Differential Amplifier Confusion

[JacobEdward]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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

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

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]

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]

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]

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]

I get that part, I really do

No, really, you don't...

[w2aew]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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.