Author Topic: Differential Amplifier Confusion (Read 13867 times)

JacobEdward · « **on:** March 22, 2015, 08:47:48 pm »

I'm on the differential amplifier part of this book

http://www.allaboutcircuits.com/vol_3/chpt_8/9.html

and I'm confused about the circuit and wanted to clarify something...

Shouldn't there be a diode in front of the second resistor connected to the V1 line so that V1 doesn't impact the output voltage of the op amp but still allow the output to provide negative feedback? Also, since the negative feedback will be combining with V1, that would mean in order to calculate what the output voltage contributes to the negative feedback, you would need to use something like the Mesh current method for the two voltage sources, right?

T3sl4co1l · « **Reply #1 on:** March 22, 2015, 10:05:53 pm »

A diode doesn't magically block intent.

For the circuit to behave as described, none of the resistors must come disconnected at any time. A diode has a very peculiar sort of behavior, something absolutely undesirable for an amplifier that you want as linear as possible -- it might block under certain conditions (namely, when the voltages are such that the diode becomes reverse biased), but that then disconnects the resistors and the amplifier ceases to be configured as a differential amplifier! Even when forward biased, the diode adds undesirable voltage drop (which varies somewhat with current flow), which would unbalance the amplifier.

The amplifier output is assumed to be a perfect voltage source, so it is not affected by currents into or out of that node; in fact, if you run the analysis, you will find that such a constraint is not even necessary.

Tim

JacobEdward · « **Reply #2 on:** March 23, 2015, 12:05:21 am »

Quote from: T3sl4co1l on March 22, 2015, 10:05:53 pm

A diode doesn't magically block intent. For the circuit to behave as described, none of the resistors must come disconnected at any time. A diode has a very peculiar sort of behavior, something absolutely undesirable for an amplifier that you want as linear as possible -- it might block under certain conditions (namely, when the voltages are such that the diode becomes reverse biased), but that then disconnects the resistors and the amplifier ceases to be configured as a differential amplifier! Even when forward biased, the diode adds undesirable voltage drop (which varies somewhat with current flow), which would unbalance the amplifier.

The amplifier output is assumed to be a perfect voltage source, so it is not affected by currents into or out of that node; in fact, if you run the analysis, you will find that such a constraint is not even necessary.

Tim

I didn't follow what you said or maybe my question wasn't clear from before... I'll try rephrasing it here:

Without something there to ensure V1 will never go through the two resistors (instead of just going through the one resistor), why doesn't V1 produce a voltage directly on the output line by traveling through the two resistors and bypassing the op amp as well as traveling to the op amp? Since there is nothing to prevent that from happening, it seems like you could only calculate this circuit by considering multiple voltage sources that travel in opposite directions through the negative feedback line (using something like the mesh current method) along with multiple voltage sources combining in the same direction on the output line (the voltage produced by the op amp would be enough to give feedback, but then combines with the voltage coming from V1 into the op amp... as in it seems like voltage should be traveling in both directions on the same line at the same time without a diode, which would either mean voltages cancel each other out or they just avoid each other while traveling)... lolz like wtf am I missing?

Hopefully what I said makes more sense...

T3sl4co1l · « **Reply #3 on:** March 23, 2015, 12:17:37 am »

Voltage doesn't "go through" things, it's dropped across them. Current is the through-going property. Resistors drop voltage, given by Ohm's law. So there's no problem, the voltages distribute from V1 to input to output, and same for V2.

Tim

IanB · « **Reply #4 on:** March 23, 2015, 12:22:30 am »

Voltage is a potential difference that occurs between two points.

Voltage doesn't "travel through things" in the sense of something flowing, so any attempt to understand how a circuit works by thinking of voltages flowing through the circuit will lead to failure. Electricity just doesn't work this way.

In the circuit schematic you have referenced, the output voltage is set and controlled by the op amp. Within limits, that voltage is exactly what the op amp sets it to be. There is of course current flowing in the resistors between V1 and Vout, and so in some sense Vout is "affected" by V1. But this is necessary and obvious from the function of the circuit, since Vout = V2 - V1. If Vout was not affected by V1 then this equation could not hold and the circuit could not work.

