Hi all, ...
If a simple op-amp circuit is described as:
-> being in a "voltage-follower with-gain" configuration, and
-> having a resistor ratio of 9:1
...Where are the two resistors positioned with respect to the op-amp circuit element, and
...How, if at all, doe they come in to play when calculating the configuration's "closed-loop gain"?
Thanks heaps!
If you use two resistors with a 9 to 1 ratio, you can get various gains out of the op-amp, but if it's a "voltage follower", I assume you want a positive gain and I'll assume it's intended to be a gain of 1/10th.
In that case the input goes into the 9 times resistor, which is connected to the positive input of the op amp. The ouput is fed back closed loop to the negative input and the positive input is also connected to the common/ground through the 1 times resistor.
So the input is effectively divided by 10 via. the two resistors and the op-amp is configured to be a voltage follower or provide unity gain of that 1/10th amplitude signal (so it's just a buffer to isolate the input impedance from the output).
(Though you could also configure things a couple different ways for a -1/10th gain or a negative 10 times gain, but I think the positive 1/10th gain is likely what was desired for this).
Confused2
27th April 2008 - 08:31 PM
Hi walkinginmud,
The thing about op-amps is that they do their best to keep the positive input at EXACTLY the same potential as the negative input. If you divide the output voltage by a factor of (say) ten and feed it back to the positive input then the output MUST swing ten times as much as the input (to the -ve amp terminal) to keep both inputs at the same potential.
-C2.
Enthalpy
1st May 2008 - 01:46 AM
Voltage follower usually means gain=1, so find a better book. But keep polite with your teacher until you get your diploma.
The sentence probably meant "gain=+10" with a resistor ratio of 9.
Then the bigger resistor is between output and inverting input, the smaller between the inverting input and grounf. The input signal goes to the noninverting input.