main bridge has diodes in series

I'm modifying a tube amp. Its existing full-wave bridge for 550V B+ uses 4 pairs of 2 diodes (8 1N4007 diodes total). It seems they used two diodes in series for each leg of the bridge (and a small 10 nanofarad 1000v cap across each diode, mounted right at the diode).

I assume the use of two diodes in series in each leg is to handle voltage spikes? Are there other advantages like any reduction in switching noise, due to the use of two lower-voltage diodes in series instead of one?

How much is the additional voltage drop across the second diode? Could I get a bit more B+ voltage if I used one higher-powered diode in each leg of the bridge, instead of two 1N4007's in series...like if I just replaced the entire mess with a bolt-mounted 'bridge rectifier module' with 4 diodes epoxy-potted in an aluminum module. How much? Of course, I'd put small caps between the terminals again to still suppress the switching noise.

I'm trying to squeeze everything I can out of this power supply.
 
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COnsidering my previous interaction with you, is this a guitar amp?

Two diodes in series will help prevent over-voltage failures. It also helps a lot in protecting the circuit from a shorted diode. If one of a pair of diodes is shorted, you might never know it for years.

How much is the additional voltage drop? Same as for the other diode. MAybe a volt at full current. Is your 500v supply is producing a full ampere? At typical tube B+ currents, probably a little less than a volt will be dropped. SO the difference between single diodes and series pairs will be 550v versus 550.8v. COnsidering that with 120vAC mains, your 550v B+ will vary about 4.5v for every single volt the mains changes. SO if your mains drops 2 volts, your B+ will drop about 9v. SO I think that less than a volt extra diode drop will vanish into the distance. It is a little more than 1/10 of 1%.

Just my opinion, but if you are trying to maximize some amp, this is about the last place I;d look. How about filtration? Messing with screen resistors? Looking to more eficient iron, etc etc.
 

Sorento

Member
2008-03-12 9:59 pm
There is technical reason behind having 2 x 1N4007s in series:
your AC secondary is at least 400Vrms, sine peak voltage 400*sqrt(2)=565v
you have 550v B+ at the cap on the positive terminal of the bridge
when AC swings to -565v, there's 1115v across one leg of the bridge which is already above 4007's rating
and I bet when tubes are still cold both AC and DC are even higher ...
so 2 in series is a necessity rather than just nice to have
 
There is technical reason behind having 2 x 1N4007s in series:
your AC secondary is at least 400Vrms, sine peak voltage 400*sqrt(2)=565v
you have 550v B+ at the cap on the positive terminal of the bridge
when AC swings to -565v, there's 1115v across one leg of the bridge which is already above 4007's rating
and I bet when tubes are still cold both AC and DC are even higher ...
so 2 in series is a necessity rather than just nice to have

Umm...depends on the type of full wave rectifier setup.

The 2 diode / ct secondary type: this is very true.

The 4 diode / "single winding" type will only see a few volts above the rectified output DC.

But for fault tolerance the two diodes is a good idea.

:)
 

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Of course, I'd put small caps between the terminals again to still suppress the switching noise.

I'm trying to squeeze everything I can out of this power supply.
The small caps were not put there to suppress any switching noise, they were put there to swamp the junction capacitance of the diodes, the capacitors form a voltage divider which forces the diode string to share the reverse voltage drop, this is a common arrangement for high voltage rectifiers.
 
The small caps were not put there to suppress any switching noise, they were put there to swamp the junction capacitance of the diodes, the capacitors form a voltage divider which forces the diode string to share the reverse voltage drop, this is a common arrangement for high voltage rectifiers.
Does the reverse current, or leakage current, force voltage sharing?
 
FWIW I used to work for a place that built low current HV power supplies and amplifiers. Nothing approaching the power levels of audio stuff, just lab bench instrumentation. We used a lot of 1N4007s and I've seen a fair number of them fail due to line transients and overloads when somebody shorted an amp briefly. Obviously those are design issues that can/must be addressed but IMO, the part is still a wimp and many designs need more safety margin. For a few extra dimes you can get a much more robust 2-3 amp part and higher voltage.
 

Mark Johnson

Member
Paid Member
2011-05-27 3:27 pm
Silicon Valley
IMO, the (1N4007) part is still a wimp and many designs need more safety margin. For a few extra dimes you can get a much more robust 2-3 amp part and higher voltage.

To name one example: Mouser sells the BY255 (1300V, 3A) for 19 cents in quantity 1; they drop the price to 10 cents if you order quantity 1000 (price and specifications link) .
 
Does the reverse current, or leakage current, force voltage sharing?
No It is the dV/dt across the capacitor which causes a current to flow through the string of capacitors. I= C dV/dt.
Some implementations use resistors, some use both capacitors and resistors and some use none. I used to build these with controlled avalanche diodes and resistors. The resistors were not strictly necessary with controlled avalanche diodes but I needed a bleeder resistor and this served that purpose.
 
Parallel caps are one way to take care of the voltage sharing but this does not offer overvoltage protection. Caps have to have enough capacitance, 0.1uF should be in most cases sufficient. Efficient protection (and no further need to equalize voltage sharing) is offered by using varistors of suitable voltage ratings (MOV).
But by far the easiest way to build a highvoltage stack is to just use a stack of controlled avalanche diodes.
 
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