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For the 2nd circuit for silicon diode the forward voltage drop is 0.7-v and for the zener diode Vz = 5-v. We need to draw transfer characteristics curve here.

Thanks for any help.

 

What you are doing is basically going from -100 volts to 100 volts across the input in the first circuit, at a step you choose and analyze the circuit at each step.

In part b, of that first one, you are doing a few cycles of a triangle wave, ramping up to 100 volts and then down to -100 volts at a 1kHz rate.

The second circuit is a clamp and the zener. Do the same thing, say from -20 to 20 volts at the input stepping it at a step rate your comfortable with. Post your results when your done.

 

Originally posted by CrktMan@Dec 1 2005, 02:04 AM
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For the 2nd circuit for silicon diode the forward voltage drop is 0.7-v and for the zener diode Vz = 5-v. We need to draw transfer characteristics curve here.

Thanks for any help.

[post=12112]Quoted post[/post]​

ahhhhhh!

Can anybody give me some concrete answers with numbers?

 

Place some values on all the components if you want concrete numbers.

This is an exercise to see the general shape of the reults for the information given.

The circuit with the zener is going to look something like the zener curve .... except you have that clamp.

The other circuit has offsetting biases ... those other power sources that will affect the Vo.

 

Originally posted by CrktMan@Dec 3 2005, 12:08 AM
ahhhhhh!

Can anybody give me some concrete answers with numbers?

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Only if we get the A.

 

Originally posted by Brandon@Dec 3 2005, 12:45 PM
Only if we get the A.

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I do not have any answer for it, because these two questions are from an exam. If you need to assume any parameters, please do so to solve the circuits or to draw transfer characyeristics.

Thanks.

 

CktMan,

What work have you done to solve these problems?

Can you solve the Vi / Vo transfer curve for the attached problem for Vi ∫ -20 to 20?

 

Originally posted by JoeJester@Dec 6 2005, 12:14 AM
CktMan,

  What work have you done to solve these problems? 

  Can you solve the Vi / Vo transfer curve for the attached problem for Vi  ∫ -20 to 20?

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Hey Geek thanks for the answer!!!

