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Old 6th July 2009, 09:34 PM   #11
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Originally posted by Ty_Bower


What is the typical failure mode for a silicon diode? Open, or short? In other words, should I parallel two diodes for reliability, or wire them in series?
I have not yet wrecked one that was run well within its PIV ratings. Failure of that diode is nothing I'd give much worry to...power tubes are quite another matter.
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Douglas
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Old 6th July 2009, 10:23 PM   #12
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In my experience with small signal stuff, diodes go bad with smoke and fire. Aka - open. But you would still have a really bad smell to investigate !! Then I'd worry about the life expectancy of the other one...

if you parallel them make sure both of them can take the full load. One will win with a better Vf and do all the work.

Sounds like you found yet another tube that will work in the SE... I'm interested in hearing what you think...
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Old 6th July 2009, 10:52 PM   #13
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+1 on the parallel silicon. If one won't do it, then two seem to fail just as easily probably for the reasons Strat mentioned (one will do the work, fail, then the second takes over and also fails). If there is some series resistance in the circuit, then splitting that resistance between the two diodes is probably enough to allow them to share the load a bit better.
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Old 6th July 2009, 11:56 PM   #14
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I have blown up way too much silicon not to have an answer here. A silicon diode will almost always fail to a short first. Then if the short circuit current is high enough, it may blow open. Look at the IXYS diode failures in the Simple SE's. The symptoms were always instantly blown fuses, due to a shorted diode. The same goes for BJT's. Mosfets often fail due to gate oxide damage.

The reasons that a silicon diode will fail are varied. They blow due to over voltage, over current, or over temperature. Often more than one of the above. In the power supply section of a tube amp most diodes are killed by excessive voltage spikes.

The power transformer is a magnetic device. The core is capable of storing considerable energy for brief periods of time. Ordinarilly the energy is transferred from primary to secondary in a controlled manner by application of a sine wave. If the power is interrupted suddenly by turning off the power switch exactly at the peak of the sine wave, the voltage can rise to extremely high levels as the magnetic energy stored in the core gets turned back into electrical energy. This is exactly the same principle used in automobile ignition coils to generate 20,000 volt sparks from a 12 volt battery.

This voltage spike usually contains little energy. In a tube amplifier where there are multiple secondaries all connected to loads, this energy is usually dissipated harmlessly in the tube filaments, leaving little to cause harm. It is still possible for destructive voltage levels to appear on the high voltage secondary. This is directly related to how the power transformer is made and the order in which the windings are placed on the core. It does seem that the current production transformers produce a lot more "spike" energy than older transformers. I have measured 2400 volt spikes on the HV secondary winding of a Hammond 274BX. Where does the 2400 volts go?

All silicon diodes have a reverse breakdown voltage spec. This is usually quite conservative. As the breakdown voltage is exceeded one or more electrons will jump through the depletion region this usually causes an "avalanche effect" leading to a lot of electrons breaching the depletion region. If the current density is high enough the diode will be permanently damaged. Some diodes are designed to operate in the avalanche region. We know them as zener diodes. Other diodes are designed and characterized such that they can withstand a certain amount of energy under avalanche conditions. They are avalanche rated.

FRED diodes are a special type of diode (Fast Recovery Epitaxial Diode) that produce far less noise than conventional diodes so we like to use them in our audio designs. The early ones like the IXYS diode were not avalanche rated, but worked OK until they changed the recipie. The Fairchild "Stealth" diode is avalanche rated and I have yet to see one fail. They absorb the spike energy and dissipate it as heat.

SO.....to sum this up use a Fairchild ISL9R8120P2 (in stock at Mouser) in your tube circuits including the Simple SE and they won't blow up. Use two in series if you really want to be sure.

If you are going to switch a Simple SE over to fixed bias (it's possible I have done it) you won't need a cathode resistor. Put a 100 or 125 mA fuse in series with a 10 ohm resistor where the cathode resistor used to be. The fuse will save your tubes and the 10 ohm allows for bias current measurement.
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Old 7th July 2009, 01:53 AM   #15
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I have blown up way too much silicon not to have an answer here. A silicon diode will almost always fail to a short first. Then if the short circuit current is high enough, it may blow open.
That's been my experience also.
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Old 7th July 2009, 02:19 AM   #16
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Originally posted by tubelab.com
If you are going to switch a Simple SE over to fixed bias (it's possible I have done it) you won't need a cathode resistor. Put a 100 or 125 mA fuse in series with a 10 ohm resistor where the cathode resistor used to be. The fuse will save your tubes and the 10 ohm allows for bias current measurement.
I can take out the cathode bypass caps (C12, C22) too, can't I?
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Old 7th July 2009, 02:32 AM   #17
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Ok, just curious here, but if the current is enough to blow the diode then why wouldn't it be enough to open the short?

Normally I don't investigate the death throws of a diode, but I'm curious... under what conditions would a diode die, but leave a short circuit?

adding to my analog book of knowledge - which is painfully incomplete.
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Old 7th July 2009, 02:35 AM   #18
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Originally posted by oldmanStrat
Ok, just curious here, but if the current is enough to blow the diode then why wouldn't it be enough to open the short?
The diode shorts first, then the fuse blows. You did put a fuse on the hot side of the primary, didn't you?
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Old 7th July 2009, 03:12 AM   #19
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I couldn't wait... I just stuck them in without making any changes. These are some seriously nice tubes. If you happen across a couple, give 'em a try.

Click the image to open in full size.
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Old 7th July 2009, 03:14 AM   #20
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Quote:
Originally posted by oldmanStrat
Ok, just curious here, but if the current is enough to blow the diode then why wouldn't it be enough to open the short?
If memory serves, break-down occurs (secondary breakdown?) at the junction that causes permanent damage to the depletion layer, or something like that. So the first failure mode is for the diode to become a short. Then it goes up in smoke and becomes an open circuit if enough current can pass through it. In a bridge or full wave rectifier, this current path will be through one of the other diodes, destroying it as well. In a power supply, the dead-short on the transformer will hopefully blow the fuse on the other side.
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