.. heavy duty contactor for capacitor switching,,, been mounting them in big power factor corrector system
they have small contacts on top, (which will connect big oilpaper caps to grid) trough integrated resistors.
and then comes the big contacts
amazing stuff
they have small contacts on top, (which will connect big oilpaper caps to grid) trough integrated resistors.
and then comes the big contacts
amazing stuff
One shouldn't be calling someone an idiot for pointing out a mistake. That doesn't qualify as educated.
Sometimes it's worth thinking about the things someone else wrote before calling him an idiot.
Ohm's law holds. For a steady-state situation, e.g. the load current. At the moment of switching on, all the filter caps are empty and have zero volts across them.
During the initial charging spike, the full voltage is across the resistor! That is the reason for the inrush current this discussion is about.
Of course this is only a short transient. And as the caps charge, the voltage drops fast, just as someone said above. But it is too true to call someone an idiot, don't you think?
Rundmaus
That shouldn't be a problem for the relay should it? There's no switching going on with a NO relay, until the time delay tells it to close the contact, at which time the voltage across the resistor should have dropped to acceptable levels.
Using a 22k 5 watt resistor, it can easilly withstand the full voltage across is. Calculations give 2.4W so I'll use 5W to be safe.
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Well,
at least the relay should be designed to withstand the full voltage spike in the off state - requiring adequate contact spacing.
22k sounds a bit high for the limiting resistor. High value power resistors are particularly prone to failure due to the very thin resistance wire.
Rundmaus
at least the relay should be designed to withstand the full voltage spike in the off state - requiring adequate contact spacing.
22k sounds a bit high for the limiting resistor. High value power resistors are particularly prone to failure due to the very thin resistance wire.
Rundmaus
Yes, it may be a problem. Even though it's not switching, if the relay isn't designed for high voltage DC, it can arc. It'll act just like a spark gap.
I still think a relay and inrush current limiter on the primary side is a much better idea. Even if you need a separate filament transformer.
If you must switch the secondary, consider using a MOSFET or IGBT. (If you have to go all tube I suppose a thyratron would work. But, I wouldn't do it that way.)
Thank you for saying what I'm thinking, just in a much politer way than I would have said it.
The power relays I'm looking at have up to 1000Mohm @ 500VDC insulation between contacts, so they should be fine.
I didn't know that about power resistors. I'll consider different values/wattage.
There have been lots of great ideas and insights the last 5 pages, so I'll test some of them for the sake of experimentation. I might end up with a relay in the primary as suggested.
And I'm not at all looking for all-tube. My PSU is of the MOSFET regulated with feedback type.
All-tube is fun sometimes, but at serious expense of money, weight and complexity.
Rundmaus
PS. I remember reading that information about power resistors somewhere in Morgan Jones' book 'Valve Amplifiers'.
I only have copies of a few important pages, maybe someone else can look it up.
Rundmaus
PS. I remember reading that information about power resistors somewhere in Morgan Jones' book 'Valve Amplifiers'.
I only have copies of a few important pages, maybe someone else can look it up.
look in resistor datashett of max operating voltage, even if you not will exceed watts
I'm aware most resistor will do only 250V to 350V. For this purpose, I'm using Xicon 5W wirewound cemented power resistors. No voltage issues when not exceding the temparature or power ratings.
The datasheet states:
"No evidence of flashover, mechanical damage, arcing or insulation break down"
You do realize that a 22K and 1000uF make for a time constant of 22 seconds, right? Which is about 1 to 2 minutes to charge up. That assumes no external load... Even 10ma of load causes that resistor to drop 220V. Something is telling me that 22K is much too high... If that's on the primary, it only gets worse. I don't even think the filaments will even heat up enough to get above a volt or two..
Yeah you're right. Although, I don't intend to fully charge the caps before bypassing the inrush resistor. It's just to limit the inrush for the first portion of charge when it's the worst. But still...
If I'm going for the 'resistor in primary' solution I'll probably go for two 15Kohm 5W paralled. Gives me 7.5K 10W, which is more than enough. Plus it has another benifit: I can digitally monitor the charge time. If one resistor fails, chargetime doubles, on which I can shut everything down and make it go into 'protection mode'.
