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10v rated diode for clipping stage (safe?)

Started by chromesphere, December 11, 2013, 10:36:59 AM

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chromesphere

Hey, I got some 10v rated germanium diodes, Russian, they look a bit like 1n34a's "detector diode 10va 40ma" is the description. They are rated for 10v max REVERSE voltage.  So im wondering if these would die in clipping section of a pedal or anywhere else, as 10v rating is cutting it fine.
Cheers
Paul
Pedal Parts Shop              Youtube

teknoman2

Did you bought the D9B ones?
I bought the D9L ones and the Vf was 330-340mV
What was the D9B Vf?
Did you measured it?

chromesphere

Yes! that's the one.  I was just looking for the page, here it is:

http://www.ebay.com.au/itm/200x-RARE-D9B-10V-40mA-USSR-Soviet-Military-GERMANIUM-Ge-DETECTOR-DIODE-9-/350946139328?pt=LH_DefaultDomain_0&hash=item51b6051cc0

I also just finished testing them all, all work with a forward voltage drop of between .28 - .3, with a few above and below.
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teknoman2

I don't know if you will face problems using the B ones
but with the L ones I didn't have any issues.
Also they are closer to the 350mV so they are a very good sub for 1n34A.

chromesphere

Yeah thtas what I was hoping to use them for but I got the wrong ones lol I still suspect they will be ok though.  Just have to remember not to use them in circuits that have over 10v...even then their probably still going to be ok, but of course run the risk...
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Cortexturizer

I am pretty embarrassed to say...but I have never tested diode forward voltage drop. Mainly because I order from a reputable supplier and I never had to doubt the diodes [given that I liked the sound of my pedals? haha] but what are your ways of testing forward voltage drop? I don't have a meter that does that automatically mind you. A schem or a method would be greatly appreaciated, thanks.
https://kuatodesign.blogspot.com - thoughts on some pedals I made
https://soundcloud.com/kuato-design-stompboxes - sounds and jams

midwayfair

Paul, the reverse breakdown voltage on a diode is simply the voltage at which it reverse conducts. It's not a "fail point" in the sense that you'll break something. Zener diodes, for instance, are designed have specific reverse conductances like 3.3V, 4.7V, 9.1v, etc.

Since antiparallel diodes in a clipping stage are already conducting at a Fv drop in both directions (almost all the time), you will never, ever see a 10V swing to result in reverse conductance in a distortion pedal ... never mind that you'd need a 24v supply to produce a signal that large without clipping the amplifying device even if you only had a single diode.

Cortexturizer: 9V > 100R > Anode > Cathode. Measure the voltage at the anode. Subtract from the supply voltage. It's approximate. Does your multimeter not have a continuity setting? That's usually also the diode tester. I'm not sure I've ever seen a multimeter for sale that didn't do that, not even the super cheap ones. Even my analog meter has it.

Cortexturizer

Quote from: midwayfair on December 11, 2013, 01:58:20 PM
Does your multimeter not have a continuity setting? That's usually also the diode tester.
If that's true then I might be an imbecile
https://kuatodesign.blogspot.com - thoughts on some pedals I made
https://soundcloud.com/kuato-design-stompboxes - sounds and jams

chromesphere

Quote from: midwayfair on December 11, 2013, 01:58:20 PM
Paul, the reverse breakdown voltage on a diode is simply the voltage at which it reverse conducts. It's not a "fail point" in the sense that you'll break something. Zener diodes, for instance, are designed have specific reverse conductances like 3.3V, 4.7V, 9.1v, etc.

Since antiparallel diodes in a clipping stage are already conducting at a Fv drop in both directions (almost all the time), you will never, ever see a 10V swing to result in reverse conductance in a distortion pedal ... never mind that you'd need a 24v supply to produce a signal that large without clipping the amplifying device even if you only had a single diode.

Thank you so much for the explanation Jon!  I get most of it I think.  I remember reading about zener diodes, they 'open up' when they hit a certain rated voltage then close again when the voltage goes below that rated voltage.  I assume this is why this particular diode is called a 'detector diode'.  So its not a break down voltage as such which is good news! I'm assuming you've suggested a 'opening up' voltage of 24 volts because the guitar single is ac?  I believe the 10v rating is referring to DC.

Glad I asked the question, I think I've learnt something today :D
Paul
Pedal Parts Shop              Youtube

midwayfair

Quote from: chromesphere on December 11, 2013, 10:26:58 PMI assume this is why this particular diode is called a 'detector diode'.

Not quite ... detector as in radio detector.

http://en.wikipedia.org/wiki/Diode_detector

*WARNING this link poses a Wikisurfing danger*

In radios, lower Fv makes for a better detector. Before Schottky diodes were available or in widespread use, Germanium diodes were the way to get a low Fv. The ones with lower Fv were better ... this is part of why 1N270s are more expensive than 1N34As.

The Russians stuck with germanium in their semiconductors longer than western Europe and the U.S., so they tend to have a lot of Ge parts that are particularly good. Their detector didoes (the "D" series, followed by a number [usually 9 or 2] and a letter [A, B, V, G, D, E, J [actually the sound is a y-glide], K]) all have very good Fv, with almost all below .35, and a huge cluster around .27V. I have a pile of them below .2V, but the package is larger so they're a bit awkward to use in stompboxes.

I specified 24V because you have to think in terms of peak-to-peak when discussing the audio signal (the circuit runs on DC, but your audio is still AC! it's just not swinging across 0V but some "fake" 0V, like 4.5V in an op amp :)). In a 9V circuit, your power rails are 9V and 0V, with your signal swinging between them; rarely are you going to see more than 6V on one side of the swing. I suppose 18V is enough to have a 12V+/6V- split with asymmetrical biasing, minus some loss on either side, so 24V might have been an exaggeration.

chromesphere

No problem that makes perfect sense, thanks Jon, your responses are always intelligible!
Paul
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