Perpetually confused about Op-Amps

Thread Starter

gerases

Joined Oct 29, 2012
186
If I put a voltage on the base of Q1, Q1 draws current which raises the emitter voltage of both transistors causing the collector voltage on Q2 to rise. If I do the same to Q2 the emitter voltage on both transistors will rise and the collector voltage of Q1 will rise. The collectors of Q1,2 are the outputs of the differential amplifier. In an OP-Amp these outputs continue on to a push-pull output stage that has a single ended output.
I think this schematic has the potential of making it clear to me. Thank you for showing it to me. Just a couple questions then:

So, voltage to Q1 raises the collector current of Q2 and vice versa and we "read" the voltage from those collectors somehow connected into one output. That's cool. I can see the differential amplification! The only thing I still don't see is how the first "golden" rule explained by this? Namely:

The output attempts to do whatever is necessary to make the voltage difference between the inputs zero.
 

kubeek

Joined Sep 20, 2005
5,794
This applies only for negative feedback, i.e. the higher the output, the higher the voltage on the negative input, which in turn lowers the output and makes the output such that the difference between positive and negative input is essentialy zero.

For positive feedback this doesn´t work and the output goes hard to one of the supply rails.
 

patricktoday

Joined Feb 12, 2013
157
In _some_ op amp designs it works like this: there is a little man living inside the chip. He has two Venetian blinds on his front wall sitting on either side of his TV. When he sits down to watch his favorite show, Rawhide, with Clint Eastwood, he wants to see a balanced view. If one blind is opened more than the other he gets entirely stressed out, it simply ruins his enjoyment of the show and basically all the cattle rustling antics of the day are for naught. The one on the right is marked (+) and the one on the left is marked (-).

He got tired of always having to get up and adjust the (-) blind interrupting his show at one point so he finally devised an ingenious plan by extended a long cable to the (-) blind's cord; that way he could now adjust the blinds while still sitting on the couch! Now, being ever a stickler for proper aesthetics in his living room he couldn't just hang this big long cord across his living room so he attached a pulley to a coat rack and pushed it off to the left side of the room. Now he could give it a tug or release some slack to properly balance the blinds.

One day he was outside mowing his front yard and he noticed someone had adjusted the (+) blind (probably the pesky neighborhood kids) and he had to go all the way inside and pull in some cable to get it back right. He found that incredibly inconvenient. So he devised an even smarter plan to extend the cable through the loop of the blind's cord and tied it to his mailbox. Now he could shorten or lengthen the cord from the outside, retie it and adjust his blind from there!

The only problem was, a couple of the wily neighbor kids liked to come play cowboys and indians at his house at times when they knew he'd be watching Rawhide and would probably be much too engrossed in Clint Eastwood's cinematic magic to chase them off. The rope on the mailbox was really too tempting for the young lads to play with so they ended up pulling it in or sometimes letting out some slack and retieing it. But luckily, he had managed a simple way to remedy this problem from the comfort of his own couch and his attitude pretty much became, at that point, "go ahead punk, make my day."

Now, many advancements in technology have been made and I believe the last design to actually use this technique was the classic 741 chip, but the principle still remains the same. ;)

OK, I apologize for the _truly_ convoluted metaphor :) But the length of rope outside is the input voltage, the blind is the (-), the coat rack's position is the feedback resistor size and the length of rope in the house is the output voltage :)
 
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ramancini8

Joined Jul 18, 2012
473
I believe that the statement in the first post is incorrect/misleading, "Whenever an op amp senses a voltage difference between its inverting and noninverting inputs, it responds by feeding back as much current / voltage through the feedback networks as is necessary to keep this difference qual to zero. (V+ - V_ = 0). This rule only applies for negative feedback." While this is a generally accepted statement it is not correct because the gain will drive the output until it nulls the voltage difference. The null or input voltage will almost never be zero volts.

A good lab expirement I used to teach this principle employs an older op amp, UA741 or LM324 in a gain of 10 to 50 configuration. Measure the input voltage ((Vin+)-Vin_)at dc, then at 10Hz, increasing in decade steps to 100KHz. As frequency increases available op amp gain decreases, and the input voltage increases to maintain status quo.
 

Thread Starter

gerases

Joined Oct 29, 2012
186
Thanks, patricktoday for a great analogy. I need to think about it carefully. Thanks for the time it took to think it over too!
 

