Could someone demystify op amps for me?

[MonkeySage]

I have continually tried to ask my prof to explain this stuff, and he just acts like I'm an idiot for not understanding. He refuses to directly answer my questions, and instead he interrupts me when I try to explain the problem I'm having. He assumes he knows what I'm having trouble with, and he goes off on a tangent explaining things I didn't even ask about.


And that has left me staring at the same op amp problem for the past several hours.

Apparently Vs and Vn are not the same, which makes no sense. Vn is supposed to equal Vp which seems redundant. In is supposed to equal zero for reasons that are never explained in my book, and yet it doesn't?

I hate when the answer is seemingly given to me, and then when a question is asked about it and I give the answer I was given earlier it turns out to be completely wrong. It means I read something wrong.

Ok. So I've been given an inverting amplifier circuit. On the negative input terminal, I have a voltage source (2V) and a resistor (4kΩ).

On the positive input terminal, I have nothing.

I have a resistor across the amp, I'm guessing Rf, its kΩ.

-VCC=-15v, VCC=15v

And Ro = 6kΩ

I am supposed to figure out which constraint determines In and what value to give for In.


The op amp is "ideal", whatever that means.

I'm supposed to find a value for (Vp-Vn) and what ideal op amp constraint determines that value, even though I've been lead to believe that Vp=Vn.

And finally, I'm supposed to calculate Vo


None of this makes any sense to me, because my teacher refuses to take the time to actually explain it. He'd rather berate me for not understanding.

[gomipile]

My eyes glazed over when you started using variables without defining them first. Do you have a labeled circuit diagram for this?

[MonkeySage]

http://i1016.photobucket.com/albums/af282/anthonysherer90/20171112_2025392_zpsx4qliakj.jpg

[AsteriskAmp]

http://i1016.photobucket.com/albums/af282/anthonysherer90/20171112_2025392_zpsx4qliakj.jpg
For those who instead get the photobucket message:

http://oi65.tinypic.com/vcx5pd.jpg

[gomipile]

I don't see a Vs in that photo.

[MonkeySage]

Voltage source, 2v, negative input terminal.

[crayzz]

Ok this is the 4th time I'm trying to write this out; I kept misreading you and having to change tact, and then I accidentally reloaded the page.

"Ideal" means "makes the math easy." For an op-amp, we basically want 2 qualities (well, more really, but for here anyways): that no current flow into the inputs, and that the difference between the positive (Vp) and negative (Vn) inputs be 0.

So, for input current: it's just governed by a simple ohm's law equation: In = Vn/Rn, where Rn is called the input resistance, or input impedance. If we want In to be 0, the input resistance must be infinite. An ideal op-amp has infinite input resistance.

(Vp-Vn) = 0 is slightly trickier.

The output of an op-amp is given by this equation

Vo = G(Vp - Vn)

where G is called the Gain of the op-amp (also called the voltage gain or open loop gain). If we rearrange, we find that the difference between the input voltages isn't actually 0:

Vo/G = Vp - Vn

But for an ideal op-amp, we take Vp - Vn = 0 as a given. To make this work, the gain of an ideal op-amp must also be infinite.

I don't think you're being asked to find either In or (Vp - Vn); you're just supposed to recognize that they're 0 in an ideal op-amp, and explain what mathematical constraints we assume to make that work (infinite resistance and gain).

Now, to find Vo: this is actually fairly simple circuit analysis, especially since you're dealing with an ideal op-amp. Since Vn = Vp, and since Vp is hooked up to ground (i.e. at 0V), Vn = 0V. Since there's no current lost into the negative input, that means you basically have one solid branch (i.e. just one current) running from Vs to Vo. Since the current is the same along the branch:

I = (Vs - Vn)/4000ohms = Vs/4000ohms
I = (Vn - Vo)/12000ohms = -Vo/12000ohms

I = I

Vo = -3Vs = -6V

[factotum]

crayzz explained it beautifully. Obviously in the real world an op-amp can't have infinite gain or input impedance, but you're being asked to assume the op-amp in your circuit is ideal, so you don't need to worry about that.

