# Diode Formula (Equation)

Discussion in 'General Electronics Chat' started by Livingwaters, Mar 5, 2008.

1. ### Livingwaters Thread Starter Member

Mar 4, 2008
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Hello Group,

What is the standard formula for determining the diode value of any given circuit? (Using Volts @amps to determine values needed)

Please make this as simple as possible to determine. I'm on this site to seek expert advoce. My electronics is pretty basic.

Thanks In Advance for the help!

Livingwaters

2. ### Audioguru New Member

Dec 20, 2007
9,411
896
Diodes are selected for a circuit by their voltage, current, power dissipation and speed ratings.

3. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
You should look at the Ebers-Moll/Shockley equations. They are pretty much the same equations expressed in different terms describing current-voltage relationship of a diode.

Dave

4. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Dave, I think you need to reread his question.

Livingwaters, diodes don't have values, like resistors do. They have part numbers, and each part has a set of specifications such as current rating, breakdown voltage, recovery time, etc.

5. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
The notion of "values" is ambiguous and the OP states merely I and V, where Ebers-Moll defines the I-V characteristic.

The differential of Ebers-Moll is ∂I/∂V - the reciprocal of which at a given voltage will give a resistance at that voltage. Naturally the resistance changes with voltage.

Dave

6. ### scubasteve_911 Senior Member

Dec 27, 2007
1,202
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I suppose if you want to be picky, everything has a part number and you cannot associate anything to an exact value. Ie. A resistor isn't just a component with a value of resistance, it has an associated power dissipation, inductance, thermal noise voltage characteristic, tolerance, and thermal coefficient.

Steve

7. ### Ron H AAC Fanatic!

Apr 14, 2005
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I guess it is ambiguous, but his statement,
led me to believe that Ebers-Moll is probably not what he is looking for. If you're correct, I will apologiz(s)e profusely.

8. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Seems a fair assessment, perhaps clarity is required from the OP. Apologies not required

And its apologise!! (As the lonesome Brit I'd be out-numbered on this one!)

Dave

9. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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OK, Mr. Spock.

10. ### Audioguru New Member

Dec 20, 2007
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Hi Dave,
You are a Brit?

I was in London one time. I remember Picadilly Circus, a foreign language called Cockney and cars driving on the "wrong" side of the road.

11. ### Dave Retired Moderator

Nov 17, 2003
6,960
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I am indeed, but I'm from the "uncultured" north (which is only 200 miles away from London and probably seems like next-door to you guys in the US and Canada, but to us 50 miles is a long way). We drive on the right side, its everyone else that's got it wrong! And I don't understand cockney either!

Dave

12. ### Livingwaters Thread Starter Member

Mar 4, 2008
16
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Hey Guys,

Thanks for all of the input. This is a great forum.

Maybe I am going about this incorrectly. Let me try to explain what we are trying to do.

We are conducting experiments on Internal combustion engines. These experiments are aimed at creating a plasma spark at the electrode of the spark plug while maintaining a constant current at the electrodes through the use of a power inverter, 110 volts @ 20 amps. Now the stock ignition of the car is left intact. The ignition coil output is app. 34,000 volts at 0.83 amps. DC. We are aiming at providing the above conditions, since we are basing our experiments on previous work done in this area. The firing of the 34kV is happening of course as the rotor makes an arc within the dist. cap.

Based on the above, would any of you have any ideas how this can be done without frying the power inverter being used for the 110 volts @ 20 amps AC, and also preventing a back feeding of the 110 v to the dist. ????

Any help in this area is really appreciated!
Livingwaters

13. ### nomurphy AAC Fanatic!

Aug 8, 2005
567
12
If I understand correctly you want to feed the 110V into the spark plug during the 34KV, while not getting the 34KV back into the 110V. And, to do this by using a blocking diode from the 110V into the 34KV.

If so, you would essentially need a diode with a minimum PIV of 35KV. But, 50KV would be safer and more robust.

Try getting ten (SDI) 1N5603JAN, and stacking them:

http://www.surplussales.com/semiconductors/Diodes-1.html

14. ### Livingwaters Thread Starter Member

Mar 4, 2008
16
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Thanks nomurphy for the help. This is what we'll do.

Livingwaters

15. ### nomurphy AAC Fanatic!

Aug 8, 2005
567
12
I suggest you look around on the web for other diodes that are high-voltage and can handle the current you want to pass through the 110V. They're going to be relatively expensive, unless you can find surplus from welders or something (which are usually high-current, more than high-voltage).

16. ### Livingwaters Thread Starter Member

Mar 4, 2008
16
0
Thanks, You have been a Big Help!

Livingwaters

17. ### beenthere Retired Moderator

Apr 20, 2004
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If you get this thing to work, watch for flying parts. The 34 KV @ .83 amps is 28,220 watts. That's just over 37 horsepower of electrical energy going into a cylinder!

18. ### Livingwaters Thread Starter Member

Mar 4, 2008
16
0
Been There,

The 34Kv @ .83 amps is the standard current going to the spark plug. This is what is happening everyday on an older model car.. Modern cars have an ignition system supplying even more volts and higher amps with no "flying parts". On the order approaching 85KV and up. We are aiming for around 24KV @ app 5-6 amps DC half rectified.

I will keep you posted on the progress.

Thanks for your concern. I do appreciate it. We can't be overly cautious when dealing with this amount of electrical power.

Best Regards,
Livingwaters

19. ### scubasteve_911 Senior Member

Dec 27, 2007
1,202
1
Something doesn't really add up with the 34KV@0.83A.. I can only see this happening momentarily for a very short time through a condenser/capacitor. 28KW isn't exactly small potatoes!!! Especially since most batteries can supply about 600A momentarily(most go to starter motor) at 14.4V, which is 8.64KW... I found it odd that rectifiers were picked for this specification as if it were continuous. You can easily get away with much less if it is for a surge, which it must be.

Steve

20. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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I am no expert on ignition systems, but here are some thoughts anyway.
I think that the 34kV is the voltage required to break down the spark gap, and the 0.83A is the current that flows during the "burn time". During that time, the voltage is on the order of 1 to 2kV when firing under compression, but much lower if at atmospheric pressure. Now we're talking about power on the order of 1kW, which is a little (but not much) more believable. However, from what I can gather, the "burn time" is typically 1 to 2mS. This means that energy on the order of 1.5 Joule is being dumped, if that 0.83A is flowing while under compression. Typical Kettering system energy output seems to be in the range of tens of millijoules, from what little info I could find on the Internet. Perhaps the 0.83A was measured at atmospheric pressure, where the voltage was lower, or perhaps even into a short circuit.
Primary energy storage on both Kettering and CDI systems seems to be on the order of 100mJ, give or take. The secondary output will be a fraction of that. This leads me to believe that, as I said above, either the current was measured at very low secondary voltage, or there is a slip in the decimal point (0.083A?).