# Infinite voltage on capacitor terminal

Discussion in 'General Electronics Chat' started by earlgrey, Feb 1, 2015.

1. ### earlgrey Thread Starter New Member

Feb 1, 2015
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In this schema :
- Ve is generated by a ne555 astable multivibrator
- Vcc is DC 15 volts alimentation
- Z is a (optional )zener diode

Question 1 ) What explains the infinite Vc voltage ?
Question 2 ) How does the Z zener diode solve the problem ?

Has someone good explanations ( with charges on capacitor terminals, electric fields inside the capacitor, electrons moves, etc... ) ???

( I need this wiring for triggering a ne555 (pin 2 ) from another ne555 ( pin3 ) )

Last edited: Feb 1, 2015
2. ### earlgrey Thread Starter New Member

Feb 1, 2015
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Oscilloscope figure above is wrong, good one is this one, sorry but I cant edit my previous post ?

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3. ### kubeek AAC Fanatic!

Sep 20, 2005
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That is not infinite voltage, it is 2xVcc. Or do you think that the bottom points in your picture are negative infinity?

4. ### earlgrey Thread Starter New Member

Feb 1, 2015
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No -

I would like to physically understand how the Vc potential goes from Vcc to ( as you said ) 2 * Vcc ; what are the charges movements ?

Anyway, thank for having answering.

5. ### kubeek AAC Fanatic!

Sep 20, 2005
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When Ve is at 0V, the capacitor charges to Vcc. When Ve suddenly jumps to Vcc, the capacitor is still having Vcc accross its terminals, so the voltage at Vc is now voltage at Ve plus voltage across the cap, which is Vcc + Vcc.

6. ### MikeML AAC Fanatic!

Oct 2, 2009
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Not quite infinite.

I think this is what you are asking about. The first circuit is a simple differentiator. The second has added a clamp diode, and is the one frequently used to trigger a second 555 from the first.... Note the action of the clamp diode. It prevents the output of the differentiator from going more than one diode drop above the 15V rail, which a Bipolar 555 input might survive, but a CMOS input wont.

Zener not required.

Last edited: Feb 1, 2015
7. ### ScottWang Moderator

Aug 23, 2012
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For a RC differentiator, if the input voltages is 0V~+15V, then the output voltage will be as -15V,0V,+15V, from your attached image that it seems not right, did you shift the voltage?

8. ### MikeML AAC Fanatic!

Oct 2, 2009
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As the TS asked, the top of the resistors is tied to +15V. Think about the differentiator that is used to trigger a 555 where the trigger pulse is longer than the desired period of the 555.

9. ### ScottWang Moderator

Aug 23, 2012
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You said that was the pulse problem, but I was asked the problem of voltage level from your image.

10. ### MikeML AAC Fanatic!

Oct 2, 2009
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What do think will be the voltage at V(out1) and V(out2) if the simulated 555 output stops high (at ~15V)? low (near zero V)?

11. ### ScottWang Moderator

Aug 23, 2012
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You can see the attached waveform, right side has the separated waveform, it could make you more easy to understand, if you still can't find the answer then I will explain for you.

Last edited: Feb 1, 2015
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12. ### earlgrey Thread Starter New Member

Feb 1, 2015
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
@ScottWang : all figures are not exact ( scales etc... ), I just draw approximately what I see on oscilloscope. I do not care to values since my problem is about an infinite one.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Lets get out the discontinuity problem by replacing the square wave with a triangle wave.

On the oscilloscope, I can see the Vc potential growing continuously as the Ve potential is rising.

So if I imagine an infinite rising slope ( we get the square wave back ), the Vc potential immediately jumps high - to 2 * Vcc, like kubeek said , but no reason for it to take an infinite value.

Thus there is no infinite potential, no diode is needed ( like the flyback diode, for inductances ).

13. ### MikeML AAC Fanatic!

Oct 2, 2009
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Earl,

Did you understand the difference between the circuit I posted and the one posted by ScottWang?

Did you understand what the clamping diode does?

Did you notice that I did not use an infinite rise time in the squarewave?

14. ### ScottWang Moderator

Aug 23, 2012
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I was tried to tell you the basic theory of the differentiator circuit, the following is same as yours, but the voltage in the image is not the same with the real 555, the output voltage of 555 should be reduce about 1.4V, so the output voltage of the differentiator should be reducing about 2.8V, the output voltage will be as 1.4V,15V,28.6V, the theory is the same with 11#, but the voltage level is different.

Edit : Modified the output voltage.

Last edited: Feb 1, 2015

Feb 1, 2015
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