Differential amplifier with zener diodes

[Femme_physics]

The "-I₁R₅" term in your KVL equation would have to be "-R₅(I₁ + I_z)", since both currents contribute to the voltage drop in R₅.

Really? Despite the fact I1 already contains Iz?

 

[gneill]

Really? Despite the fact I1 already contains Iz?

Okay, I take it back! After reviewing the previous posts I see where you've calculated a value for I1 that is independent of I2.

For some reason I was thinking that your I1 and I2 were concerned with separate subcircuits. That'll teach me to pay closer attention!

 

[Femme_physics]

Okay, I take it back! After reviewing the previous posts I see where you've calculated a value for I1 that is independent of I2.

For some reason I was thinking that your I1 and I2 were concerned with separate subcircuits. That'll teach me to pay closer attention!

Good So my V+ is correct?

And that takes me back to my original idea that I posted today:

http://img703.imageshack.us/img703/6555/1stsj.jpg

http://img189.imageshack.us/img189/6773/2ndzs.jpg

Whereas the only problem is that I have 4 unknowns and 3 loops for I3 and I4.

 

[gneill]

Your value for I₂ looks a bit high, possibly a calculation slip? And since V+ is measured with respect to ground it should be equal to the potential across R₄, not R₃.

For your second figure you also know that V- will be equal to V+, so you can find a Vout that will make this happen. Vout - I₃R₂ = V-. You'll find that I₃ is actually flowing from the 8V zener junction through R₁ and then R₂ towards Vout, and not in the other direction.

 

[Femme_physics]

Your value for I2 looks a bit high, possibly a calculation slip? And since V+ is measured with respect to ground it should be equal to the potential across R4, not R3.

Gotcha, will work on that!

For your second figure you also know that V- will be equal to V+, so you can find a Vout that will make this happen

How do I know it without calculating?

 

[I like Serena]

How do I know it without calculating?

Perhaps you can apply this:

Just apply the following rules:

1. Treat the op-amp as if its inputs + and - are not connected to the circuit.
That is, they draw no current in KCL, and they are not part of any loop in KVL.

2. Set V+=V- and solve using KCL and KVL, unless you find it's impossible, or if Vout would come outside its bounds.

 

[gneill]

How do I know it without calculating?

Usually it is something that is proven in class, and is then used as a rule of thumb for op-amps with feedback from then on.

Do you wish to go through the analysis to show that V- will be driven to be equal to V+?

 

[Femme_physics]

Perhaps you can apply this:

Usually it is something that is proven in class, and is then used as a rule of thumb for op-amps with feedback from then on.

Fair enough, so for every non-comparator V- = V+ ?

Do you wish to go through the analysis to show that V- will be driven to be equal to V+?

I'm tempted to say yes, but I bet it'll be more complex than I think it is.
AT ANY RATE.
http://img20.imageshack.us/img20/3663/thehell.jpg

I'm still having issues. As you can see, my I4 turned up in the minus. And I get my Vout to be 3.8 [V]. Is that right?

 

[gneill]

Fair enough, so for every non-comparator V- = V+ ?

Well, any op-amp circuit with a feedback path from Vout to V-.

I'm still having issues. As you can see, my I4 turned up in the minus. And I get my Vout to be 3.8 [V]. Is that right?

If you keep a few more decimal places in you values of V+ and V- you should be good (the amplification factor for the circuit due to R1 and R2 makes it sensitive to small differences, so keep a few more decimal places in intermediate results).

 

[technician]

Is this question still trying to calculate Vout for the circuit posted at the start or has it become something else??
V at + input = 6.66V (potential divider)
V at - input = 6.66V
V across R into - input = 8 - 6.66 = 1.34
Current through Rin = 1.34/5k current through Rf = 1.34/5k
Voltage across Rf = 10k x 1.34/5k = 2 .68 V
Therefore Vout = 6.66 - 2.68 = 4.00V
This is the way to calculate Vout !

