Differential amplifier with zener diodes

[Femme_physics]

I think I'm getting better at this ((thanks to gneill, ILS, ehild and technician I'm quite sure I got it right using the easy methods (formulas). I'm not too sure about my "check up" to my solution using KCL and KVL...I didn't solve the check-up all the way because I wanted to see if my formulas are correct first.

http://img46.imageshack.us/img46/6050/helpwithit.jpg (IGNORE the mini-circuit drawn in (B)!)

 

[technician]

I think you have V+ incorrect. The 10V from that Zener is connected across a potential divider...(5k and 10k)...I am sure you can sort that out !

 

[Femme_physics]

Wait...I do't get it...do you see a mistake?

 

[gneill]

Voltage divider:

 

[technician]

Can you now see what V+ is?...Then what V- MUST be...then...

 

[technician]

The way they have drawn that potential divider is very sneaky. If it had been drawn directly across the zener diode I think you would have seen it

 

[Femme_physics]

According to the simulator I actually got the right answer using the first method (Vout = 4V), my only problem is using the verification

http://img233.imageshack.us/img233/650/checkup.jpg

 

[gneill]

For (III), isn't the path indicated in blue the one you're describing?

If so, what happened to R4?

 

[technician]

This is a strange one FP...you got 4V as Vout, I got 4.1V. There is a nasty coincidence in the numbers which means you got the right answer but by the wrong numbers.
Here are my calculations...see what you think.
1) Potential divider means V+ = 6.7V (I have rounded 6.666 to 6.7)
2) this means V- = 6.7V
3) This means the V across the 5k into the - input = 8-6.7 = 1.3V
4) Current through that 5k = (1.3)/5 mA This also equals current through feedback 10k
6) Voltage across feedback R = 10 x (1.3)/5 = 2.6V
7) left hand end of feedback R ( the - input) voltage = 6.7V therefore Vout = 6.7 - 2.6 = 4.1V

Can you see it ?
Hope this is not judged to be too much help !

 

[gneill]

This is a strange one FP...you got 4V as Vout, I got 4.1V. There is a nasty coincidence in the numbers which means you got the right answer but by the wrong numbers.
Here are my calculations...see what you think.
1) Potential divider means V+ = 6.7V (I have rounded 6.666 to 6.7)
2) this means V- = 6.7V
3) This means the V across the 5k into the - input = 8-6.7 = 1.3V
4) Current through that 5k = (1.3)/5 mA This also equals current through feedback 10k
6) Voltage across feedback R = 10 x (1.3)/5 = 2.6V
7) left hand end of feedback R ( the - input) voltage = 6.7V therefore Vout = 6.7 - 2.6 = 4.1V

Can you see it ?
Hope this is not judged to be too much help !

Keep a few more decimal places in your intermediate calculations (particularly the V+ voltage). Round when the smoke clears

 

[technician]

If you want to set yourself a challenge change resistor to 5k so the potential divider is 5k and 5k in series. If you apply your formula to this you will not get the correct answer.
You got an answer of 4V because the difference between 10V and 8V is 2V and the gain with 10k as feedback resistor and 5k as input resistor = 2. coincidence

 

[technician]

The difference in the 2 answers is not due to rounding off numbers !

 

[gneill]

The difference in the 2 answers is not due to rounding off numbers !

Sure it is. Regardless of whether one was obtained via correct method or not, one answer was 4.0V and the other 4.1V. The difference is 0.1V.

With a few more decimal places kept your result would be 4.0V also, and the difference would be 0V. Of course, this still doesn't address the issue of whether or not an answer was obtained via an incorrect procedure Just sayin', that's all.

 

[technician]

Sorry... we must disagree! we need a 'correct' method. If the 2 numbers are the same as a result of a coincidence there is something wrong.
Change the potential divider to 5k and 5k in series (V+ = V- = 5V)
The output voltage is not 2 x (10-8) (that is the formula that FP used)

 

[gneill]

Sorry... we must disagree! we need a 'correct' method. If the 2 numbers are the same as a result of a coincidence there is something wrong.

Oh no, we are in complete agreement there! I was just pointing out that the difference in the result, 4 vs 4.1, was due to rounding. The fact that the '4' was obtained via an incorrect method is another issue.

Change the potential divider to 5k and 5k in series (V+ = V- = 5V)
The output voltage is not 2 x (10-8) (that is the formula that FP used)

Again I agree.

