Rectifier diodes: 4 vs 6.

[leadfootdriver]

I see some amps will use 3 pairs over 2 to get your DC power. Are they any significant sound differences?

[Chuck H]

?

Three pairs of diodes configured how? Can you post circuit examples?

[Enzo]

Like an old Twin Reverb or something, each leg of the PT had three diodes in series.

http://bmamps.com/Schematics/fender/...b763_schem.pdf

[Chuck H]

I thought the same thing based on the title. Then I opened the thread and read "three pairs" and didn't know how to reply. Wouldn't the TR rectifier represent two trios? And I just couldn't think of a rectifier arranged in three pairs.

[Alan0354]

If it it Twin type, I think two in series is plenty. Now a days, you get 1000V diode easily. Get a 3A diode instead of the 1N4007!!! 1N5408 will be my choice. Even if you use one or even bridge rectifier, it should be a whole lot more reliable than 1N4007.

[Merlinb]

Are they any significant sound differences?

No. Diodes may be connected in series to increase their combined withstand voltage, but this has nothing to do with sound.

[teemuk]

Indeed. In 2 vs. 3 diodes series in a rectifier line the increased series resistance and forward voltage drop are in a totally negligible range. What you do gain is increased reverse voltage rating, which increases in much more notable scale.

I do agree though, that these days you can easily just pick a single diode, which has greater ratings overall than a bunch of series connected older diodes that were available in the 60's - 70's.

[R.G.]

More to the point, simply putting diodes in series does not necessarily give you added reverse withstanding voltages. The voltage across each diode must be less than that individual diode can withstand, and that is set by the junction and stray capacitances for changing voltages, and the reverse leakage currents for DC voltage. If they are un-matched enough for a reverse voltage to pop one diode and the others can't take the resulting increase in voltage, they pop in sequence, weakest first, then the next, then...

1000 volt diodes are marginal in tube amps all by themselves. With near-500V supplies, the voltage across a diode in a full-wave-CT circuit is about twice the DC, so a FWCT is right at the edge for 1000V devices. Two 1kV diodes are better if you get any sharing, but a suspicious person like me would put in paralleled high value resistors and small value caps to force sharing if I was using series diodes.

A much simpler thing is to get modern diodes with higher voltage ratings. There are fast+soft recovery diodes like the FREDS easily available at 1200 and greater voltages that will withstand the reverse voltages as well as not needing snubbers to keep radiated diode turnoff noise from being heard as a buzzing hum.

[leadfootdriver]

Thanks for the replies.

I have used 1N5408 diodes in a build, but found that the switching noise in a 1N4007 added favorably to the sound.

That said, I have another amp that does have a lot of buzzy background on noise. Ma be I should put the 5408's in that amp!

[Alan0354]

More to the point, simply putting diodes in series does not necessarily give you added reverse withstanding voltages. The voltage across each diode must be less than that individual diode can withstand, and that is set by the junction and stray capacitances for changing voltages, and the reverse leakage currents for DC voltage. If they are un-matched enough for a reverse voltage to pop one diode and the others can't take the resulting increase in voltage, they pop in sequence, weakest first, then the next, then...

1000 volt diodes are marginal in tube amps all by themselves. With near-500V supplies, the voltage across a diode in a full-wave-CT circuit is about twice the DC, so a FWCT is right at the edge for 1000V devices. Two 1kV diodes are better if you get any sharing, but a suspicious person like me would put in paralleled high value resistors and small value caps to force sharing if I was using series diodes.

A much simpler thing is to get modern diodes with higher voltage ratings. There are fast+soft recovery diodes like the FREDS easily available at 1200 and greater voltages that will withstand the reverse voltages as well as not needing snubbers to keep radiated diode turnoff noise from being heard as a buzzing hum.

Two diodes in series definitely increase the voltage standoff. The way the diode works is it is actually a Zener diode ( with less accurate breakdown voltage). A 1000V diode just guaranty it will not zener until it passes 1000V.

Diode don't just burn the moment you pass the breakdown voltage, it just starting to conduct current like a zener. It's the power dissipation that burn the diode. If you have a second diode in series, even if you think one diode take up all the voltage ( never happens as there's always leakage current), the moment it start to conduct, it will put the voltage across the second diode and prevent the current from rising that destroy the diode. Two diodes in series will extend the breakdown to almost 2000V!!!

We do that all the time in HV design for protection. In fact, we count on voltage standoff sharing. We protect a 5KV supply by stacking 15 transorb ( TVS) diodes in series and it withstand all the stringent test we dished out.


I forgot, diodes in series do not change the sound. The voltage drop is only 0.7V and the on impedance is so low it does not effect the sag. The equivalent series resistance is 25/I ohm. If you conduct 10mA, the resistance is only 2.5 ohm. So you can put 2 diodes in series and it will add less than 10 ohms to the +B. This is much lower resistance than the winding of the PT.

Also, I apologize, For Twin type of PT, you need two 1000V diode is series. I was thinking 2 X 450V = 900V which is within 1000V. But at no load, the voltage can go as high as 560V, that would be 1120V, that will burn the diode if there's only one.

[R.G.]

Two diodes in series definitely increase the voltage standoff. The way the diode works is it is actually a Zener diode ( with less accurate breakdown voltage). A 1000V diode just guaranty it will not zener until it passes 1000V.

Yes - it definitely increases the voltage standoff. The question is then - how much? And what happens if one of them zeners?

It gets very unclear exactly what happens.

Diode don't just burn the moment you pass the breakdown voltage, it just starting to conduct current like a zener. It's the power dissipation that burn the diode. If you have a second diode in series, even if you think one diode take up all the voltage ( never happens as there's always leakage current), the moment it start to conduct, it will put the voltage across the second diode and prevent the current from rising that destroy the diode. Two diodes in series will extend the breakdown to almost 2000V!!!