[JeenLeen]

If the above didn't explain it to you, talking to your professor outside of class may help. Perhaps bring up the theoreticals crayzz mentioned, but note that you are curious about real-world applications and/or want to clarify if he is asking about idealized settings or not.

Or, maybe don't. If the guy isn't listening to you in class, he may not out of class, and any frustration you have may tick him off, which doesn't help grades. I guess use your judgement about risk of ticking off prof vs. potential gain of understanding.

[wumpus]

crayzz explained it beautifully. Obviously in the real world an op-amp can't have infinite gain or input impedance, but you're being asked to assume the op-amp in your circuit is ideal, so you don't need to worry about that.

Reality check: a quick check shows that a typical opamp has a gain*bandwidth product ~1MHz (https://www.digikey.com/products/en/integrated-circuits-ics/linear-amplifiers-instrumentation-op-amps-buffer-amps/687?k=opamp) [oddly, this apparently hasn't improved since my intro EE class in the 1990s, although other parts have improved]. If you need a gain of a million, your signal had better be below 1Hz (actually I'd be shocked if it could force Vp and Vn that close, but you get the idea).

The other issue is that Vo must be between +/-15, unless the thing has an internal power supply (and expect to pay extra if you need to actually get to +/-15, they are called "rail to rail" op-amps, others only go to +/- 14 or less).

As long as you don't push them to their limits, a real op-amp (available for $0.50 a pop on the link, or =$0.10 in quantity 2,500, and a typical package will contain 2-4 op-amps) will act remarkably close to your textbook ideal. You can typically even buy something that really pushes the "ideal nature", but typically only for one feature/requirement.

[gomipile]

Reality check: a quick check shows that a typical opamp has a gain*bandwidth product ~1MHz (https://www.digikey.com/products/en/integrated-circuits-ics/linear-amplifiers-instrumentation-op-amps-buffer-amps/687?k=opamp) [oddly, this apparently hasn't improved since my intro EE class in the 1990s, although other parts have improved].

https://www.mouser.com/Search/m_ProductDetail.aspx?Microchip-Technology%2fMCP654-E-SL%2f&qs=kJ7kJBa88dGBI6F0FGjKDQ%3d%3d

[wumpus]

https://www.mouser.com/Search/m_ProductDetail.aspx?Microchip-Technology%2fMCP654-E-SL%2f&qs=kJ7kJBa88dGBI6F0FGjKDQ%3d%3d

Yeah, I mentioned that "you can have an ideal op-amp, just pick which feature you want". Judging by how long ago that was true, you probably pick 2 or 3 "ideal" characteristics now. I'm somewhat curious where you would use such a beast. Looks like nowhere close to enough bandwidth for an input amp (amplifying everything the antenna picks up) although the output power will be non-trivial (less than a Watt, but compared to the antenna it is huge). Presumably for huge intermediate stages, and shoveling huge amounts of data (or ecc games).

I'm guessing that is Microchip's answer to "design your own amplifier if you can't get large bandwidths".

[gomipile]

Yeah, I mentioned that "you can have an ideal op-amp, just pick which feature you want". Judging by how long ago that was true, you probably pick 2 or 3 "ideal" characteristics now. I'm somewhat curious where you would use such a beast. Looks like nowhere close to enough bandwidth for an input amp (amplifying everything the antenna picks up) although the output power will be non-trivial (less than a Watt, but compared to the antenna it is huge). Presumably for huge intermediate stages, and shoveling huge amounts of data (or ecc games).

I'm guessing that is Microchip's answer to "design your own amplifier if you can't get large bandwidths".

A 50MHz GBWP isn't "enough" compared to 1MHz?

[wumpus]

A 50MHz GBWP isn't "enough" compared to 1MHz?

It always depends on the application. Lots (most?) interesting radio signals are higher than that, so you can't use it to amplify directly from the antenna. It should be useful for anything past that. To be honest, I think I'm so impressed by the "ordinary" op-amp to be all that crazy for the improved op amp (1MHz is pretty good, just under where a cheap scope has problems, and you can get away with simple scope probing techniques). Then again, I avoid RF like the plague, so ask someone who works in that range.