amps when another word, like "optimally", is probably what he wants. \$\endgroup\$ – Scott Seidman Dec 26 '13 at 19:49

The basic error you made is in your understanding of bias currents: -

Bias currents are not something that YOUR input signal is responsible for so, the direction or polarity of input signals are irrelevant. Bias currents can be regarded as currents taken (or given) by the input stage of the OP-AMP and they will flow into or out of your input voltage sources irrespective of their polarity. They are self-generated within the op-amp and not to be seen as some current proportional to your input signal level.

  • \$\begingroup\$ Thanks Andy! So, bias currents will be present even if nothing is connected to the inputs? Are they generated as a reverse leakage currents for transistors in a differential amplifier stage? \$\endgroup\$ – Haml Dec 26 '13 at 20:07
  • \$\begingroup\$ With nothing connected to the inputs they can't be present of course but as soon as anything of a circuit appears on the inputs a current will flow. It's a dc current too (plus noise). For a BJT op-amp you can regard them as the current into the base of the transistors that biases them. For JFET op-amps, it is gate leakage currents but I can't explain them chapter and verse other than they will double every 10C rise in temperature. Maybe that's another question? \$\endgroup\$ – Andy aka Dec 26 '13 at 20:25
  • \$\begingroup\$ Sure, I meant if not input voltage is present, but inputs are connected to somewhere and not in a HighZ. Got that, thanks a lot! \$\endgroup\$ – Haml Dec 26 '13 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.