Again, the issue is - how close? Is it 1900V? 1800? 1200? It's not a silly question. At HV, you can have avalanching that causes hot spots that then retract to a much lower sustaining voltage due to localized heating. When that happens, an unknown amount of the voltage is dropped across the next device. The voltage divides in the ratio of the leakages, not the voltage ratings.

We do that all the time in HV design for protection. In fact, we count on voltage standoff sharing. We protect a 5KV supply by stacking 15 transorb ( TVS) diodes in series and it withstand all the stringent test we dished out.

That's good. However, I can assure you that your designs would be more robust if you used parts to force the sharing. That way variations in leakage currents and internal capacitance do not cause funny things to happen with high and varying voltages.

I forgot, diodes in series do not change the sound. The voltage drop is only 0.7V and the on impedance is so low it does not effect the sag. The equivalent series resistance is 25/I ohm. If you conduct 10mA, the resistance is only 2.5 ohm. So you can put 2 diodes in series and it will add less than 10 ohms to the +B. This is much lower resistance than the winding of the PT.

Also, I apologize, For Twin type of PT, you need two 1000V diode is series. I was thinking 2 X 450V = 900V which is within 1000V. But at no load, the voltage can go as high as 560V, that would be 1120V, that will burn the diode if there's only one.

And it will be a much more reliable design if you put 10M resistors in parallel with each diode and small 5% caps in parallel with that. The resistors swamp out any variations in leakage current and force the DC voltage distribution to be even, and the caps swamp out dynamic variations and force voltage sharing with rapidly changing signals. It's a power version of the frequency compensation done in scope probes.

Of course, a single 1200 to 2000V diode would be good too.

Many HV diodes are actually stacks of diodes internally, but they are matched for leakage and capacitance so they don't go into chain breakdown.

[Alan0354]

Yes - it definitely increases the voltage standoff. The question is then - how much? And what happens if one of them zeners?

It gets very unclear exactly what happens.


Again, the issue is - how close? Is it 1900V? 1800? 1200? It's not a silly question. At HV, you can have avalanching that causes hot spots that then retract to a much lower sustaining voltage due to localized heating. When that happens, an unknown amount of the voltage is dropped across the next device. The voltage divides in the ratio of the leakages, not the voltage ratings.


That's good. However, I can assure you that your designs would be more robust if you used parts to force the sharing. That way variations in leakage currents and internal capacitance do not cause funny things to happen with high and varying voltages.




And it will be a much more reliable design if you put 10M resistors in parallel with each diode and small 5% caps in parallel with that. The resistors swamp out any variations in leakage current and force the DC voltage distribution to be even, and the caps swamp out dynamic variations and force voltage sharing with rapidly changing signals. It's a power version of the frequency compensation done in scope probes.

Of course, a single 1200 to 2000V diode would be good too.

Many HV diodes are actually stacks of diodes internally, but they are matched for leakage and capacitance so they don't go into chain breakdown.

The point is a 1000V diode does not zener until at least 1000V. At the time, it want to conduct current, that will quickly develop voltage across the second diode and take away the excess voltage. This is how it works. I designed bipolar analog IC in the 80s, there is no zener diode!!! It's only a reverse bias of the emitter base junction ( base emitter diode). That's the only zener we had in the IC. All the band gap reference derived from that.

You can count on at least 1900V for two. I would not push it too close.

Also, obviously the old Fender BF and SF using 3 in series, I never see them goes bad, never heard of it goes bad. I don't even know what kind of diodes are those, but they are so old I doubted it's 1KV diodes. It's not wrong to use high value resistor or cap to divide the voltage though.

It is easy to say it's better using robust design........until you are pushing the limit of the technology. We design up to 15KV supply with 2 to 5KV stacking on top. We use TVS stacking to protect the lower voltage supply. There is no TVS that is over 500V at least at the time, we stacked like 15 of them to get protection of 5KV. It was ultra reliable, we sold many systems using this. We tested by arcing the output of the lower voltage supplies to ground while they are floating on top of 15KV. Those are big arc and we arc them like a spark plug for testing. You can smell the ozone generated by the arcing. Never ones any failed.

Over voltage break down is not close to as scary as people think, look at the data sheet, a lot of them spec a certain amount of reverse current at the given standoff voltage. It's the power that burn. Within the chip inside the diode, matching is very good, it's not like you parallel a bunch of small zener diodes and some turn on much faster and carry a lot more current and create a hot spot. I don't design diodes, but it would be very stupid if whoever design the diode not to think of this.

We do 1000V pulsing circuits using 1000V MOSFET. Sometimes we accidentally put more than 1000V and actually saw the MOSFET zener and clamp at about 1000V. They live happily afterwards!!!

In fact I designed a 5KV pulsing circuit that have rise and fall time in about 1uS driving about 70pF. It is a stack of 7 MOSFET of 1000V each. It is clamped by TVS and we count on the TVS turn on to protect the MOSFET. It turn on on every pulse. Now you talk about high switching current dI/dt is extremely high. It's amps of current switching in uS.

For low voltage like guitar amp, and no switching voltage required, you can put resistor divider or cap divider to spread the voltage. In our equipments, the HV power supplies only can source 1 or 2mA, we cannot put divider on it. Also we do HV high speed pulsing, the major power consumption is charging and discharging the capacitance of the coax alone ( 30pF/ft). How can we add caps across just to divide the voltage? We had no choice but to put 100% of the burden on the diode and TVS, they work with flying color.