Bypassing rectifier diodes with parallel capacitors

[Gary_S]

This is just to see if anyone can explain why using capacitors in parallel with ss rectifier diodes encourages equal voltage sharing among the diodes.

Again this is from Merlin's power supplies book in the rectifier section.

I'd just like to understand how it happens. Merlin says the tiny amount of leakage current through the cap causes it, but how? 

[jazbo8]

It may be easier to think of the circuit as a capacitor voltage divider, if the capacitance are equal then the voltage across each is the same, thus ensuing each diode also sees the same voltage. It is also common to use the capacitor with a resistor to form a snubber network that can get rid of the ringing caused by the diode switching, which the silicon diodes like the 1N4007 are well known for.

[Ed_Chambley]

Here is an example I have tried.  Truthfully, I did not notice any difference, but some of the guys at the Metro forum swear by it.

[Gary_S]

Thanks for the replies folks.

ED: Man! that's a pretty complicated looking rectifier with the cap and the resistor along with the diode. It'd be interesting to try that out. Thanks for putting it up.


jazbo8: That's a good way of thinking about it and makes it a bit clearer to picture in the mind's eye. I was just thinking about when the current is coming from the power supply and then hitting the diode/capacitor junction; how is the current dividing up there between the diode and the cap? I know about capacitive reactance and the cap will charge and discharge, but i'm trying to not think about that here and i'm trying to think about what component get's the current flowing through it and whether it divides up equally.

[eleventeen]

"I know about capacitive reactance and the cap will charge and discharge, but i'm trying to not think about that here and i'm trying to think about what component get's the current flowing through it and whether it divides up equally."

My view, and of course I have not empirically tested things to destruction; is that these schemes are directed at the notion of inrush current when the amp/power supply is first turned on----and not having that inrush current (which we know can be very, very large) blow up your diodes and create a cascading failure scenario. In a tube amp, we know that the diodes, even clunky 4007's, can switch very fast. Those diodes switching on OR off can create sort of unknowable conditions since they are hitched up to an inductor (the output of the PT and possibly a choke, downstream) and of course, our pals, the electrolytic caps, which take time to charge up.

Once a few cycles of AC power have gone through and the rectifier portion has reached a more-or-less steady state, the diodes turned on probably swamp (meaning short out) the action of the caps. The transients of the diodes turning off gets absorbed easily into the electrolytics. Again, I believe this is all overcautiousness for those first few turn-on cycles....and it is not completely unjustified. These are cheap parts and you or I would no doubt rather put them in during the build when it is pretty painless versus taking down the amp, blown up or flamed or with exploded cap(s) and hauling it on to the bench.

A lot of this, IMHO, is left over from the paradigm of a Twin Reverb where we see 3 diodes connected in series on each side of the full-wave rectifier. Maybe few people trusted diodes back in the early 60's. There they were, taking the place of gigantic 5U4. Now, we have completely ordinary 1N4007 diodes good for 1000 volts and 1 amp --- basically, more than TWO 5U4s...and maybe we still don't quite believe it. Nevertheless, they cost all of 3 cents each. If this stuff makes you feel better, I say do it.

[Gary_S]

Great post eleventeen and yeah the SS diode is a remarkable thing, being able with a couple of those to simulate a full wave rectifier or four in a bridge rectifier.

Tube rectifiers seem very much a blast from the past now and you wonder if they're worth the bother. Yes you can get that tube sag but according to Merlin's book you can replicate that with other things. Of course if you're building an exact replica, of an old fender for example, you'd have to have a tube rectifier for the authenticity.

[jazbo8]

There a few different things going on, the in-rush current and filter capacitor charging/dis-charging are separate issues. What Merlin showed in the book refers to voltage-sharing for situations where a single diode is under-rated for the job, e.g., if you need to have PIV (peak inverse voltage) of say 500V but you only have 250V diodes on hand, then you would need to stack two of them up to handle the 500V, so the capacitors in parallel with the diodes are there to make sure that the diode sees the same voltage (250V), so their voltage rating are not exceeded.

[eleventeen]

" if you need to have PIV (peak inverse voltage) of say 500V but you only have 250V diodes on hand, then you would need to stack two of them up to handle the 500V, so the capacitors in parallel with the diodes are there to make sure that the diode sees the same voltage (250V), so their voltage rating are not exceeded."