As you heard from Tim, a diode in the feedback loop of a linear circuit is a terribly, terribly bad idea. A diode is a very non-linear circuit element and putting one in would completely destroy the good linear properties of the system.

JacobEdward · « **Reply #5 on:** March 23, 2015, 12:25:19 am »

Quote from: T3sl4co1l on March 23, 2015, 12:17:37 am

Voltage doesn't "go through" things, it's dropped across them. Current is the through-going property. Resistors drop voltage, given by Ohm's law. So there's no problem, the voltages distribute from V1 to input to output, and same for V2.

Tim

Imagine you just remove the op amp from the circuit, V1 would still effect whatever load is on the Vout line, why doesn't that remain true when there's an op amp? If it does remain true, it would be like putting two voltage sources in parallel...

T3sl4co1l · « **Reply #6 on:** March 23, 2015, 12:33:29 am »

If you remove the op-amp, what is the nature of the voltage at that node?

Open circuit, it has the voltage of V1, yes. But is it an ideal voltage source?

Suppose you put 1mA into the node, and the resistors are 1k. Two resistors in series add = 2k, and so drop 2k * 1mA = 2V. Now instead of V1, it's V1 + 2V. Not a very good voltage source I'd say!

Tim

JacobEdward · « **Reply #7 on:** March 23, 2015, 12:41:55 am »

Quote from: T3sl4co1l on March 23, 2015, 12:33:29 am

If you remove the op-amp, what is the nature of the voltage at that node?

Open circuit, it has the voltage of V1, yes. But is it an ideal voltage source?

Suppose you put 1mA into the node, and the resistors are 1k. Two resistors in series add = 2k, and so drop 2k * 1mA = 2V. Now instead of V1, it's V1 + 2V. Not a very good voltage source I'd say!

Tim

You're still not addressing the initial confusion... why doesn't V1 directly effect the output line when there's an op amp (since there's nothing to prevent flow to go in that direction, like a diode, and you just said V1 would effect the output line when the op amp is removed from the circuit)? Also, if I am understanding this, current is flowing in both directions on the same line: V1 to the output line and op amp providing negative feedback to its input... is that correct?

Hideki · « **Reply #8 on:** March 23, 2015, 12:55:14 am »

It was a strange question. Voltages don't travel across resistors

If anything, current is the thing that travels. Current must travel from V1 through both resistors and into (or out of) the opamp output. For an ideal opamp, no current goes in or out of the inputs.
V2 does the same thing, but the current goes through two resistors to ground instead of the opamp output.

You are missing the point that the opamp actively DRIVES the output based on the difference between the + and - output. Should the opamp output be at the same voltage as V1, there is no "travelling" taking place, since the voltage across the two resistors will be zero. For example, when V1 = 1 volt and V2 = 2 volt: V2 is divided in half by the two resistors, so the + input sees 1 volt. The opamp will drive the output to 1 volt to make the - input be also be at 1 volt.

Maybe it's easier to see with a plain inverting opamp with the + input connected directly to ground. With 1 volt in, the opamp drives the output to -1 volt, so that the midpoint that's connected to the - input sits at 0 volts. There will be a "travelling" current (2 volts across two resistors), but as long as you're within the valid (voltage/current) range for the opamp, the input voltage will not magically travel to the output. The opamp output is typically a low impedance (let's say 1 ohm) and the resistors are typically much higher (let's say 10000 ohms). So unless you send crazy high voltages in, the opamp gets to decide what comes out and not the voltage you send in.

IanB · « **Reply #9 on:** March 23, 2015, 01:03:32 am »

Perhaps it is a good idea to start at the beginning?

http://www.allaboutcircuits.com/vol_1/index.html

Quote from: JacobEdward on March 23, 2015, 12:41:55 am

You're still not addressing the initial confusion... why doesn't V1 directly effect the output line when there's an op amp (since there's nothing to prevent flow to go in that direction, like a diode, and you just said V1 would effect the output line when the op amp is removed from the circuit)? Also, if I am understanding this, current is flowing in both directions on the same line: V1 to the output line and op amp providing negative feedback to its input... is that correct?

