1
\$\begingroup\$

I was confused about op-amps and asked this question: In an op-amp, why does Vn approach Vp until the difference is delta V / A? Why doesn't Vn just equal Vp? and received really helpful answers.

I now know why in a real-world op-amp, Vn has to differ from Vp; Vn can't just equal Vp or else the output voltage will equal 0. However, how does a real-world op-amp ensure that Vp and Vn differ by a small amount? How does it make sure that Vn doesn't increase to equal Vp? How does it ensure that the difference between Vp and Vn is Vp - (Vp - Vn) / A instead of 0?

\$\endgroup\$
3
  • \$\begingroup\$ Given that a real amplifier has finite (but large) gain, what would the output voltage be, when the input voltage is exactly 0 V? \$\endgroup\$
    – The Photon
    Oct 30, 2016 at 19:26
  • \$\begingroup\$ @ThePhoton If the input voltage is exactly 0, then output voltage is 0. That's why Vn doesn't increase until it hits Vp. However, in the real-world (and perhaps this is a hardware question), HOW does it actually ensure that Vn doesn't reach Vp but instead stops at some threshold (so that output voltage ISN'T 0)? \$\endgroup\$
    – laura
    Oct 30, 2016 at 19:33
  • \$\begingroup\$ physically, the input voltage causes the output voltage, not the other way around. We just often calculate backwards because it makes the math simpler. \$\endgroup\$
    – The Photon
    Oct 31, 2016 at 16:20

1 Answer 1

2
\$\begingroup\$

A very simple model for an opamp is used in the example below. A voltage-controlled voltage source with a gain of 1000 is used to build a non-inverting amplifier.

enter image description here

As you can see a difference of about 2mV is needed to get a voltage of about 2V at the output (since the gain is 1000). For a smaller difference at the input the output voltage is reduced and the voltage at the inverting input gets smaller. So it's actually the external negative feedback that takes care of a non-zero differential voltage and not the opamp itself. It just provides high gain.

\$\endgroup\$
4
  • \$\begingroup\$ This makes so much more sense! One quick question: In a voltage follower though (an op-amp without the resistances in the negative feedback loop), where would the non-zero differential voltage come from? \$\endgroup\$
    – laura
    Oct 30, 2016 at 19:49
  • \$\begingroup\$ Same principle. With 1V at the non-inverting input we need 1V at the output so a 1mV difference is required at the input (for a gain of 1000). \$\endgroup\$
    – Mario
    Oct 30, 2016 at 19:51
  • \$\begingroup\$ I follow what you're saying but I don't understand the order of it -- you're saying we know we need a certain voltage at the output and this tells us what the voltage differential is (therefore we know what the voltage at the inverting input is). This is saying that output voltage is what determines input voltage at the inverting terminal. However, isn't it the opposite case? Isn't it input voltage (and the differential) which determines what output voltage is? \$\endgroup\$
    – laura
    Oct 30, 2016 at 20:19
  • \$\begingroup\$ That's where the feedback mechanism comes into play. The output is fed back to the input of the amplifier and this results in the required input voltage to satisfy the conditions set by the feedback network. A good strategy to understand the concept of feedback is to imagine a undesired/unrealistic situation. E.g. differential voltage slightly too high, too low or even zero. For a well behaved feedback loop this should result in a condition that brings the circuit back to a stable condition. \$\endgroup\$
    – Mario
    Oct 30, 2016 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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