I myself would never do that. Not even close. If you are going to build tube amps, IMHO you do yourself a favor & go buy 100 qty 1N4007 1000 volt/1 amp diodes for under $5 and you are done. Then you can blow up 2 dozen of them, use 2 dozen of them on your 12 volt projects and not even notice.

[Katie 77]

???
why would anyone go thru that rather than just dropping in a pair of UF5408s?
this is a real question. why?

[jazbo8]

Guys - Gary asked for an answer specifically on Merlin's book - it is not about what you would do (I think we all would prefer to do it the right way) but trying to explain how the circuit works.  BTW, the ultra-fast recovery diodes generate less noise than the standard silicon diodes such as the 1N4007, so for a bit more money, they are really worth it IMO.

[Gary_S]

Guys - Gary asked for an answer specifically on Merlin's book - it is not about what you would do (I think we all would prefer to do it the right way) but trying to explain how the circuit works.  

Thanks jazbo8, yeah that's really what it was. I know folks can use fast switching diodes and that would be more ideal, but with Merlin discussing the possibility of using the caps to parallel the diodes i just wondered about the details of the ratio of how much of the current would split between cap and diode and how having the cap in parallel would cause equal voltage sharing between all the diodes in the rectifier.

I suppose the answer to; why would anyone use that method if it's not the best solution? would be, maybe if they had those diodes to hand and if they did the job adequately then why not use them with the parallel caps?  

[jjasilli]

Guys - Gary asked for an answer specifically on Merlin's book - it is not about what you would do (I think we all would prefer to do it the right way) but trying to explain how the circuit works.  BTW, the ultra-fast recovery diodes generate less noise than the standard silicon diodes such as the 1N4007, so for a bit more money, they are really worth it IMO.

Good point.  My lingering puzzlement is that these bypass caps (and resistors, if any) pass AC around the diodes, thus defeating the rectification process.  I do not see this issue addressed anywhere.  I suppose the answer is that the caps block enough AC so that whatever gets through, along with the usual ripple, gets suppressed by the filter cap stages.

[Gary_S]

Yeah i don't really understand that bit, because the caps will pass the AC normally but block any DC, yet in the book Merlin says there's a very small leakage current from the cap and that's enough to equalize the voltage across the diodes of the rectifier.

I suppose it's like most things; you're better off just taking it on faith rather than to get overly technical about how it actually works. But i always want to know!!  

[sluckey]

Yeah i don't really understand that bit, because the caps will pass the AC normally but block any DC, yet in the book Merlin says there's a very small leakage current from the cap and that's enough to equalize the voltage across the diodes of the rectifier.

Caps don't just pass AC like a switch. They offer resistance that is inversely proportional to frequency.

A .01µF cap offers 26.5KΩ of resistance to 60Hz AC.

[Gary_S]

Yeah, i know they don't pass it like a switch. There is a capacitive reactance involved, depending on the size of the cap and the frequency, but they DO pass AC, whereas they don't pass DC. In fact they actually don't physically pass AC, it's a displacement of current around the system but it happens so fast that to all intents and purposes it looks like it's passing current but actually it isn't.

[PRR]

Everything is ratios. Ratios is everything.

Ignoring the rectifier: you have a 0.01uFd cap into a 40uFd cap. *MOST* of the AC appears across the 0.01uFd cap (97.5%).

With the diode in place, half the time the 26K cap impedance is bypassed with 1 ohm of diode impedance, so it doesn't matter.

[Gary_S]

Cheers PRR, yeah it makes sense when you lay it out in those terms; current wanting to always take the path of least resistance.

[beny84]

@Gary_S, a rectifier can be used along with a rectifier for so many reasons but voltage division might not be ne the most common one. Remember a capacitor acts as an open circuit to a DC value so there is not much in it for it to be considered for voltage division. However, along with a rectifier, a capacitor can be used to smooth out the voltage signals that are coming. It's a little similar to the circuit where we use a crystal oscillator with a microcontroller and use capacitors on both ends, just to smooth out the edges in a signal. There are other such applications too but all dealing with signals, not DC voltage levels.