Think of the op amp as a regulated power supply. It actively regulates its output voltage until the voltages on the input terminals are equal.

If you have a 12 V regulated power supply and you attach a load like a bulb between the terminals, does the 12 V change? We could say that the load has connected the output to ground and so it is affecting the voltage. Surely the output voltage will be sucked down towards zero by the load? But if you measure the voltage with a meter it probably won't have moved (much). Ask yourself why not? (If we take away the power supply and just have the bulb, then what happens to the output voltage? Why does the power supply make a difference?)

JacobEdward · « **Reply #10 on:** March 23, 2015, 01:17:18 am »

Quote from: Hideki on March 23, 2015, 12:55:14 am

You are missing the point that the opamp actively DRIVES the output based on the difference between the + and - output. Should the opamp output be at the same voltage as V1, there is no "travelling" taking place, since the voltage across the two resistors will be zero. For example, when V1 = 1 volt and V2 = 2 volt: V2 is divided in half by the two resistors, so the + input sees 1 volt. The opamp will drive the output to 1 volt to make the - input be also be at 1 volt.

I get that part, I really do, but imagine you just remove the op amp from the circuit, V1 would still effect whatever load is on the Vout line, why doesn't that remain true when there's an op amp?

w2aew · « **Reply #11 on:** March 23, 2015, 01:20:03 am »

Quote from: JacobEdward on March 23, 2015, 12:41:55 am

You're still not addressing the initial confusion... why doesn't V1 directly effect the output line when there's an op amp (since there's nothing to prevent flow to go in that direction, like a diode, and you just said V1 would effect the output line when the op amp is removed from the circuit)? Also, if I am understanding this, current is flowing in both directions on the same line: V1 to the output line and op amp providing negative feedback to its input... is that correct?

It's all because of negative feedback with the op amp. A *tiny* voltage difference at the input of the op amp causes the output of the op amp to move a lot - in other words, it has a LOT of gain. The end result is that with negative feedback, the op amp's output will "do whatever it can" to keep the two inputs of the op amp at the same voltage. Maybe watching my video on the basics of op amps will help. If the operation of an op amp with negative feedback isn't clear to you, then you're not going to understand the differential amplifier:

Hideki · « **Reply #12 on:** March 23, 2015, 01:21:53 am »

Then you don't get that part, you really don't

Maybe you can think it of this way: The opamp SETS the output to whatever it wants, overriding V1, which is only weakly connected through some resistors. Opamp wins, resistors lose.

IanB · « **Reply #13 on:** March 23, 2015, 01:25:38 am »

Quote from: JacobEdward on March 23, 2015, 01:17:18 am

I get that part, I really do

No, really, you don't...

w2aew · « **Reply #14 on:** March 23, 2015, 01:25:45 am »

Quote from: JacobEdward on March 23, 2015, 01:17:18 am

Quote from: Hideki on March 23, 2015, 12:55:14 am
You are missing the point that the opamp actively DRIVES the output based on the difference between the + and - output. Should the opamp output be at the same voltage as V1, there is no "travelling" taking place, since the voltage across the two resistors will be zero. For example, when V1 = 1 volt and V2 = 2 volt: V2 is divided in half by the two resistors, so the + input sees 1 volt. The opamp will drive the output to 1 volt to make the - input be also be at 1 volt.

I get that part, I really do, but imagine you just remove the op amp from the circuit, V1 would still effect whatever load is on the Vout line, why doesn't that remain true when there's an op amp?

Without the op amp, V1 will affect the output simply because it is connected through resistors. With the op amp, the output voltage is set by the op amp's low impedance output stage, and the negative feedback is causing the op amp to adjust the output voltage until the voltages at the op amp inputs are basically equal.

Your question is similar to: "If a wheel on a car rolls uphill when the car is driven that way, why doesn't the wheel roll uphill when you remove the rest of the car?"