 

[Femme_physics]

Is this question still trying to calculate Vout for the circuit posted at the start or has it become something else??
V at + input = 6.66V (potential divider)
V at - input = 6.66V
V across R into - input = 8 - 6.66 = 1.34
Current through Rin = 1.34/5k current through Rf = 1.34/5k
Voltage across Rf = 10k x 1.34/5k = 2 .68 V
Therefore Vout = 6.66 - 2.68 = 4.00V
This is the way to calculate Vout !

Thanks but I'm trying to verify it using KVL and KCL alone.

If you keep a few more decimal places in you values of V+ and V- you should be good (the amplification factor for the circuit due to R1 and R2 makes it sensitive to small differences, so keep a few more decimal places in intermediate results).

But no matter how many numbers I keep after the decimal point the result of I4 will still turn out minus. Did I do something wrong?

 

[gneill]

But no matter how many numbers I keep after the decimal point the result of I4 will still turn out minus. Did I do something wrong?

No, nothing wrong It just means that Vout is less positive than V-. And as you found out, V- is about 6.667V and Vout about 4.0V. So all's good.

 

[technician]

Sorry FP... I had forgotten that your emphasis was on KVL etc..

 

[Femme_physics]

Technician, outside the realm of KVL and KCL I have no problem to accept your method but I learned two formulas for differential amplifier which seems to result in Vout = 0

Them being

Vout = A (V1 - V2)
A = Rf/R1Are they false? Or am I misusing it? (I used it in my original post)

 

[gneill]

Technician, outside the realm of KVL and KCL I have no problem to accept your method but I learned two formulas for differential amplifier which seems to result in Vout = 0

Them being

Vout = A (V1 - V2)
A = Rf/R1


Are they false? Or am I misusing it? (I used it in my original post)

Careful there, FP. Don't mix up the open-loop gain of the op-amp (A) with the closed-loop gain of a circuit with feedback resistor Rf. Open loop gains for modern op-amp chips are typically on the order of 10⁶. Closed loop gains for a single stage are generally kept less than about 1000 for reasons of circuit stability. More typical are stage gains of about 10 to 100. If more amplification is required, stages are cascaded.

When multiplicative values like a million show up in the equations of circuits it is often akin to dealing with ratios of large numbers in mathematics (limits), where the result as A → ∞ is often some nice finite number with practical application (i.e. a rule of thumb).

For circuits with feedback from Vout to V- we use the rule of thumb that V- = V+ because it can be shown that the large value of A (the open-loop gain of the op-amp), will force them to be very, very close in value, usually closer than measurement error or component tolerances would let you see. If you need convincing on this point we can go through the mathematical demonstration.

While it is certainly possible to analyze every circuit from scratch with KVL, KCL, and so on, it is generally expedient to use the rule of thumb where applicable.

 

[technician]

FP... from outside the realms of KVL and KCL..!
I have attached another circuit diagram with the hope of impressing upon you the close link between Vin, Rin, Rf and Vout and especially the importance of the -input
You are perfectly correct to write A = Rf/Rin. (should be -Rf/Rin...inverting amplifier)
BUT you must be sure that this means the V across Rf = A x the V across Rin.
The -input can be seen as a pivot on a lever. Push Vin down and Vout goes up by a magnified amount (magnification = -Rf/Rin)
The battery or voltage at the + input 'lifts' the pivot (by 2V in my diagram)...
So Vout = -A x (V1 - V2) + V2 ... it is easy to forget to add V2 !
As an example if A = 2 and Vin = +5V then Vout = -2 x (5-2) + 2 = -4V
If Vin = +4V then Vout = -2 x (4-2) +2 = -2V
If Vin = -4V then Vout = -2 x (-4 -2) +2 = +14V (if supply voltage allows)
and so on.
If V2 = 0 ... it usually does ... then you have the straight forward equation:
Vout/Vin = - Rf/Rin The -sign indicates an inverting amplifier.
Hope this does not make life more confusing for you

 

[Femme_physics]

Careful there, FP. Don't mix up the open-loop gain of the op-amp (A) with the closed-loop gain of a circuit with feedback resistor Rf.

