# Circuits with diodes - II

Discussion in 'Homework Help' started by PsySc0rpi0n, Jun 24, 2015.

1. ### PsySc0rpi0n Thread Starter Senior Member

Mar 4, 2014
1,446
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Morning... I'm back with some problems...

I'm trying to draw Vout for the attached circuit.

I was trying to use the nodal analysis but I don't know how to write the current for the middle vertical branch.
I should consider the ideal model for the diode (short circuit).

If I say that 3 currents exits the Vout node, I can write:

I1 + I2 + I3 = 0

(Vout-V(t))/R1 + ???? + Vout/100 = 0

I just don't know what to place on the ??? of the above equation!

PS: Ignore the 400 ohm of R3 resistor. It's 100 ohm.

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2. ### ericgibbs Moderator

Jan 29, 2010
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hi Psy,
You know the Rate of rise/fall of the Triangular tooth source voltage.
Calculate the Rate/s at the junction of R1 and R3, you are give Vb =5V and its an ideal diode.
So determine the conduction period thru the diode and calc the RMS or Average current over that conduction period.
E
The 1N4007x diode model I have used, has been set for Vfwd very close to 0V.

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3. ### PsySc0rpi0n Thread Starter Senior Member

Mar 4, 2014
1,446
8

Is that all needed to find the output equation in function of Vin?

4. ### WBahn Moderator

Mar 31, 2012
24,553
7,691
It shouldn't be.

All you need to do if determine the voltage transfer characteristic for the circuit, i.e., a plot of Vout vs. Vin, which should be very simple since you've just done the same thing for a more complex multi-diode circuit. Then it's a simple matter applying that characteristic to the Vin signal.

5. ### ericgibbs Moderator

Jan 29, 2010
8,512
1,711
Hi Psy.
Look at the V(vd) plot on the image I posted, that is the voltage transfer function of that circuit, so post the equation which produces that plot.

6. ### PsySc0rpi0n Thread Starter Senior Member

Mar 4, 2014
1,446
8
Ok, I would say that D1 is ON for Vout >= 5.7V and D1 is OFF for Vout < 5.7V.

So,

$
\displaystyle {V_{out} = \left\{\begin{matrix}
\frac{1}{3}\cdot V_{in}\,, V_{in} < 5.7V\\
5.7V\,, V_{in} \geq 5.7V
\end{matrix}\right.}
$

Mr. EricGibbs, I'm assuming that diode is a 0.7V battery!
So when D1 is OFF we have a voltage divider between R1 and R2 for V_in value.

Last edited: Jun 29, 2015
7. ### ericgibbs Moderator

Jan 29, 2010
8,512
1,711
No,
I did post: The 1N4007x diode model I have used, has been set for Vfwd very close to 0V.

8. ### PsySc0rpi0n Thread Starter Senior Member

Mar 4, 2014
1,446
8
Ok, nevermind. It's all the same but instead of Vin < 5.7V should be Vin < 5V and instead of Vin >= 5.7V should be Vin >= 5V, right???

Is the rest correct?

9. ### ericgibbs Moderator

Jan 29, 2010
8,512
1,711
hi Psy,
Consider your transfer function post #56.
Is it possible for you to plot the transfer function, [as per the image in post #52], using pencil and paper, for that particular circuit, using your function.???

What is missing.?

E

10. ### PsySc0rpi0n Thread Starter Senior Member

Mar 4, 2014
1,446
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Well, I've done it again but I can't find the first condition in function of Vin!

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11. ### WBahn Moderator

Mar 31, 2012
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To find the critical value of Vin (the value where the diode just changes from being ON to OFF (or vise versa), pick the operating point for the diode that has it at that critical point where it is both ON and OFF at the same time. Namely, it has 0V across it but ALSO 0A through it. With that constraint, what is the corresponding value of Vin and Vout?

12. ### PsySc0rpi0n Thread Starter Senior Member

Mar 4, 2014
1,446
8

Have you seen what I posted just before your post? I have done it all again in paper. And considering the ideal diode as I have stated at the beginning to try to be consistent.

Then Mr. EricGibbs asked what was missing, and I think he's referring to the fact that is missing a third equation for the descending side of the input sawtooth wave form. But I don't know how to find it because the condition that turns the diode OFF is already stated for the rising part of the input waveform!

Anyway, answering to you, WBhan, that critical point should be Vin = 5V but I think I could also say that it is Vout = 5V.

Mar 31, 2012
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14. ### WBahn Moderator

Mar 31, 2012
24,553
7,691
Yes, I saw what you posted just before my post, where you attached the diagram with the following transfer characteristic at the bottom:

This shows the critical point being at Vin=15V and Vout=5V

15. ### PsySc0rpi0n Thread Starter Senior Member

Mar 4, 2014
1,446
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Ok, so, what's missing there that Mr. EricGibbs was talking about? Was what I was saying about the descending part of the waveform?
And what about the condition I said that I couldn't find? I marked it red on my handmade work!

16. ### ericgibbs Moderator

Jan 29, 2010
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hi Psy,
We have said that you should always check your answer by using result in some way to verify the answer.
The manual plot of post #10 shows its OK.

Vd=5V when Vi/3 => 5V [diode On, conducting]
else
Vd = Vi/3 [diode Off, not conducting]

17. ### PsySc0rpi0n Thread Starter Senior Member

Mar 4, 2014
1,446
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I think you have selected 2 zeners instead of one zener and one diode.

And yes, I have downloaded the whole list of standard.xxx available at LTSpice wiki page, so I have plenty diodes, zeners, etc in my library!

18. ### ericgibbs Moderator

Jan 29, 2010
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Woops!
I posted the same file twice!. I will correct that and repost tomorrow.

You could change one asc sim from zener to diode and re-run it yourself.
E

EDIT:
My advice would be not to use batteries as an equivalent circuit for diode or zener.

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19. ### PsySc0rpi0n Thread Starter Senior Member

Mar 4, 2014
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Sure no problem... I'll change the .asc file.

You say I should not use batteries to replace the diode/zener. Why and what should I use? Regular voltage sources?

20. ### ericgibbs Moderator

Jan 29, 2010
8,512
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OK, I did post an edit.
The reason why not, is because a battery or voltage source will not give the correct response/result in your remodelled circuit.

Why would you want to use a battery or voltage source to replace a zener diode model.?

Last edited: Jul 1, 2015