JacobEdward · « **Reply #15 on:** March 23, 2015, 01:26:00 am »

Quote from: IanB on March 23, 2015, 01:03:32 am

If you have a 12 V regulated power supply and you attach a load like a bulb between the terminals, does the 12 V change? We could say that the load has connected the output to ground and so it is affecting the voltage. Surely the output voltage will be sucked down towards zero by the load? But if you measure the voltage with a meter it probably won't have moved (much). Ask yourself why not? (If we take away the power supply and just have the bulb, then what happens to the output voltage? Why does the power supply make a difference?)

I have started at the beginning, I'm not sure you're understanding the question... imagine you just remove the op amp from the circuit, V1 would still effect whatever load is on the Vout line, why doesn't that remain true when there's an op amp? If it is true, then you have to add whatever output is driven from the op amp with V1 after traveling through the resistors like it would if the op amp was removed.

IanB · « **Reply #16 on:** March 23, 2015, 01:30:53 am »

Quote from: JacobEdward on March 23, 2015, 01:26:00 am

I have started at the beginning, I'm not sure you're understanding the question... imagine you just remove the op amp from the circuit, V1 would still effect whatever load is on the Vout line, why doesn't that remain true when there's an op amp? If it is true, then you have to add whatever output is driven from the op amp with V1 after traveling through the resistors like it would if the op amp was removed.

Consider what Alan just said in a previous post. You have a car, it can drive down the road at 60 mph. Imagine you just remove the engine from the car, the wind resistance will slow the car down. So why doesn't that remain true when there is an engine?

w2aew · « **Reply #17 on:** March 23, 2015, 01:31:59 am »

Quote from: JacobEdward on March 23, 2015, 01:26:00 am

Quote from: IanB on March 23, 2015, 01:03:32 am
If you have a 12 V regulated power supply and you attach a load like a bulb between the terminals, does the 12 V change? We could say that the load has connected the output to ground and so it is affecting the voltage. Surely the output voltage will be sucked down towards zero by the load? But if you measure the voltage with a meter it probably won't have moved (much). Ask yourself why not? (If we take away the power supply and just have the bulb, then what happens to the output voltage? Why does the power supply make a difference?)

I have started at the beginning, I'm not sure you're understanding the question... imagine you just remove the op amp from the circuit, V1 would still effect whatever load is on the Vout line, why doesn't that remain true when there's an op amp? If it is true, then you have to add whatever output is driven from the op amp with V1 after traveling through the resistors like it would if the op amp was removed.

One more time... ...because the negative feedback of the op amp is causing the effect of V1 to be cancelled out at the inverting input of the op amp. In other words, the current flowing through the input resistor will be equal to the current in the feedback resistor - and the voltage at the inverting input of the op amp will be equal to the voltage at the non-inverting input (which is determined by a simple voltage divider).

JacobEdward · « **Reply #18 on:** March 23, 2015, 01:34:26 am »

Quote from: IanB on March 23, 2015, 01:25:38 am

Quote from: JacobEdward on March 23, 2015, 01:17:18 am
I get that part, I really do

No, really, you don't...

Could you explain why the op amp would completely block the effects V1 would have on the load the op amp (the effects that would be there if the op amp wasn't there)? The op amp is just amplifying it's output to make the two terminals equal, it's not determining what happens to the load... just the output pin and the negative feedback line.

JacobEdward · « **Reply #19 on:** March 23, 2015, 01:36:50 am »

Quote from: IanB on March 23, 2015, 01:30:53 am

Consider what Alan just said in a previous post. You have a car, it can drive down the road at 60 mph. Imagine you just remove the engine from the car, the wind resistance will slow the car down. So why doesn't that remain true when there is an engine?

It seems like that is a false analogy because there is still V1... you wouldn't be removing the whole engine, just the jet turbine you attached to the hood of the car...

w2aew · « **Reply #20 on:** March 23, 2015, 01:38:29 am »

Quote from: JacobEdward on March 23, 2015, 01:34:26 am

Quote from: IanB on March 23, 2015, 01:25:38 am
Quote from: JacobEdward on March 23, 2015, 01:17:18 am
I get that part, I really do

No, really, you don't...

