Capacitance questions

[g1]

Some of the big cans had cardboard sleeves over them, Ampeg and Garnet are a couple I've seen. But this was to avoid electrocuting the user/service person from what I recall. I think they also had an insulating base-plate where they connected to the chassis?

[Justin Thomas]

Yep, I've pulled a few of those insulating boards out of Ampegs to reuse if possible. Sometimes the can I'm replacing the old one with doesn't need it. I've got a couple of those cardboard shells, too.

Justin

[Gingertube]

Just a bit of a caution - I have never fully got my head around the 2nd cap bank , the one after the standby switch.

Stacked caps will share voltage according to their capacitance. The same charge current flows thru' them so small cap charges to larger voltage and large cap charges to smaller voltage. So in simple terms I would expect 220/220+50 = 81% of the voltage across the 50uF and 19% across the 220uf.

THis complicated twice -
The 2nd 50uF partially isolated by the 1K resistor
AND
The 220K + 220K voltage share resistors which will be working overtime and will NOT be able to achieve equal sharing.

The take away from this is that if you keep the circuit as shown on the schematic I would expect maybe 2/3 (butt pluck figure) of the voltage to be across the 50uF caps, make sure you use the highest voltage rating for them.

Has anyone done measurements of the voltages at that point?

Cheers,
Ian

[Chuck H]

Ah! Yes. I hadn't noticed the different cap values on the later node at a glance. Kinda stupid design IMHO. Why not just make all three caps 100uf and call it macaroni!?! Fender has a couple of designs that totem different value caps. Makes no sense. If you're an engineer and know the math it's possible to calculate the right value for balancing resistors. I'm not an engineer. And clearly the guys that designed that power supply weren't either since there's no consideration given to resistor values across the different cap values.

Three 100uf/350V caps would be my answer.

[Randall]

I'm not an engineer, and don't have mastery over the math. This is a dead stock piece that was recently paid top dollar for because of that, so I am not inclined to attempt to re-design the power supply, regardless of how much sense it makes. We all know how that goes. I just need to give it back the same as it was, within reason, and those down stream can decide what to do from there.

[Helmholtz]

As capacitors can't conduct net/average DC current (ignoring leakage), they don't influence average DC voltages in steady state. In other words average DC voltages are solely determined by the resistances (including leakage) in the circuit.
The 50µ cap will see more ripple voltage amplitude than the 220µ, though.

[Dave H]

If you're an engineer and know the math it's possible to calculate the right value for balancing resistors.

You don't need to be an engineer

Without balancing resistors the capacitors would share the (DC) voltage according to their leakage currents, the capacitor with the larger leakage current receiving the lower voltage. The resistor value is chosen to make sure that the current through the resistors is much larger than the capacitor leakage current in which case the (DC) voltage across the capacitors is defined by the resistor ratios. Two 220k resistors would share the voltage equally between the capacitors regardless of the differing capacitor values.

[Tony Bones]

You don't need to be an engineer

Without balancing resistors the capacitors would share the (DC) voltage according to their leakage currents, the capacitor with the larger leakage current receiving the lower voltage. The resistor value is chosen to make sure that the current through the resistors is much larger than the capacitor leakage current in which case the (DC) voltage across the capacitors is defined by the resistor ratios. Two 220k resistors would share the voltage equally between the capacitors regardless of the differing capacitor values.

Right. The AC (ripple) voltage is shared according to the inverse of the capacitance (1/C). What are the balancing resistors there for? To balance DC, not AC.

[Chuck H]

But, but...

We did have a discussion on the asymmetrical totem arrangement here before. g1 links the circuit in question in post #21, Malcolm makes a good point on the matter in post #30 and Ian takes over in post #33. There is some further discussion on the topic after that.

https://music-electronics-forum.com/...ght=totem+pole

[Tony Bones]

Ian makes a good point that the initial charging current (at power up) is the same for both caps, so the DC voltage that they initially charge to is not shared equally, but split according to 1/C. That means that with unequal C's the smaller one gets more than half of the total. It might take a while for the balancing resistors to fix that.

[Helmholtz]

Ian makes a good point that the initial charging current (at power up) is the same for both caps, so the DC voltage that they initially charge to is not shared equally, but split according to 1/C. That means that with unequal C's the smaller one gets more than half of the total. It might take a while for the balancing resistors to fix that.

THat's why I restricted my explanation above to steady state. Initial turn-on is a transient event, causing relative voltage over-shoot across the smaller cap.

[Old Tele man]

The REASON for eqaul-sized "balancing resistors" is to "EQUALLY" divide the voltage across the two caps at FULL/peak voltage and NOT during the initial "charge/pump" up when the voltages are smaller (but quickly getting larger). Operating capacitors at voltages below their rated maximums is OK, it's at/above maximum ratings that things get dicey.

[Dave H]

Just a bit of a caution - I have never fully got my head around the 2nd cap bank , the one after the standby switch.

Stacked caps will share voltage according to their capacitance. The same charge current flows thru' them so small cap charges to larger voltage and large cap charges to smaller voltage. So in simple terms I would expect 220/220+50 = 81% of the voltage across the 50uF and 19% across the 220uf.

THis complicated twice -
The 2nd 50uF partially isolated by the 1K resistor
AND
The 220K + 220K voltage share resistors which will be working overtime and will NOT be able to achieve equal sharing.

I've been thinking about this
And I think it goes like this -

The 220k balancing resistors will divide the voltage equally for the DC steady state condition.

For the switch on state we could could say the top two 50u caps are in parallel to make 100u so at switch on the voltage will quickly rise until there is 166V across the 200u cap and 333V across 100u cap (for a 500V supply). The 220k resistors will then slowly equalise the voltages until there is 250V across each cap. The caps are rated at 350V so it should be OK.