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Zener diodes in parallel

2/12/2005 11:13 PM
Tim Gagan
Zener diodes in parallel
Howdy folks,  
Does anyone around here know what affect, if any, placing zener diodes in parallel does? I'm trying to build RG Keen's MOSFET tremelo circuit, and his design calls for calls for two 1n5281B's in series. Apparently those are obsolete 100v 1/2 watt (? I'm not sure about the 1/2 watt--RG's diagram doesn't say), so I got what Mouser provides as a substitute: NTE 5060A. Well, those turn out to be 200 volt half watt. The 200 volt part works, but I don't know if the half-watt is enough dissipation. So, I'm wondering if doing some series-parallel-ing will work, or if two 200v zener diodes in parallel still regulates the voltage to 200v.  
PS RG's article can be found at  
scroll down the page and click on 02/05/02 MOSFET Heresies Again!
And now, a word from our sponsors:

2/13/2005 10:38 AM

Parallel zener diodes usually don't work because one will have slightly lower voltage and hog all the current. You could install a resistor in series with each diode to promote current sharing but that sort of defeats the idea of using a zener in the first place. Series zeners all get the same current and small differences in actual voltage just mean slightly different dissipation.  
To figure a zener's dissipation you must know the worst case current going through it and simply multiply by the voltage. As with any semiconductor, some derating may be necessary to insure reliable operation. Try to find a data sheet for the actual part you are using.  
If you can find the part at Mouser or Allied there is usually a link on their websites to a data sheet. If RG is using two 100V 1/2W zeners in series, a 200V 1W should work no problem.
2/13/2005 12:12 PM
Dave H

"Well, those turn out to be 200 volt half watt. The 200 volt part works, but I don't know if the half-watt is enough dissipation."
You can work out the power in a 200V zener. (All the zener current has to pass through the 330k resistor.)  
2/13/2005 12:53 PM
Tim Gagan

Thanks guys.  
About calculating the zener dissipation, will it be the same as the dissipation of that 330K resistor? And will that depend on the current draw of the circuit thatís using the zener-regulated voltage? Or can it be determined simply looking by the voltage drop across that 330K resistor even if thereís no load beyond the zener? If so, then even with RGís max recommended drop of 220 volts, thatís just .15 watt, unless my math is fuzzy (220v divided by 330K ohms = .00066666666w times 220v = .14666652).
2/13/2005 8:25 PM

A zener zenes at its rated voltage. Putting another one next to it doesn't change that. You cannot parallel them to get other voltages. You can make them share current, but what would that get you?  
Mouser lists the 1N4764A, is that no longer available?  
But since you want to series them to get 200v, isn't the 200v NTE what you want anyway? The zener dissipates for itself. The current through the circuit does not go through the zener. You calculate the limiting resistor to set the dissipation through the zener. You figure how much current the zener needs, and how much the circuit it serves needs, then add them and calculate the resistor needed to provide it.  
If you look up 0A2 at the front of the RCA tube book, you wuill find a nice explanation of how to use voltage regulator tubes. These are directly equivalent to zeners and are used the same way.  
It is entirely likely the current through the zener will be greater than the current the circuit needs.
2/13/2005 9:31 PM
Don Symes

As mentioned by Enzo, DaveH and loudthud - diodes (rectfiers, FREDs, LEDS, Schottkys, Zeners, any diodes) don't parallel well.  
You _could_ use 2 660k resistors, 2 1k resistors and 2 zeners if you had to - just don't connect the 2 zener cathodes together.  
In the single-zener case, your power would be ((Your_B+ - Vzener)/330k) * Vzener.  
Using R.G's specified range, worst-case power at 420V becomes ((420 - 200)/330k) * 200 = 133mW.  
With a half-watt Zener, I think you'll be fine.  
2/13/2005 10:44 PM
Tim Gagan

Thanks Don for doing some math here. So it looks like the zener current is the same as the current going through the limiting resistor with no load other than the zener, is that correct?

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