Could you explain why the op amp would completely block the effects V1 would have on the load the op amp (the effects that would be there if the op amp wasn't there)? The op amp is just amplifying it's output to make the two terminals equal, it's not determining what happens to the load... just the output pin and the negative feedback line.

This is where you're mistaken. The op amp DOES determine what happens at the load.

IanB · « **Reply #21 on:** March 23, 2015, 01:39:55 am »

Quote from: JacobEdward on March 23, 2015, 01:34:26 am

Could you explain why the op amp would completely block the effects V1 would have on the load the op amp (the effects that would be there if the op amp wasn't there)? The op amp is just amplifying it's output to make the two terminals equal, it's not determining what happens to the load... just the output pin and the negative feedback line.

Like I said, the op amp is a power supply. It's not just moving some voltage around, it is actively adding or removing power in the system to force things to be the way it wants them to be. It's got muscles.

Go back to the car analogy. How can the cruise control in a car completely block the effects of wind resistance or gradients? After all, it is just adjusting the fuel flow to make the speed equal the cruise control setting...

IanB · « **Reply #22 on:** March 23, 2015, 01:43:27 am »

Quote from: JacobEdward on March 23, 2015, 01:36:50 am

It seems like that is a false analogy because there is still V1... you wouldn't be removing the whole engine, just the jet turbine you attached to the hood of the car...

Does your car have a jet turbine attached to the hood?

JacobEdward · « **Reply #23 on:** March 23, 2015, 01:47:11 am »

Quote from: w2aew on March 23, 2015, 01:31:59 am

because the negative feedback of the op amp is causing the effect of V1 to be cancelled out at the inverting input of the op amp. In other words, the current flowing through the input resistor will be equal to the current in the feedback resistor

So if I understand this correctly, the current from the negative feedback cancels any current flowing from V1 to that negative feedback line?

JacobEdward · « **Reply #24 on:** March 23, 2015, 01:49:41 am »

Quote from: IanB on March 23, 2015, 01:43:27 am

Quote from: JacobEdward on March 23, 2015, 01:36:50 am
It seems like that is a false analogy because there is still V1... you wouldn't be removing the whole engine, just the jet turbine you attached to the hood of the car...
Does your car have a jet turbine attached to the hood?

That's not really helping... sure, if in this analogy you have two sources of power like V1 and the op amp, then you have to add another power source that is vastly more powerful than the other, which would be something like a jet turbine and a internal combustion engine.

Yes I have a jet turbine on my car, it's fucking badass too.

Hideki · « **Reply #25 on:** March 23, 2015, 01:53:15 am »

Quote from: JacobEdward on March 23, 2015, 01:34:26 am

Could you explain why the op amp would completely block the effects V1 would have on the load the op amp (the effects that would be there if the op amp wasn't there)?

Because the effects of the opamp is like connecting a power supply directly to the load. The power supply does its best to maintain a certain voltage and basically ignores the V1 you connect to it through the resistors.

Quote

The op amp is just amplifying it's output to make the two terminals equal, it's not determining what happens to the load... just the output pin and the negative feedback line.

It completely determines what happens to the load. And in this case, since nothing else is connected, the resistors back to V1 _IS_ in fact the load.

dannyf · « **Reply #26 on:** March 23, 2015, 01:54:15 am »

think of any (ideal) opamp as a gain block with two important attributes:

1) it has infinite gain;
2) its input terminals have infinite impedance so no current flows into them.

Take that to your circuit and you can quickly work out the math that links the output to the inputs.

Conceptually, the opamp will affect the output so that the potential on the inverting input is the same as that on the non-inverting input. aka it has isolated V1 from the output.

JacobEdward · « **Reply #27 on:** March 23, 2015, 01:56:49 am »

Quote from: Hideki on March 23, 2015, 01:53:15 am

Quote
The op amp is just amplifying it's output to make the two terminals equal, it's not determining what happens to the load... just the output pin and the negative feedback line.

It completely determines what happens to the load. And in this case, since nothing else is connected, the resistors back to V1 _IS_ in fact the load.

