Gavin,
The purpose of bleeder resistors in PSUs employing cap. I/P filters is to discharge the caps. when the unit is powered down.
R3 will discharge C4, at a minimum. Put the 220 K part across C3, as that will surely discharge C1, C2, and C3.
BTW, please explain the purpose of a current source connected from B+ to ground. I'm scratching my head.
The purpose of bleeder resistors in PSUs employing cap. I/P filters is to discharge the caps. when the unit is powered down.
R3 will discharge C4, at a minimum. Put the 220 K part across C3, as that will surely discharge C1, C2, and C3.
BTW, please explain the purpose of a current source connected from B+ to ground. I'm scratching my head.
Hi,
Basically that depends on how safe you want to play it...
As you have a choke and some resistors in series with the B+ line, I'd put a bleeder across all the caps.
By doing so you'd still discharge the B+ reasonably fast enough not to get killed by unexpected high voltage present on one of the caps.
As you want to use a 220K resistor and you have 4 caps, I'd wire a 1M resistor across each of the caps.
Cheers, 😉
Basically that depends on how safe you want to play it...
As you have a choke and some resistors in series with the B+ line, I'd put a bleeder across all the caps.
By doing so you'd still discharge the B+ reasonably fast enough not to get killed by unexpected high voltage present on one of the caps.
As you want to use a 220K resistor and you have 4 caps, I'd wire a 1M resistor across each of the caps.
Cheers, 😉
Eli Duttman said:
Duncan's PSU designer won't let you put a resistor in parrallel with the load. So I put in a current tap to simulate a 220K resistor in parallel. At least that's what I had in mind.
Hi,
Eli,
R3 being the load, i.e. the circuit itself, the value goes up as the unit is powered down.
In all fairness it will likely discharge the lot quite fast enough under normal circumstances but...
I've seen people pulling the tubes out of their units, powering down and, without bleeder resistors put where they belong, being quite amazed at seeing their soldering iron fried by touching a closeby cap with it....and that's the happy ending for you...😉
Cheers,😉
Eli,
R3 being the load, i.e. the circuit itself, the value goes up as the unit is powered down.
In all fairness it will likely discharge the lot quite fast enough under normal circumstances but...
I've seen people pulling the tubes out of their units, powering down and, without bleeder resistors put where they belong, being quite amazed at seeing their soldering iron fried by touching a closeby cap with it....and that's the happy ending for you...😉
Cheers,😉
Um...
What's the purpose of everything past C2, aside from the loads?
No wonder you want a bleeder resistor. For that much capacitance (suitable for filtering a good 500mA) you'll need almost as much as your 12k load resistor there, maybe 20k. Such will reach 99% of operating voltage in about a minute (five time constants), assuming your load turns off instantly (which it doesn't. Adding that in, it might take 20 to 30 seconds.)
Tim
What's the purpose of everything past C2, aside from the loads?
No wonder you want a bleeder resistor. For that much capacitance (suitable for filtering a good 500mA) you'll need almost as much as your 12k load resistor there, maybe 20k. Such will reach 99% of operating voltage in about a minute (five time constants), assuming your load turns off instantly (which it doesn't. Adding that in, it might take 20 to 30 seconds.)
Tim
If you are planning on biasing up your heaters with dc from your B+, the voltage divider that you use will act as your bleeder. Just be sure that it's at the end of your B+ supply.
As long as the primary will sustain the voltage... here's a great way to bleed it quickly that I've used on low voltage supplies
http://www.reed-electronics.com/ednmag/contents/images/70501di.pdf
or
http://www.reed-electronics.com/edn...le&articleid=CA90760&spacedesc=bestofdi_power
http://www.reed-electronics.com/ednmag/contents/images/70501di.pdf
or
http://www.reed-electronics.com/edn...le&articleid=CA90760&spacedesc=bestofdi_power
Eli Duttman said:
Other than dischaging the caps, the bleeder resistors main purpose should be providing the minimum load current for chokes.
Minimum Current= 2*1.414*Vin /6Pi *pi *f *L
Choke can not operate correctly if the current falls below the min value.
You are entirely correct,
However, nothing precludes using both approaches when appropriate.
Allows selection of minimum bleed resistor to maintain minimum load current for the choke (when a choke is used),
and
A lower resistance to safely and quickly discharge upon turn off.
However, nothing precludes using both approaches when appropriate.
Allows selection of minimum bleed resistor to maintain minimum load current for the choke (when a choke is used),
and
A lower resistance to safely and quickly discharge upon turn off.
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