Thread Starter

gerases

Joined Oct 29, 2012
186
I believe that the statement in the first post is incorrect/misleading, "Whenever an op amp senses a voltage difference between its inverting and noninverting inputs, it responds by feeding back as much current / voltage through the feedback networks as is necessary to keep this difference qual to zero. (V+ - V_ = 0). This rule only applies for negative feedback."
That quote is straight out of "Practical Electronics for Inventors". Let those guys defend for themselves :)

Yesterday I started playing with the first op-amp circuits. I got a successful inverting and non-inverting amplifier going! It was exciting.

I used a 6V dual power supply (from battery packs), a 741, a 1K, a 10K pot (for the Rf) . Instead of the pot, I first used a 5K, and when it worked I decided to put a pot in to see more action on my scope :)

I set the function generator to 0.5V and the frequency to 1 kHz. It was cool to see the output being clipped if it was getting above about 11V peak-to-peak.

At some point I put the input voltage at 1V, turned the pot to get 5V out and started increasing the frequency. Already at 30 kHz, 50 kHz, the output dropped about 1 volt. I thought that wouldn't happen until about 1MHz?

Regardless, I enjoyed it. Hoping to go through many more experiments this week.
 

WBahn

Joined Mar 31, 2012
29,979
Very likely.



Is Vout an independent voltage source or is it a result of Vin? Do they common ground? Sorry, I don't really need to be challenged more than I already am.
You can think of them both as indepedent voltage source -- hell, think of them as two batteries. Yes, they have a common reference point which we can call ground.
 

Thread Starter

gerases

Joined Oct 29, 2012
186
You can think of them both as indepedent voltage source -- hell, think of them as two batteries. Yes, they have a common reference point which we can call ground.
Ok, if two wires, one having one voltage (V1), the other another (V2) and both are referenced to the same ground, than the voltage between them will be V1 - V2?
 

WBahn

Joined Mar 31, 2012
29,979
Ok, if two wires, one having one voltage (V1), the other another (V2) and both are referenced to the same ground, than the voltage between them will be V1 - V2?
Correct. Now, what is the current that is flowing? Be sure to indicate which direction it is flowing. Now, what is meant by this is let's say you come up with some expression of the form I = (x-y)/z (or whatever). If you say that this is the current from from V1 to V2, what this means is that if, after plugging in numbers and coming up with a number as a result, that result is positive that the current is flowing from V1 to V2. If it turns out to be negative, then current is actually flowing from V2 to V1. The point is that you don't have to know what direction the current will actually be flowing, you just need to decide which direction the current will be flowing if that current turns out to be positive.
 

Thread Starter

gerases

Joined Oct 29, 2012
186
Correct. Now, what is the current that is flowing? Be sure to indicate which direction it is flowing. Now, what is meant by this is let's say you come up with some expression of the form I = (x-y)/z (or whatever). If you say that this is the current from from V1 to V2, what this means is that if, after plugging in numbers and coming up with a number as a result, that result is positive that the current is flowing from V1 to V2. If it turns out to be negative, then current is actually flowing from V2 to V1. The point is that you don't have to know what direction the current will actually be flowing, you just need to decide which direction the current will be flowing if that current turns out to be positive.
Hmm, didn't you already answer it for me? :)

(V1 - V2) / R, where R is almost nothing because there are no resistors. Since we assigned no values to either V1 or V2, we can't say whether the I will be negative or positive. Is that correct?

Can this help in understanding negative feedback? I'm not being sarcastic at all. Just curious what you have in mind.
 

WBahn

Joined Mar 31, 2012
29,979
(V1 - V2) / R, where R is almost nothing because there are no resistors.
How are you getting that R is almost nothing. Read it again:

f you have two resistors in series, Ro on the left and Rf on the right, what will the voltage at the junction between them be if a voltage Vin is applied to the left end of the string and a voltage of Vout is applied to the right end of the string.


Vin------Ro-------<v>--------Rf------Vout


What is the voltage at <v>, the junction of the two resistors?
 

Thread Starter

gerases

Joined Oct 29, 2012
186
How are you getting that R is almost nothing. Read it again:

f you have two resistors in series, Ro on the left and Rf on the right, what will the voltage at the junction between them be if a voltage Vin is applied to the left end of the string and a voltage of Vout is applied to the right end of the string.
I messed up completely. Now, that I see the drawing, I can imagine a bit better.

Is going to be: (Vout - Vin) * Rf / (Ro + Rf) ?

It could also be (Vout - Vin) * Ro / (Ro + Rf) if we measure the voltage between <v> and Vin? (going for cover)
 
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