# Help Understanding Electronics Theory (Op Amps & Open/Closed Loop Gain)

Discussion in 'Electronic Basics' started by Adrian Jackson, May 26, 2005.

I'm in an electronics based job & did my training nearly 4 years ago.
Although to complete my Modern Apprenticeship I need to complete 4 BTEC
Assignments, one of which is an Electronic Principles one. As my training
what a while ago I'm struggling to remember some of the theory & was
wondering if anyone could help.

Part of the assignment is on Operational Amplifiers.

The circuit is an op amp with a voltage on the positive leg & a feedback
loop fed from the output into the negative leg of input. It then gives you
an input voltage & open loop gain then asks what the output voltage is.

Whats I've got is:-

Open loop gain (Ao) = Vout/Vin

So I transposed that to Vout = Ao*Vin

Is that right or would they want the output voltage including the feedback?

Also it gives me a few bits of data that I'm unsure of:-

Common Mode Rejection Ratio (CMMR) = 85db
Slew Rate = 0.5v/us

What are these?

As you can probably tell I'm very rusty on this.

2. ### Rich GriseGuest

Well, what's the circuit presented along with the question? What was the
exact wording of the question? I'm just guessing, but if they _gave_ you
the open-loop gain, they're probably asking, what's the gain of the entire
loop?
CMRR is, in laymens' terms, how much a common-mode voltage affects the
output. Common-mode voltage is, c'mon. I know this is "basics", but ...
Common-mode voltage is the voltage that's applied to both input pins
"in common". Like, as if they were strapped directly in parallel.

Slew rate is just, how fast can the output slew, or change. Very much
like rise-time.
Yes, but as I've noted, this is the "basics" group - please forgive
me if I've been a little harsh with you here. (there _is_ kind of
an ambience amongst us old farts that we don't like answering
homework questions very much, as you've probably been able to tell.)

Here's a hint that really, really helped me when I was just learning
opamps. The opamp (assuming infinite open-loop gain) does whatever it
needs to do to its output to make the voltage at the inverting input
(by way of the feedback network) equal to the voltage presented at its
non-inverting input.