 

[technician]

I am genuinely pleased about that gneill. This is a horrible question with nice round numbers chosen which make a trap for a student. it is easier to get the correct answer by the 'wrong' method than it is to get the right answer, and any student will learn absolutely nothing.
As you can probably see, FP did not find out the value of V+ (6.66V) and fell into a terrible trap.
I am certain we have the same mind set... I love physics and love teaching, I have made some mistakes in my contributions and what I like about this forum is the variety of approaches to problems... I have learned a lot about Kirchoffs laws (which I have never been comfortable with !)
Cheers
Tech

 

[Femme_physics]

As you can probably see, FP did not find out the value of V+ (6.66V) and fell into a terrible trap.

What? What trap? Hmm...I might see it for myself, but heading off to bed.Also,

http://img233.imageshack.us/img233/7803/16951367.jpg

 

[technician]

FP:
The person who made up this circuit happens to have picked values for resistances that mean you got the correct answer for the wrong reasons... it is not your fault !
You calculated the gain of the amplifier to be Rf/Rin = 10k/5k = 2.
There is nothing wrong with that but what it means is that the voltage across Rf is 2 x the voltage across Rin... it does not tell you directly what Vout is !
You took Vin to be the difference between 10V and 8V = 2V (the zener diodes) and a gain of x2 made you come up with 4V as the Vout.
Can you see that V+ = 6.66V due to the 5k and 10k potential divider??
Therefore V- = 6.66V
Therefore V across Rin = 8-6.66V = 1.34V therefor V across Rf = 2 x 1.34 = 2.68V
The voltage at the left hand end of Rf = 6.66V and therefore the voltage at the right hand end of Rf (Vout) = 6.66 - 2.68 = 3.98V ( I am going to say 4V !)
This is the voltage you got but by a different way. And this is the 'trap' produced by the question maker (probably unintentional)
I imagine this will give you nightmares (when you wake up)

It would have been better for you if the resistance values in this question had been more 'random'

If you are not totally knocked out by this try the calculation with the potential divider across the 10V zener as 5k and 5k
I will do a solution for you and see if we agree. Sleep tight
Cheers
Tech

 

[gneill]

 

[technician]

I agree with
10V - (R3 + R4)*I2 =0
which means that I2 = 10V/(R3 + R4)
The voltage at V+ of the op amp is the voltage across R4 = I2 x R4 = 10V x R4/(R3 + R4)
V+ = 10 x 10/15 for the values in this circuit
i.e V+ = 6.666...=V-

 

[technician]

FP
If your original circuit had been drawn like this do you feel that you would have seen the potential divider that connects to V+ more easily?
(doing this to also see if I have discovered how to include diagrams

 

[Femme_physics]

OK, I amended my mistakes but I still fall short. Sorry it took me so long, was busy with other stuff. I understand the loops and the current splits in my opinion. Oh, and I also got a new laptop with HD webcam, so the pics will turn out clearer


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

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

 

[technician]

Hello FP!
Did you see that V at the +input is not 10V?...it is a fraction of 10 because of that potential divider?

 

[Femme_physics]

Hello FP!
Did you see that V at the +input is not 10V?...it is a fraction of 10 because of that potential divider?

So I can't use this loop? I thought it's a legit KVL loop.

http://img7.imageshack.us/img7/6804/image201112150003.jpg

 

[technician]

There is current flowing through the Zener diode...that is giving the 10V Across R3 and R4

 

[Femme_physics]

But the zener diode is not a part of the loop I chose, why should I include it?

 

[technician]

FP:
I am a little ashamed to admit that I do not find KVL very interesting !
I know it is important but have never needed to resort to using it in op amp appications.
I am certain the usual contributers will be able to give a lot of help when they see your post... I admire and enjoy reading their responses in this area.
You expression looks OK to me because you have realized that some current goes through the
diode. You have I1 through R5 and I2 through R3 and R4.
The current through the diode is therefore I1-I2.
Cheers!

 

[gneill]

But the zener diode is not a part of the loop I chose, why should I include it?

Hi FP. The zener is part of a loop which impinges on the loop you've chosen. As such, it has an influence on your loop. In particular, the current through the zener must come via R5. So R5's total current must include the zener current.

Here's the circuit redrawn to emphasize the loops and their interaction:

The blue line represents your chosen KVL path. Note that what I've indicated as I₁ must be made up of I₂ AND I_z.

 

[Femme_physics]

Yes, that's exactly what I did! I split the currents in my KLV whereas

I1 = I2 + Iz


I don't see how what you say contradicts what I did.

 

[gneill]

Yes, that's exactly what I did! I split the currents in my KLV whereas

I1 = I2 + Iz


I don't see how what you say contradicts what I did.

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.

 

[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?