Thanks, I was using the word 'load' incorrectly, but just to confirm, does that mean the feedback resistor would always be considered the load, just on a parallel line? Does load just mean whatever is attached to the output pin of the op amp in this case?

w2aew · « **Reply #28 on:** March 23, 2015, 02:02:00 am »

Quote from: JacobEdward on March 23, 2015, 01:47:11 am

Quote from: w2aew on March 23, 2015, 01:31:59 am
because the negative feedback of the op amp is causing the effect of V1 to be cancelled out at the inverting input of the op amp. In other words, the current flowing through the input resistor will be equal to the current in the feedback resistor

So if I understand this correctly, the current from the negative feedback cancels any current flowing from V1 to that negative feedback line?

The current is the same in both resistors because the op amp input draws neglible current. The op amp drives the output voltage (and whatever load is connected to it) such that the voltage at the (-) input equals the voltage at the (+) input.

Hideki · « **Reply #29 on:** March 23, 2015, 02:09:55 am »

Quote from: JacobEdward on March 23, 2015, 01:56:49 am

Thanks, I was using the word 'load' incorrectly, but just to confirm, does that mean the feedback resistor would always be considered the load, just on a parallel line? Does load just mean whatever is attached to the output pin of the op amp in this case?

Well, yes. The value of the feedback resistor does matter in real circuits. A real opamp can't push unlimited current, so any real circuit needs to take into account not only the load it's driving but also the current going through the feedback resistor (and through whatever the feedback resistor connects to).

Ideal opamps have no such limitations of course. They are perfect beings that can output unlimited current without the output voltage moving the slightest.

albert22 · « **Reply #30 on:** March 23, 2015, 03:10:56 am »

I think that you need to review the inverting a amplifier configuration for an op amp. Be able to understand how it works and derive the equations. Also the non inverting. And then go for the differential.
Here is another explanation that seems more clear than the tutorial that you are following:
http://www.electronics-tutorials.ws/opamp/opamp_2.html

JacobEdward · « **Reply #31 on:** March 23, 2015, 04:42:15 pm »

Quote from: w2aew on March 23, 2015, 02:02:00 am

The current is the same in both resistors because the op amp input draws neglible current. The op amp drives the output voltage (and whatever load is connected to it) such that the voltage at the (-) input equals the voltage at the (+) input.

I dont understand why just ensuring the two inputs are the same will negate the effects V1 has on the load (the effects it would have had if you removed the op amp from the circuit and it was just one series line from V1 to the load... imagine there's a 10kohm resistor on the Vout line that is connected to ground that will act as the load, that would mean without the op amp, V1 is connected to three resistors before reaching ground).

Are you saying the output of the op amp is canceling the effect of V1 through the negative feedback line (as in a current of 5A going left will combine with a current of 5A going right to make 0A or 4A going left combined with 5A going right will equal 1A going right)?

That doesn't make sense since, even though any subtraction in current through the negative feedback line will be added to the input from V1 creating the proper voltage level, there is still nothing that would prevent the output from the op amp to continue up the V1 line, thereby subtracting any current coming from V1, which makes V1 useless...

Why does V1 suddenly not have an effect on the load with an op amp there and no diodes, it's still connected to the load in parallel with the op amp?

IanB · « **Reply #32 on:** March 23, 2015, 05:44:55 pm »

Quote from: JacobEdward on March 23, 2015, 04:42:15 pm

Why does V1 suddenly not have an effect on the load with an op amp there and no diodes, it's still connected to the load in parallel with the op amp?

V1 does have an effect on the output. In mathematical terms we have:

Vout = V2 - V1

Or in other words,

Vout = f(V2, V1)

This says that the output voltage is a function of the input voltages. If the input voltages did not affect the output voltage the whole circuit would not work!

Hideki · « **Reply #33 on:** March 23, 2015, 05:49:57 pm »

V1 has no direct effect on the load because no current passes from V1 to the load. It all goes into (or out of) the opamp.

Assuming all resistors being 1k and 0 volt at the noninverting input, then V1 = 1 volt makes the opamp output -1 volt. The current from V1 into the opamp output is 1 - (-1) = 2 volt divided by 1k + 1k = 2k. 2V / 2000 ohm = 1 mA. That current goes from V1, through the resistors, into the opamp output and comes out of the negative supply pin.

