# back-to-back zener diode problem

Discussion in 'The Projects Forum' started by studystudystudy, Jun 17, 2010.

1. ### studystudystudy Thread Starter Member

May 21, 2010
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0
the back to back zener diode is used to limiter the voltage across. but i don't understand why resistor is needed here. please see the attachment.

the explaination i found is the resistor is served to impede the return charge drain current rate when one or both zener diode are operational.

please explain how the resistor can impede the return charge drain.

i try to understand it, but still don't understand

your help will be deeply appreciated. thank you

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2. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,402
1,227
Where did you get that schematic? That certainly isn't limiting the voltage.

3. ### BillB3857 Senior Member

Feb 28, 2009
2,402
348
As drawn, the diodes would essentially short the signal source whenever the top lead is positive with respect to the bottom lead of the generator. When the polarity reverses, the diodes would not conduct at all since the breakdown voltage (combined) is only 10 volts. 6 volts 1.414 is only 8.484 volts. In many simulation circuits, a ground must be provided "somewhere" in order for the simulation to run. Usually, one side of the signal source is connected to ground. Can you better clarify the problem and also verify the drawing?

4. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,792
771
that is a really stupid diagram.
The 16VAC is always shorted either way.
The diodes would blow, according to the diagram

Last edited: Jun 20, 2010
5. ### BillB3857 Senior Member

Feb 28, 2009
2,402
348
How do you figure the diodes are shorted either way? With 6VAC (assuming RMS value), one polarity of the signal will indeed forward bias the zeners as in normal diode mode, BUT, when the polarity reverses, it will not have sufficient amplitude to break over the combined 10 volt Zener voltage. What am I not seeing?

6. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,792
771
Oh !!! Shooot..

It's a 6V source. I missed it..sorry

Then my comment is that the 6V is not enuf to bias the zener.
so there won't be a current flowing in it.

7. ### eblc1388 Senior Member

Nov 28, 2008
1,542
102
R!f@@ is not that far off in a real life case.

Just give the circuit enough time to fail in the forward direction.

8. ### BillB3857 Senior Member

Feb 28, 2009
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No biggie, R!f@@ ! ! I thought I had missed something. For sure, I have in the past!

9. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,792
771
Coool...

who would wire that circuit that way any way

{Ed}
it was a typo

10. ### BillB3857 Senior Member

Feb 28, 2009
2,402
348
Like I said in my first post, if it was laid out in a simulation package, such as Circuit Maker, a ground in required "somewhere". Putting a resistor to ground would satisfy the simulator whether it made sense or not. Again, normally, the signal source will have one of its leads grounded. We'll have to wait for clarification from the OP on this one, I'm afraid.

11. ### BillB3857 Senior Member

Feb 28, 2009
2,402
348

True, assuming the source has zero impedance.

12. ### Potato Pudding Well-Known Member

Jun 11, 2010
684
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I would guess that one of the zeners is meant to be turned around. Anode to Anode or Cathode to Cathode.

That would provide something over 5V limiting both ways - except... They will blow up.

A current limiting resistor is normally needed. Load and the Zeners should both be on the other side of a resistor or possibly two, with one for each AC side. I don't see it but possibly it is there as a series resistance in the voltage source.

The ground resistor zener connection that the OP asked about is to drain charge trapping between the Zeners that would work as an unstable capacitative effect. That should not be an issue at 60Hz with normal Zeners. It might be noticed with some of the surge protection types of Zeners that have very wide and very capacitative junctions.

I am stretching for this, so I am only giving myself 50% chance that I have gotten something important badly wrong.