I thought that maybe I did something wrong,it wouldn't be that hard to make mistakes at the level I am now

You can calculate the adjust range using the resistor values in the voltage divider and the value of the voltage reference.

 

Hello,

Real quick note here.


You see the important relationships between R5 and the three associated transistor and the drive to the lower transistor. In addition, the emitters of the two differential transistors must not be allowed to change too much or it may not act like a differential amplifier anymore.

A design like this requires certain checks and balances which you the designer have to calculate and check. If you can get everything right with the current design then you dont have anything to worry about, but if not, you may have to go to a higher gain transistor, or change the design.

Ok that wasnt as quick as i thought it would be

Very special thanks to Bertus for posting a COMPLETE schematic so we dont have to download and open a pdf file just for one little schematic. It's so easy to post a schematic drawing i dont see why anyone would want to post a pdf unless it contained a lot of additional info or something, but even then it's much nicer to be able to see the schematic right away. Thanks.

 

Snipped the contents of answer ......

This is homework. We're supposed to let the student figure it out as much as possible.

 

This is homework. We're supposed to let the student figure it out as much as possible.

Hi,

What you think that is too much information?
What is too much information to you?
He has to have some kind of hint or he will NEVER figure it out right?
It doesnt make sense to keep saying, "no that's not right, go back and do it again".
But i can delete part of that if you tell me what is too much information you think.
Part of it has already been deleted.

 

What you think that is too much information?
What is too much information to you?
He has to have some kind of hint or he will NEVER figure it out right?

If you read post #18, it's clear that the student doesn't know how the components interact and is guessing:

I guess I should make some computation on the circuit and stop guessing..

It would be far better for the OP to be made to understand component interaction by being given some guidance. You just posted the interaction. Unless you're going to help him do the rest of his homework and help with his tests, you're doing more harm than good. If he's completely clueless, it would be better that he learn from his instructor how the components interact.

Any number of us could have corrected his design, but he's not going to learn that way. That's ignoring the fact that having someone else do his homework is cheating.

 

If you read post #18, it's clear that the student doesn't know how the components interact and is guessing:


It would be far better for the OP to be made to understand component interaction by being given some guidance. You just posted the interaction. Unless you're going to help him do the rest of his homework and help with his tests, you're doing more harm than good. If he's completely clueless, it would be better that he learn from his instructor how the components interact.

Any number of us could have corrected his design, but he's not going to learn that way. That's ignoring the fact that having someone else do his homework is cheating.

Hello,

Sounds to me like you caught the hypercritical bug. Better get that checked out.
I've helped far too many students and others with their circuits and their success spoke for itself when their grades came in.
Why dont you ask some that i have helped.

Of course you dont work out the details, but a hint or two is always a good idea to get them going in the right direction.
What you are talking about is probably too advanced for this student. Their multiple replies attest to this.
But hey, why dont you take it from here and i'll watch.
I wish you all the best.

Oh yeah, BTW, nice work on the schematic.

 

Hello,

The used comparator is basicaly a long-tailed pair differential amplifier.
See this page for some theory about it:
https://wiki.analog.com/university/courses/electronics/text/chapter-12

Bertus

 

I've helped far too many students and others with their circuits and their success spoke for itself when their grades came in.
Why dont you ask some that i have helped.

I've seen a number of your posts in Homework Help where you hand out answers. A lot of students are looking for that type of help (i.e. aren't above cheating).

Of course you dont work out the details, but a hint or two is always a good idea to get them going in the right direction.

Without knowing what the learning objective was, you have no idea whether you were being helpful or not. It could be that the instructor gave a starting circuit that wouldn't satisfy the requirements and wanted to see if the students understood enough about the circuit to address any limitations.

But handing out answers is less likely to ever be helpful.

Homework help often proceeds at a glacial pace because the students who understand the material don't come here.

 

I've seen a number of your posts in Homework Help where you hand out answers. A lot of students are looking for that type of help (i.e. aren't above cheating).

Without knowing what the learning objective was, you have no idea whether you were being helpful. It could be that the instructor gave a starting circuit that wouldn't satisfy the requirements and wanted to see if the students understood enough about the circuit to address any limitations.

But handing out answers is less likely to ever be helpful.

Hello again,

Yes but if you really cared that much you would have PM'd me not say all this right in the thread that contained the questions from the OP.
Also, you seem to be the one that 'decided' that someone 'handed out' the 'answers'. But didnt you correct the schematic?
If you really want to discuss this, then talk in PM's where it wont bother this thread. I'll start a conversation in a minute or two.

 

But didnt you correct the schematic?

I only trimmed and removed white space. Circuit topology and component values were unchanged.

 

I only trimmed and removed white space. Circuit topology and component values were unchanged.

Hi again,

But shouldnt you have waited and let him see that it should have been drawn better rather than you correcting it yourself?
Please talk in PM's about this matter as it will be better for the overall thread. I sent you a PM

 

Mod Note:

If you read a Thread that you consider should be in the Homework Forum or a fellow member who in your opinion is giving too much help to a Student, please Report it to Moderation.

The Moderation team will decide on the appropriate action, please do not discuss another members posts on the open forums.

If you wish to discuss this type of problem with the Moderation team , do so using the Conversation feature.

Eric

 

Hello,

Let's do some calculations on your circuit:



Looking at transistor Q12 : as the base of Q12 is fixed by the zener D1 at 6.8 Volts, what is the current in R3?
You may assume the Vbe = 0.7 Volts.

Bertus

 

Looking at transistor Q12 : as the base of Q12 is fixed by the zener D1 at 6.8 Volts, what is the current in R3?
You may assume the Vbe = 0.7 Volts.

I applied Kirchhoff on the loop formed by D1,Q12 and R3.I calculated the current on R3 and the result of the calculus is 8.97mA.
Then I was thinking of applying Kirchhoff on the loop formed by R8,R10,Q13 and R3.I didn't know exactly how to take the value of the potentiometer R10,I mean when it is set to 0 or 1(I'm thinking '0' means minimum resistance and '1' maximum resistance),so I made the calculations with 2kohms(set to 1) and found the current on the branch with R8 and R10 I=0.4mA and with 0(set to 0) and calculated the current I=0.4533 mA.I will try to find other loops where I can calculate voltages and currents.

 

Hello,

When you read the page given in post #27, you want to have balance between the collector voltages of Q12 and Q13.
You can calculate the voltage on the collector of Q13 using the output voltage and the voltages accross the base-emittors of Q16 and Q17.
When you know the collector voltage of Q13, you can calculate the collector resistors of Q12 and Q13, as the voltage accross these resistors is the input voltage minus the collector voltage.

Bertus