this is the first question I have received that does not state "Assume that all op amps are ideal". Should I assume that in this situation or is this all together different?
How did you arrive at that conclusion? I would think that for an ideal op-amp the voltage across the input +/- terminals would be 0V.Say we had 10V @ Vin. that would mean that +5V would have to be at +/- op amp terminals.
thats correct...How did you arrive at that conclusion? I would think that for an ideal op-amp the voltage across the input +/- terminals would be 0V.
If Vin = 10V then wouldn't the positive side be +5V and the negative side be -5V? for a total peak to peak of 10V?In post #4, you used 10V as the input. In post #6, you say that the op amp +input will be 5V. Which is it?
The negative side is connected to ground. Ground is your reference, so is defined as zero volts.If Vin = 10V then wouldn't the positive side be +5V and the negative side be -5V? for a total peak to peak of 10V?
Right! so that means that the positive side of the power supply is automatically +10V?The negative side is connected to ground. Ground is your reference, so is defined as zero volts.
Yes.Right! so that means that the positive side of the power supply is automatically +10V?
I understand... I was just assigning an arbitrary value for Vin for the sake of demonstration.Hi notoriusjt2,
You seem to be confusing input voltage with the op-amp supply voltage.
In this case the op-amp supply voltage is not stated as it is irrelevant for the purposes of solving the problem.
If the op-amp were capable of running from either a single or dual polarity supply the outcome would be the same for this problem.
No. See the attachment.allright
if
node1 = 1V
then
node2 = 1V
now that I look at it, node3 and node4 are the same point
and node6 and node2 are also the same point
in order for this to work correctly
node4 would have to be 4V
node5 would be 2V
node6 = node2 = 1V
no?
i see.... how did you calculate those voltages @ the resistors?No. See the attachment.
I started with these facts:i see.... how did you calculate those voltages @ the resistors?