0
\$\begingroup\$

I'm trying to increase my knowledge of electric circuits and their major components. My current understanding is very basic, which will explain why my question may seem laughable to most people on here. Regardless, I'm hoping someone can water down an explanation for me regarding operational amplifiers. After a bit of reading and looking for explanations, the common thread is that:

In an ideal amplifier, current does not flow into or out of the inputs (there is 0 input bias current). However, I keep hearing that you essentially "put in" a voltage. I'm unclear as to how one can input a voltage while supplying no current.

I understand how there can be voltage without current flow (say, if a point charge is present but not moving in space). But in amplifier inputs, the important part seems to be "inputing" a voltage. I just don't understand how you can supply or input a voltage without moving some charge through the input (at least initially).

If my understanding is so far gone that you do not understand my question, please let me know (and perhaps provide a few resources that you feel describe amplifiers or electronics in an easily digestible way).

\$\endgroup\$
1
  • \$\begingroup\$ When they say "you input a voltage" they mean "your circuit sets the voltage (however you want it to), and then the amplifier looks at the voltage". It's "input" as in an input of information, not an input of electricity. \$\endgroup\$ Commented Sep 15, 2016 at 2:41

2 Answers 2

1
\$\begingroup\$

An op-amp has what is known as an input impedance. In the most simplistic terms this can be thought of as a resistance to ground. When you connect a voltage to the input current flows through that resistance to ground.

The "ideal" op-amp has an infinitely large input impedance, which means that an infinitesimally small current will flow - i.e., no current. That never happens though.

The thing to remember is that current is the movement of electrons around a circuit, but voltage is simply the difference in electrical potential between two points in a circuit. Nothing needs to flow for there to be a difference in potential.

Take a simple AA battery, for example. With it disconnected there is no circuit, yet the + terminal is 1.5V higher than the - terminal. The input of the op-amp's input is simply measuring that voltage. But of course, just like in quantum physics, the act of measuring it changes it, because the input impedance isn't infinite, and so some current flows, which then has an effect on the circuit (and hence the voltage) that you are measuring.

\$\endgroup\$
1
  • \$\begingroup\$ So if I have this right, would it be good (though not perfect) to almost think of these inputs in the op-amp as a type of voltmeter? My rationale would be that, voltmeters (like our op-amp) need to have high resistance (in the op-amps case, impedance) to reduce the current through the alternate path provided by setting up a voltmeter in parallel. So the inputs of the op-amp can detect the voltage while reducing the current through the input pins. \$\endgroup\$
    – user120818
    Commented Aug 16, 2016 at 1:27
0
\$\begingroup\$
  • Voltage without current: YES (forgetting unwanted parasitic decharging effects);

  • Current without voltage: No. In each conducting body we need an E-Field (that means: a voltage) which causes the force which is needed for moving the charges (which is "current").

  • Amplifiers: We have to distinguish between (a) voltage driven and (b) current-driven amplifiers. Classical opamps belong to the first class and have a large input resistance. Therefore, there is always a (small) input current as long as there is a voltage between the input terminals.

    However, in most cases, this current (although it does exist) is neglected if compared with the current which goes through the feedback path. This leads to the concept of an "ideal" amplfier. Something similar applies to the input voltage of an opamp (with feedback). There is always a small remaining differential voltage across the input terminals (µV range), but this voltage may be neglected in comparison to the amplifiers (opamp with feedback) input and output voltages.

\$\endgroup\$
1
  • \$\begingroup\$ A recommendation to the moderators: The process of downvoting should be connected with a justification - in the interest of the questioner. Otherwise it does not help at all - in contrary: It confuses the questioners. \$\endgroup\$
    – LvW
    Commented Oct 17, 2016 at 21:03

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.