Now the problem you have given to me is: because of zero input voltage, output will be all the time zero, right? correct me if I'm wrong!

~~~~~~~~~~~~~~~~~~~~

Now from my original 2 problems: the 2nd one with the Zener diode and a silicon diode. The zener diode will be in OFF state for positive cycle bcs it needs 5-v (Vz = 5-v given) while the silicon needs only 0.7-v to conduct! Right?
As the problem asks for only dc analysis (am I right?), we can consider the capacitor is open circuit for dc. So, for positive cycle the circuit will be with the input voltage Vi and 2 resistors, meaning the output Vo across the 2nd resistor will be Vo = Vi*R/2R = Vi/2. In the transfer curve for the positive cycle output will delay 0.7-v and will be half of the input, right?

For the negative half cycle: no diode is conducting, so Vo = Vi. One can plot this.

Now the 1st circuit: for the positivecycle (100-v dc) all 3 diodes are conducting. So making diodes short we get a network of 3 loops. Solving the loop that has the Vo, one can get: current I = (45+20-15)/18kΩ = 2.78 mA. Now Vo = -20+ 18*2.78 = 30-V, (right?).
For the reverse cycle I'm totally, entirely lost here!!!

If you guys can help me accurately to get the answers for (a) and (B) for the same circuit, as given in the problem sheet.

The first circuit is very crucial for my exam, plese help!

Thanks.

 

I wanted you to solve the problem for Vi to be from -20 to +20 volts. Plot the transfer curve.

From your words you are on the right track, but the assignment is for you to Draw the transfer curve for your second circuit ... and post it.

I'm sure you'll think of the first circuit's answer before I log on again in 10 or so hours.

 

Originally posted by JoeJester@Dec 6 2005, 03:15 AM
I wanted you to solve the problem for Vi to be from -20 to +20 volts. Plot the transfer curve.

From your words you are on the right track, but the assignment is for you to Draw the transfer curve for your second circuit ... and post it.

I'm sure you'll think of the first circuit's answer before I log on again in 10 or so hours.

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Geek,

I think you are totally wrong! Your circuit has no diodes, so it goes out of this context! Even though your circuit ouput would be an exponentially increasing curve across the capacitor until it reaches 12-V.

You did not make any comments on my deatiled solving expalanations!

 

First off, you can call me Joe. No need to tag a new name to me ...

The transfer curve I'm looking for is the one from Vi and Vo.

I said you were on track with your explaination but your assignment was to draw or sketch the transfer curve.

If you had a simple voltage dividing circuit like the one below, you have a Vi / Vo transfer curve ... and you'll notice there are no diodes in the circuit.

You did a good job thinking about that second problem and once the sketch is done, I'm sure it will be correct. Just think about all the twists and turns you encounter when drawing and what component is causing that turn.

The first problem involves a sawtooth wave .... linear from -100 to 100. Which is nothing more than you going from -100 to 100 in 500 uS and reversing for another 500 uS.

I know this is not the answer you are looking for, but I won't come right out and give you the final answer, but help you arrive at the answer.

 

Originally posted by CrktMan@Dec 1 2005, 01:04 AM
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For the 2nd circuit for silicon diode the forward voltage drop is 0.7-v and for the zener diode Vz = 5-v. We need to draw transfer characteristics curve here.

Thanks for any help.

[post=12112]Quoted post[/post]​

I'll assume the zener is 0.7V when forward biased also.

Notice from 0 to .7 volts both diodes will not conduct. They can be viewed as open circuits. So Vo = Vi. Once Vi gets to .7 the diode will begin to conduct. At this point the equal resistors will divide the input by 2. This will continue until the zener conducts which will clamp the voltage at 5. I think you can see that as the voltage goes negative the zener will be forward biased and clamp the voltage too.

Good luck.

 

Originally posted by aac@Dec 7 2005, 03:04 PM
I'll assume the zener is 0.7V when forward biased also.

Notice from 0 to .7 volts both diodes will not conduct.  They can be viewed as open circuits.  So Vo = Vi.  Once Vi gets to .7 the diode will begin to conduct.  At this point the equal resistors will divide the input by 2.  This will continue until the zener conducts which will clamp the voltage at 5.  I think you can see that as the voltage goes negative the zener will be forward biased and clamp the voltage too.

Good luck.

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Thank you aac and Joe:

I have tried to study a little further on these 2 problems. My latest results are as follows:

For the 2nd circuit with Zener diode I have the Vi/Vo transfer curve as below:

The first circuit a) I have the circuit solution and Vi/Vo curve as given. The first part of the circuit is when didode does not conduct until Vi reaches 45-V. Solving the circuit I got I = (-45+15-20)/(18+18) = 1.39 mA. Vo = -20 +1.39*18 = 5-V. So the slope is tan(5/45), θ = 6.3 degrees.

When Vi exceeds 45-V and reaches 100-v I have the circuit as shown. Solving for current I have I = (45+20-15)/18 = 2.78 mA, and Vo = 30-v, angle θ = tan-1 (30/100) = 16.7 degrees.

During the negative half cycle, only the battery 20-v works for Vo. So Vo = -20-V, as shown.

For the part B) the input (Vi) triangilar wave vs time and output (Vo) vs time curves are shown.

Please leave your opinion whenever you get a chance. Thanks

 

cktman,

The 2nd problem [with the 5 volt zener] is correct. Good job.

The first problem, when you did from -100 to 100, has a minor error. This caused your second answer to have the same error. Review the Vo again.

Hint: Place a meter on each branch ... try to view graphically what each would do from Vi = -100 to 100.

 

Originally posted by CrktMan@Dec 3 2005, 01:08 AM
ahhhhhh!

Can anybody give me some concrete answers with numbers?

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Joe, you mean part (a) or part (B) of the first problem?

I do not have the lab facilities to test that way. Otherwise it would have been nice!

 

CKTman,

My mistake. You are correct on both problems.

Congratulations. There error I was seeing was all mine, I had the wrong potential for battery 3.

Good job.

 

Originally posted by JoeJester@Dec 10 2005, 11:17 PM
CKTman,

My mistake. You are correct on both problems.

Congratulations. There error I was seeing was all mine, I had the wrong potential for battery 3.

Good job.

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Thank you Joe. Much appreciated. I hope and believe the results we have come with are 100% correct! Because this would be crucial for my exam. I did not find any text books that work with the circuit like the first one. But your your assurance make me confident enough now.

 

Cktman,

Attached are the circuit simulations concerning the two questions. You can file it for future reference.

Good work.

 

Originally posted by JoeJester@Dec 11 2005, 09:26 PM
Cktman,

Attached are the circuit simulations concerning the two questions. You can file it for future reference.

Good work.

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WOW!!!
Excellent Joe! What program did you use to draw the circuit and transfer curves?

 

I've been using Tina ... home page: http://www.tina.com

There are others, like:

superspice, at http://www.anasoft.co.uk/

multisim, at http://www.electronicsworkbench.com


I know these programs have student versions.