This is without any additional load.

If you add a load from the opamp to ground, say a 1 kohm resistor, then, since the output is still -1 volts, the load will force the opamp to drive one more milliamp to the negative supply.
If you add a load from the opamp to ground, say a 1 ohm resistor, then, since the output is still -1 volts, the load will force the opamp to drive one more amp to the negative supply.

JacobEdward · « **Reply #34 on:** March 23, 2015, 07:08:46 pm »

Quote from: IanB on March 23, 2015, 05:44:55 pm

V1 does have an effect on the output. In mathematical terms we have:

Vout = V2 - V1

I dont understand why just ensuring the two inputs are the same will negate the effects V1 has on the load (the effects it would have had if you removed the op amp from the circuit and it was just one series line from V1 to the load... imagine there's a 10kohm resistor on the Vout line that is connected to ground that will act as the load, that would mean without the op amp, V1 is connected to three resistors before reaching ground)

?? Load Resistor = (V1-voltage drop for input resistor+voltage drop for feedback resistor)+Op Amp output Voltage ??

V2 is not directly connected to the line connected to the Load Resistor (since it only goes to the Op amp and ground) so it wouldn't cause an effect.

IanB · « **Reply #35 on:** March 23, 2015, 07:15:26 pm »

Quote from: JacobEdward on March 23, 2015, 07:08:46 pm

I don't understand why...

Frankly, you are not going to understand why, today, no matter how much anyone tries to explain it to you.

You will need to mull over the problem for a few weeks, or months, or even years until eventually it clicks. This is how learning works. Failure of understanding is caused by knowledge gaps in other areas, leaving holes in the picture. Eventually those other holes will be filled as you learn more, and then you will wonder how you ever could have been puzzled by this. But I don't think it is going to happen today, or you would have seen the answer already.

JacobEdward · « **Reply #36 on:** March 23, 2015, 09:23:09 pm »

Quote from: Hideki on March 23, 2015, 05:49:57 pm

V1 has no direct effect on the load because no current passes from V1 to the load. It all goes into (or out of) the opamp.

Why? I've attached a picture of the path I think current travels without diodes to prevent flow... could you explain why that is incorrect (preferably visually since we seem to be getting nowhere with just text).

JacobEdward · « **Reply #37 on:** March 23, 2015, 09:24:18 pm »

Quote from: IanB on March 23, 2015, 07:15:26 pm

Quote from: JacobEdward on March 23, 2015, 07:08:46 pm
I don't understand why...
Frankly, you are not going to understand why, today, no matter how much anyone tries to explain it to you.

You will need to mull over the problem for a few weeks, or months, or even years until eventually it clicks. This is how learning works. Failure of understanding is caused by knowledge gaps in other areas, leaving holes in the picture. Eventually those other holes will be filled as you learn more, and then you will wonder how you ever could have been puzzled by this. But I don't think it is going to happen today, or you would have seen the answer already.

Is there a way to block profiles on this forum? If not, can you simply choose to not comment next time... it's not my fault you're so bad at explaining how these concepts work...

JacobEdward · « **Reply #38 on:** March 23, 2015, 09:41:01 pm »

Quote from: Hideki on March 23, 2015, 05:49:57 pm

That current goes from V1, through the resistors, into the opamp output and comes out of the negative supply pin.

I've attached a picture of what I am visualizing when I read this... let me know if I am correct

JohnnyBerg · « **Reply #39 on:** March 23, 2015, 09:51:07 pm »

Quote from: JacobEdward on March 23, 2015, 09:41:01 pm

if I am correct

No

JacobEdward · « **Reply #40 on:** March 23, 2015, 09:53:13 pm »

Quote from: JohnnyBerg on March 23, 2015, 09:51:07 pm

Quote from: JacobEdward on March 23, 2015, 09:41:01 pm
if I am correct

No

That was just so mind boggling helpful I am so glad you decided to take the time to post that response... like seriously, what would I do without you

katzohki · « **Reply #41 on:** March 23, 2015, 09:55:31 pm »

If V- is greater than Vo, current flows from V- to Vo. If V- < Vo current flows from Vo to V- ...

