audio-talk

On the Pass single ended mosfet amps I'm constructing the power supply has some 63v caps that I'm using in series across the rail as the 63v is too low for my needs.

So I need resistors across the terminals.

On my valve amps 220K would be about right, but for 60v would 30K be more suitable, gives about 2ma, is that enough ?

Sounds good.

Hopefully it would be way more than the leakage current of the capacitor(s), if any, which is the reason for having them; I'd try 100k and see how that goes, if the junction is exactly half Voltage when measured then OK then. Job done.

Surely, if you use two identical value resistors the voltage between the two caps will always be half the positive rail voltage, regardless of current ?

Or, have I misunderstood you Mike.

pre65 wrote: ↑Sun Feb 19, 2017 1:49 pm Surely, if you use two identical value resistors the voltage between the two caps will always be half the positive rail voltage, regardless of current ?

Or, have I misunderstood you Mike.

Not if the cap leakage is of the same magnitude as the resistor current.

I would use this as a reference. https://www.illinoiscapacitor.com/pdf/P ... istors.pdf

For example, 6800uf using their rule of thumb would suggest 1.47k. Watch the dissipation of the resistor.

If you look up the leakage current you can calculate a more accurate value. I would have probably gone for 3k3 2W

All I'm saying is, the current can be small, so the resistors can be large value, at least that's what I've found.

Of course you can have it as 2mA if you really want to.

Thanks Mike & Nick.

I'll give it some thought.

I was also thinking of resistance to bleed the cap charge when the amp is shut down, but I may well fit what Duncans PSU designer call a current tap (1K) for that purpose. The transformer is not short on VA so won't be a problem.

Ah capacitor balancing resistors can do that too.

How many pairs of capacitors need them?

Yes Mike, I was aware of that, hence my thinking that 2ma might be OK.

Nicks suggestion (3K3) would be 9ma, so that would be better ?

I have 4 of 22000uf @ 63v, two pairs in series across the rail then those two pairs in parallel, giving 22000uf @ 126V.

The 4 of 6800uf @ 160v are in parallel, giving 27200uf @ 160v

OK, so I make it around 33k for each resistor, somewhat under 2mA. Or could be 22k each for 2.8 mA. I'd try the latter.

HTH

While I am in danger of being grumpy and pedantic, that link I posted tells you how to calculate the resistor. Assuming the 22000uf 63v caps are PEH200, the spec sheet for then gives the leakage current at 8.158 mA, using the formulae, that gives a equivalent resistance of 7k7, so the recommended balancing resistance is 770R, very different from 30k. Its fine to use what works when you build it, but you also have to consider that the leakage current will be more in a years time.

Interesting stuff, Nick - thanks. I hadn't realised the leakage current was so high !

Mine are PEH169 22000uf @ 63v, and the data sheet gives leakage current as

Leakage Current
I = 0.003 CV + 4,000 (µA)
C = rated capacitance (µF), V = rated voltage (VDC). Voltage applied for 5 minutes at +20°C

Not sure I understand that.

{ 0.003 x22000 x 63 ( =4158 ) + 4000 } uA
ie. 8158uA
= 8.158mA

IslandPink wrote: ↑Sun Feb 19, 2017 5:36 pm { 0.003 x22000 x 63 ( =4158 ) + 4000 } uA
ie. 8158uA
= 8.158mA

Thanks Mark, I was just beginning to understand.

So, Nicks calculation of 7.72K divided by 10 was correct.(not that I doubted it)