I got this equation for a summer circuit output voltage fed into an inverting opAmp. Both opAmps have supply voltage of ±15 V.

Vout = -6*V1-30*V2-60*V3

I got Vout = -390 mV.

I know that the ± 15 volts is there to tell a range of possible voltages; Vout cannot be out of the supply range.

However, I don't understand whether the range for possible Vout is 0-30 V or -15V to +15 V.

Here is the question.enter image description here

  • \$\begingroup\$ The most obvious problem with your answer is that the inputs are 10 mV, 1 mV, and 5 mV, not volts. So your calculated output should be -390 mV, not -390 volts. \$\endgroup\$ – WhatRoughBeast Aug 19 '16 at 1:39

The output voltage of an op amp is limited by its supply voltages. So best case (rare and expensive but possible with ideal op amps), the output voltage could range between +15 V and -15 V if those are the supply voltages. In real op amps the output can usually only approach the power "rails" as they're called, so maybe +13 to -13V for this example.

An op amp has two power supply pins, one positive and one negative. The notation +/-15 V indicates the positive supply is 15V and the negative supply is -15V.

Note there is no (power) ground pin on an op amp (although one or more pins may be grounded, depending on the circuit configuration), which certainly struck me as odd when I first encountered it :)

Also note there is such thing as a so-called "single supply" op amp, which is designed to be useful when the positive supply is, say +12V and the negative supply is 0V (ground).

If op amp 1 in your circuit can put out between +/- 15V and op amp 2 in your circuit can put out between +/- 15V, I think you have the basic ideas you need to reason through your question :)

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  • \$\begingroup\$ So if I am given V1= V2 = 1 V and V3=0, then Vout = -15 V despite my expression giving Vout = -36 V. That is because opamps don't create voltage. Right? \$\endgroup\$ – captain Aug 19 '16 at 2:07
  • \$\begingroup\$ When an op amp (two words, not one) gets to its maximum output voltage, it is said to have "saturated". So yes, if the calculated voltage is -36V then the op amp would saturate and only put out -15V (the value of the negative power rail). This in the case of ideal op amps, which these seem to be. A real op amp might be -13V or so and this value could be found on the data sheet for the device. \$\endgroup\$ – scanny Aug 19 '16 at 4:19

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