Op-Amp Basics Clarification - Page 1
Forum section: electronics / beginners

author: pmbrunelle
author_rank: Regular Contributor
timestamp: December 13, 2015, 04:51:33 pm
content: So for pretty much every reference I read, for an open-loop op-amp, it generates an output voltage (lets say at DC) according to the following idealized formula:VOUT= AOL(VNON-INVERTING- VINVERTING- VOFFSET)So then in theory, you would wrap around your feedback network, and see to what value the output is stabilized to...But... What is VOUTrelative to? Is VOUTmeasured relative to some point between the supply rails? I am baffled... Is an op-amp an integrator, so therefore it would have no concept of a nominal "midpoint"?

author: c4757p
author_rank: Super Contributor
timestamp: December 13, 2015, 04:56:05 pm
content: Yes, in the ideal case. In a real opamp there will be some reference point determined by the internal circuitry, which could be totally arbitrarily based on one supply rail or the other, or both.

author: dom0
author_rank: Super Contributor
timestamp: December 13, 2015, 05:51:33 pm
content: Even a real op amp has no concept of a reference potential, just look at the structure of a 741-like op amp.To answer your question, Vout is referred to the same reference potential as Vnoninv and Vinv. This is an arbitrary choice the op amp isn't concerned with at all. For linear operation of the op amp you need Vnoninv - Vinv < ~Ut (or more, depending on emitter resistance in the input stage) and Vnoninv, Vinv within the common mode input range, and Vout within the output range. Whether your reference is -8 kV below the op amps negative supply or at center between V+ and V- is not relevant.

author: pmbrunelle
author_rank: Regular Contributor
timestamp: December 13, 2015, 06:09:49 pm
content: Okay, so I am looking at a 741-style opamp:https://en.wikipedia.org/wiki/File:OpAmpTransistorLevel_Colored_Labeled.svgIn this case, if we look at the class A gain stage, it looks like there is a "reference point", which is the threshold voltage for Q15 and Q19 to begin conducting, two diode drops above the negative supply rail. An uninformed observation...If we consider AOLto be finite, and do not consider the limit as AOLtends to infinity, then the reference point does not seem arbitrary to me. Depending on the choice of reference point, you would change the amount of steady-state error

author: dom0
author_rank: Super Contributor
timestamp: December 13, 2015, 07:18:40 pm
content: The input signal of the gain stage is already the output of the differential input stage, it is already referred to the supply of the op amp. But yes, I was wrong above, since the common mode gain of the differential amplifier is non-zero, a common-mode between the op amps supplies and the (still arbitrarily chosen) reference does indeed influence the output. (And, as usual, any common mode of the signal itself to the reference has the same effect)

author: Kleinstein
author_rank: Super Contributor
timestamp: December 13, 2015, 09:39:24 pm
content: As there is usually also an input offset error, the reference point is quite arbitray. Depending on the reference point you choose - the offset error will be different. Usually supply devided by A_vol is less than the offset specifications.So who cares if a OP hat something like 490 µV or 480 µV of offset error.

author: TimFox
author_rank: Super Contributor
timestamp: December 13, 2015, 09:44:55 pm
content: Another way to consider it is that for the output of the amplifier to be within bounds (determined by the power supplies and internal circuitry), the two input terminals (inverting and non-inverting) must be almost equal.In practice, when the output is within bounds, then the difference between the two input terminals is the offset voltage.This offset voltage may depend on the exact voltage of the inputs with respect to the power terminals (power-supply rejection and common-mode rejection). Also, at higher frequencies, the gain of the amplifier falls off and the input voltage required to develop the AC output voltage must be added to the DC offset voltage at the input. Furthermore, the input bias currents flowing through the external resistors will shift the voltages with respect to the external nodes.

author: dom0
author_rank: Super Contributor
timestamp: December 13, 2015, 10:28:53 pm
content: Quote from: TimFox on December 13, 2015, 09:44:55 pmFor most op amps the gain corner frequency is very low, somewhere between <0.1 Hz and ~200 Hz. This is because most op amps are dominant pole compensated, i.e. they have a single pole dominating their open loop frequency response (in the basic form implemented by deliberate design of the Miller capacitance in the voltage gain stage). E.g. a LM358 has 100 dB of voltage gain and 1 MHz unity gain frequency. Hence the corner frequency (where voltage gain has fallen by 3 dB) is 10 Hz (100 dB of gain difference to be covered divided by 20 dB/decade roll off).

author: TimFox
author_rank: Super Contributor
timestamp: December 14, 2015, 08:48:59 pm
content: Quote from: dom0 on December 13, 2015, 10:28:53 pmYes, that is true. I assumed that the original post was concerned with DC or low frequency amplification.An exception to this low gain corner is video op amps. Seehttp://www.edn.com/design/analog/4325660/Selecting-video-op-ampsThese tend to have relatively low DC gain (compared with a 741 or 358) and a relatively high corner frequency. This can reduce phase shift in the passband of the amplifier (with feedback).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 14, 2015, 10:37:21 pm
content: An even better way to view an op-amp is as an integrator:Real ones act more as an integrator (-20dB/dec and 90 degree phase shift for  all frequencies).The result from an  integral is always "plus a constant". That constant is a free variable, and has to be defined by the feedback of your circuit.(If you've taken integral calculus, this will be rather peculiar, because feedback puts the same variable (say, a node voltage) on either side of the integral. You actually get a differential equation, which might not have a simple integral solution. Generally, these equations have solutions with exponents. Which is what you'll observe in the circuit: RC time constants decay over time as the output settles. Assuming it does settle, of course.)Without feedback, the output voltage is undefined: it could be anywhere, as long as it's within the supply rails of course. This is why trying to use an op-amp as a regular amplifier (signal to inputs, no feedback from the output), just sucks (nevermind that the offset, gain and distortion of an open loop op-amp are pretty awful, too).So when you normally wire up a circuit, the output is referenced to some other node that you've put in the circuit. For example, the differential amplifier circuit senses a differential input, and supplies an output relative to the ground terminal: but this could be any arbitrary voltage reference, not necessarily the circuit ground.For a precision circuit, such as a power supply's "remote sense" connection, you can use this principle to extend where the voltage is actually sensed. You connect one wire to that ground point (not just in the net -- that is, anything touching "ground" -- but exactly at that physical location), and then when you read that voltage at some distant location (which might not share the exact same ground, for a variety of reasons like voltage drop and AC noise), the voltagedifferenceis still exactly what it should be.Tim

author: pmbrunelle
author_rank: Regular Contributor
timestamp: December 15, 2015, 04:56:23 am
content: I think I sort of get it...At low frequency/DC, the opamp generates an output voltage relative to some internal reference potential. This output voltage is proportional to the difference between the inputs (and an offset).At frequencies above the dominant pole corner frequency, the opamp acts like an integrator, so the the concept of a reference potential no longer applies.I will have to think about all of this for a while...