I have access to some 300,000uF can-style capacitors. I was wondering if I could use them instead of 680uFs for input power caps. I know that the purpose of the caps are to reduce ripple, and wouldn't a stupidly large cap eliminate the ripple?
Why would I do this? Because I can.
Why would I do this? Because I can.
it can be bad for your transformer, diodes, fuse
big caps draw all of the required current from the xfmr sec in increasingly narrow, higher current pulses as the C goes up
you can add some series R, make a "pi" filter with more reasonable caps next to the bridge
big caps draw all of the required current from the xfmr sec in increasingly narrow, higher current pulses as the C goes up
you can add some series R, make a "pi" filter with more reasonable caps next to the bridge
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Well those would indeed be stupid, but in a fun way.
As JCX says, use a pi filter. Put a more sane cap value after the rectifier, then a resistor, then your stupid caps. That resistor will slow down the big current inrush. But you'll have the stupid big caps as the reservoir. The ripple won't get thru the resistor.
Sounds like fun.
As JCX says, use a pi filter. Put a more sane cap value after the rectifier, then a resistor, then your stupid caps. That resistor will slow down the big current inrush. But you'll have the stupid big caps as the reservoir. The ripple won't get thru the resistor.
Sounds like fun.
Well those would indeed be stupid, but in a fun way.
As JCX says, use a pi filter. Put a more sane cap value after the rectifier, then a resistor, then your stupid caps. That resistor will slow down the big current inrush. But you'll have the stupid big caps as the reservoir. The ripple won't get thru the resistor.
Sounds like fun.
So...
An externally hosted image should be here but it was not working when we last tested it.
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Um. How would they discharge unevenly, given that they're connected at the ground? They'll try to keep at the same potential, even with different ESRs. Am I mistaken here?
But - two things:
1) 680uF might be too small for the first C. Try something around an order of magnitude better.
2)2k resistor is a LOT. That'll waste I^2 * R watts of power. And with 330k of capacitance you don't need a lot to eliminate ripple. You'll need to lower that quite a bit and you'll still need to get resistors with high wattage.
It sounds like a lot of fun tho. I want to see pics of those beasts.
But - two things:
1) 680uF might be too small for the first C. Try something around an order of magnitude better.
2)2k resistor is a LOT. That'll waste I^2 * R watts of power. And with 330k of capacitance you don't need a lot to eliminate ripple. You'll need to lower that quite a bit and you'll still need to get resistors with high wattage.
It sounds like a lot of fun tho. I want to see pics of those beasts.
Um. How would they discharge unevenly, given that they're connected at the ground? They'll try to keep at the same potential, even with different ESRs. Am I mistaken here?
But - two things:
1) 680uF might be too small for the first C. Try something around an order of magnitude better.
2)2k resistor is a LOT. That'll waste I^2 * R watts of power. And with 330k of capacitance you don't need a lot to eliminate ripple. You'll need to lower that quite a bit and you'll still need to get resistors with high wattage.
It sounds like a lot of fun tho. I want to see pics of those beasts.
What about ~6800uF and 20 Ohm?
6800 is fine. You can drive a headphone amp with that alone
The resistor you can size based on how much power you can afford to lose. That depends on how much current will pass trough it. Let's say, for the sake of it, that your two channels will draw about 250mA each. That means you'd be drawing 500mA trough the resistor, which for 20 ohm will have to dissipate 10W. I'd be using 20W rated, heatsinked resistors for this, maybe more. Sounds a bit too wasteful, but on the other hand - you are putting 330k caps in this, so I'm guessing that nothing can be considered overboard
The resistor you can size based on how much power you can afford to lose. That depends on how much current will pass trough it. Let's say, for the sake of it, that your two channels will draw about 250mA each. That means you'd be drawing 500mA trough the resistor, which for 20 ohm will have to dissipate 10W. I'd be using 20W rated, heatsinked resistors for this, maybe more. Sounds a bit too wasteful, but on the other hand - you are putting 330k caps in this, so I'm guessing that nothing can be considered overboard
Let's compromise here. The caps are already insane, make them 10ohm 20W resistors (you can get away with smaller wattage if you know how much current you'll draw) and call it a day, hehe.
The design should draw about 15W; the PSU is a wall wart rated 350mA @ 48VDC.
Well then don't go too high with the resistor as Pano says, or you'd just waste all the power there. What headphone amplifier are you planing on using anyway?
The 12AU7 SSMH from pmillett. I am just toying with the design to further my knowledge of electronics.
Oh. I thought this was a low voltage solid state design. You do need to be careful with a big cap like this and 48V. It can pack a wallop. A bleeder resistor across the big cap is a good safety.
How large should the bleeder resistor be?
EDIT: Oh, I used a calculator. 100Ohm 50W would work and make it safe(ish) at 12v after ~1.5 minutes. 48VDC is on the threshold of high voltage, but not too bad.
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It gets even more fun since I understand you might put a smps before those huge caps. Ripple on an smps is high frequency, so these caps are not going to do much to fight that enemy.
2x300.000 uF would store 0.6 Coulomb of charge for every Volt. So times 48 = 28,8 Coulomb at the voltage you are working at. Which means you need 28.8 A/s to load the caps, which on a power supply that delivers 350 mA implies that the time to just charge your caps would be 28.8/.35=82.3 seconds.
The upside is that your amplifier would play for minutes after switching off power, so you could carry it around the house from outlet to outlet without any disruption.
vac
2x300.000 uF would store 0.6 Coulomb of charge for every Volt. So times 48 = 28,8 Coulomb at the voltage you are working at. Which means you need 28.8 A/s to load the caps, which on a power supply that delivers 350 mA implies that the time to just charge your caps would be 28.8/.35=82.3 seconds.
The upside is that your amplifier would play for minutes after switching off power, so you could carry it around the house from outlet to outlet without any disruption.
vac
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