If a voltage source is applied to the output of the op-amp (with NO feedback to the input), what happens to the circuit/device? Does the signal just go straight to ground or open circuit? Or does it cause the op-amp to break down because of high current, etc?
2007-07-18 09:45:34 · 2 answers · asked by M A 1 in Science & Mathematics ➔ Engineering
The output of an ideal op-amp is a voltage source, itself, so it would be similar to connecting 2 batteries of different voltage together. In series they would add or subtract, and the higher current density battery would tend to push it's current through the "weaker" battery (if the load dictated it).
The output of a normal op-amp is only good for a few milliamps of current, and has an output impedance of less than 100 Ohms (equivalent of less than 1 Ohm if feedback is used). That 100 Ohms is all that's in the way of the external voltage source from forcing too much current into the output pin of the op-amp.
In general it's not a good thing to do, as you'd probably damage the output transistors.
If you put an current "equalizing" resistor in series with the output pin and the external voltage source of > 500 Ohms (safer with > 1 kOhms) then it should work, whatever it is you have in mind.
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2007-07-18 10:28:02 · answer #1 · answered by tlbs101 7 · 1⤊ 0⤋
If its a short circuit proof op amp it should take it.
2007-07-18 11:32:50 · answer #2 · answered by Anonymous · 0⤊ 1⤋