JohnnyBerg · « **Reply #42 on:** March 23, 2015, 09:58:19 pm »

Quote from: JacobEdward on March 23, 2015, 09:53:13 pm

That was just so mind boggling helpful I am so glad you decided to take the time to post that response... like seriously, what would I do without you

You're welcome. Any time again.

katzohki · « **Reply #43 on:** March 23, 2015, 10:04:58 pm »

I'm hoping this will help you. Here is an analogy of how an opamp works:

From wikipedia (http://en.wikipedia.org/wiki/Operational_amplifier) if you read that article it might help.

Here's a look at the internal circuitry of an opamp:

From http://www.play-hookey.com/analog/op_amps/inside_741.html

The opamp is considered an active component and is actually capable of "supplying" voltage at its output, unlike inactive components (resistors, capacitors, inductors) which can't.

Hideki · « **Reply #44 on:** March 23, 2015, 10:20:27 pm »

No current flows out of the INPUT. The current flows out the negative SUPPLY connection.

So in the image right above (inside_741) it goes into pin 6 (marked OUT), through the 50 ohm resistor below, through the PNP transistor and out pin 4 (V-).

katzohki · « **Reply #45 on:** March 23, 2015, 10:22:06 pm »

OK I think I understand where your confusion is coming from. In your drawings you have drawn on the opamp the + and - input terminals and the output terminal (and that's not incorrect), but not the voltage supply terminals on the opamp. And so everyone is telling you that the opamp can supply voltage, but you must be thinking "how? Where is the voltage coming from?" Well, it comes from the power supply inputs on the opamp.

TimFox · « **Reply #46 on:** March 23, 2015, 10:24:26 pm »

Re-stating slightly what was mentioned in several replies:

For an ideal op-amp,

connected to appropriate DC power supplies
with input voltages at the differential input stage within the usable voltage range
with output voltage and current within the capability of the amplifier and power supplies
and with an appropriate DC feedback network

The amplifier will adjust its output voltage so that the two input terminals are at the same voltage. The current required out of the output pin comes from the power supplies connected to the op amp.

If you remove the op amp from its socket, or disconnect the DC power supplies, this effect will not happen and the output voltage will be different.

w2aew · « **Reply #47 on:** March 24, 2015, 02:33:19 am »

Quote from: JacobEdward on March 23, 2015, 09:41:01 pm

Quote from: Hideki on March 23, 2015, 05:49:57 pm
That current goes from V1, through the resistors, into the opamp output and comes out of the negative supply pin.

I've attached a picture of what I am visualizing when I read this... let me know if I am correct

The diagram is not correct. It shows current flowing into the op amp inverting input, and that is wrong. Essentially no current flows in/out of the op amp inputs. This has been clearly stated (and requoted by you) in several posts.

Of course V1 affects the output, but it affects it because it causes the op amp to react in such a way to adjust the output voltage until the two op amp inputs are equal.

We all understand your question. Several have given you perfectly accurate answers to your question. Maybe the piece of the puzzle you're missing is that the op amp output is a low impedance, thus can override any direct affect that V1 has on the output.

mrkev · « **Reply #48 on:** March 24, 2015, 04:19:38 am »

Hi, not that I'm positive that you'll actually get it after you've dismissed everybody that tried to help, but it might help someone else. So I drew the picture of how this actually works with an ideal oamp.

The voltage between - and + input pins of OP is 0V, due to the negative feedback. Currents to both + and - pins are also 0A, they act only as "voltage sense". OA is "balancing" output voltage so that the voltage at the node between R1 and R2 at the top branch is equal to the voltage at the node between R1 and R2 at the bottom (that is literary the same thing as the first sentence).

In other words, your problem with current from V1 is dealt with by OA.

And btw, i drew it without any load (load would be at U_O terminals), so I1 = - I_oamp (it flows to the OA).
With load connected, the OA would provide (or sink) exactly as much current, that the voltage between + and - inp. would still be 0V.


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