Well, they both appear like the same type of loop...I think that would be "open".

http://img3.imageshack.us/img3/149/votn.jpg

You are perfectly correct to write A = Rf/Rin. (should be -Rf/Rin...inverting amplifier)

My circuit is inverting? It's "differential op-amp", neither non-inverting or inverting.I admit I'm a bit confused, but I'll take it step by step. At least I made my final scan.

http://img94.imageshack.us/img94/9074/haiim.jpg

 

[gneill]

The open loop gain of an op-amp refers to the intrinsic gain of the amplifier without feedback to limit it:

The output voltage is then Vout = A(V1 - V2). A can be very high, almost always greater than 10⁵, and typically closer to 10⁶. The actual gain from chip to chip, even in the same batch of chips can vary quite a bit. But it's so large that it usually doesn't matter since op-amps are rarely used open-loop fashion, except as comparators.

When the op-amp is embedded in a circuit with feedback (thus "closing the loop"), the gain of the circuit is "restrained" by the feedback and is much more manageable and can be set precisely by the circuit components in the bias network. Hence your Rf/R1 formula.

 

[I like Serena]

I'd like to emphasize the distinction between open-loop-amplification and feedback-amplification.
$V_{out}=A_{OL}(V_+ - V_-)$ if there is no connection between $V_{out}$ and the inputs of the op-amp. $A_{OL} \approx 10^6$.
In a differential amplifier configuration, you have:
$V_{out}=A_f(V_2 - V_1)$ since there is a connection between $V_{out}$ and the inputs of the op-amp (aka feedback). $A_f = {R_f \over R_1}$.
Note that $V_+ = V_-$ in this case.

 

[technician]

FP
You have written
Vout = A(V2 - V1). Where A = Rf/Rin
This calculates the voltage across Rf !
To get Vout you must add on V+...You did this in your last analysis with the Zener diodes.

Vout = A(V2 - V1) + (V+)
Best wishes

 

[Femme_physics]

Ah...so basically those formulas I wrote don't apply since this is a closed loop gain, yes? Gotcha.Technician, can you write me the formulas you applied to solve it? I don't appear to have them written. Thanks a bunch, you're a world of help

I'd like to emphasize the distinction between open-loop-amplification and feedback-amplification.

The output voltage is then Vout = A(V1 - V2). A can be very high, almost always greater than 105, and typically closer to 106. The actual gain from chip to chip, even in the same batch of chips can vary quite a bit. But it's so large that it usually doesn't matter since op-amps are rarely used open-loop fashion, except as comparators.

Why not? What's the practical difference/advantage of closed loop to open loop?

 

[technician]

I think I have misread the various Vs... I will go through again

 

[Femme_physics]

I think I have misread the various Vs... I will go through again

What do you mean? I got the correct solution by copying your text, but I'm unsure what formulas you've used. I'm just trying to reverse-engineer the logic

 

[technician]

FP
everything is OK The Vout for the differential amplifier is = A(V1-V2) when the ratios of Rf/Rin and R2/R1 are equal.
In my last post I should delete that you must add on V+...that works its way through with the equal ratios. I will go back and edit that post so there is no confusion.
If the ratios are not the same then you cannot use the simplified formula
Vout = A(V1 - V2)

 

[Femme_physics]

Then it's all sorted out. Thanks , everyone!

 

[gneill]

The output voltage is then Vout = A(V1 - V2). A can be very high, almost always greater than 10⁵, and typically closer to 10⁶. The actual gain from chip to chip, even in the same batch of chips can vary quite a bit. But it's so large that it usually doesn't matter since op-amps are rarely used open-loop fashion, except as comparators.

Why not? What's the practical difference/advantage of closed loop to open loop?

Well, as I stated in the next paragraph the greatest advantage is precise control over the circuit gain and thus the circuit's behavior. Imagine how difficult it would be to manufacture something if no two circuits behaved exactly the same way because the gain of the components could not be controlled. You'd probably have to reject half the products that come off the assembly line.

Another reason to "close the loop" is that you can do some clever things by putting more than just resistances in the feedback path. Some powerful filters can be made by placing reactive components (capacitors, inductors) in the loop. You may cover this in your course. It's also possible to make circuits that take square roots or logs of the applied voltage! The op-amp with feedback is the building block of analog computing.