I'm afraid that given that high of a value... Charge time doesn't double. It doesn't even get to 10% of charge because you can't supply enough current to heat your filaments.
I strongly suggest using an inrush current limiter instead. They are designed for this purpose and the manufacturers have app notes that can help you choose the right one. Just keep in mind that you'll need something bigger than the app notes suggest if you're powering the heaters too.
Hey... since we seem stuck on resistors and relays ... Just did a little integration-over-time analysis ... and I confirmed what I recall hearing many years ago. Namely, that the total power dissipated in the slow-charge resistor remains essentially a constant between Vc=0 and Vc=X (some high value).
It isn't one of those obvious Laws like Ohms or Kirchoff's.
Anyway - I tried a whole bunch of different resistance values. For a 1,000uF load cap, with a 500 volt rectified power source, the best resistor ranges from 4.7K down to 2.7K. This limits inrush current to 106 to 185 mA (respectively), and yet gets the caps charged to 95% in 10 to 6 seconds (respectively). The total power dissipated at those 95% charge levels (for the resistor) is 125 joules ... so divided by charge rate = 12.5 to 20 watts over the charge-up interval. A heftier resistor, to be sure.
By way of comparison ... 90% charge is reached in 50 seconds, for an average power of 2.5 watts (peak of 10 watts up front, and about 8 watts average over the first 10 seconds).
Well there you are. I can see why you don't wait - it takes a long time to complete the charging to 99%! But using the higher value resistor allows a smaller wattage one to be deployed.
Interestingly, there is no way to cheat the power-to-dissipate-as-heat demon! Even when I modeled in a constant current source, it too dissipated 125 joules to get to full charge on the filter cap bank. Lesson learned. Need a decent sized heat sink for that bad boy.
But it sure was fast! 100 ma gives 100 V/sec ... and a FULL charge in only 5 seconds. Like a bloody 555 timer chip. Just got to get rid of the heat. Chunk of aluminum.
It isn't one of those obvious Laws like Ohms or Kirchoff's.
Anyway - I tried a whole bunch of different resistance values. For a 1,000uF load cap, with a 500 volt rectified power source, the best resistor ranges from 4.7K down to 2.7K. This limits inrush current to 106 to 185 mA (respectively), and yet gets the caps charged to 95% in 10 to 6 seconds (respectively). The total power dissipated at those 95% charge levels (for the resistor) is 125 joules ... so divided by charge rate = 12.5 to 20 watts over the charge-up interval. A heftier resistor, to be sure.
By way of comparison ... 90% charge is reached in 50 seconds, for an average power of 2.5 watts (peak of 10 watts up front, and about 8 watts average over the first 10 seconds).
Well there you are. I can see why you don't wait - it takes a long time to complete the charging to 99%! But using the higher value resistor allows a smaller wattage one to be deployed.
Interestingly, there is no way to cheat the power-to-dissipate-as-heat demon! Even when I modeled in a constant current source, it too dissipated 125 joules to get to full charge on the filter cap bank. Lesson learned. Need a decent sized heat sink for that bad boy.
But it sure was fast! 100 ma gives 100 V/sec ... and a FULL charge in only 5 seconds. Like a bloody 555 timer chip. Just got to get rid of the heat. Chunk of aluminum.
You are correct. The energy dissapated in the resistor is roughly equal to the energy stored in the capacitors: (1/2) (V^2) C
There's just one kicker: If it's in the primary, and that's where I think it should be, the heaters act as a load and the capacitors never charge much. Even on a different winding, they're effectively shunting the capacitor and should be modeled with a low value resistor. That's why resistor values in the Kohms won't work out.
if your resistor is too high, then your worst portion of charge will be the second one🙂 (again, depends on timing)
or going heat-less, 470n =6k8@50Hz (if there is full AC)
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On those pesky charging resistors...
Here's my "final (and best)" solution.
[1] lose the 22K resistor in the 500 B+ leg.
[2] insert a 2-terminal high voltage 100 ma regulator in the ground leg.
[3] give it a mid size heat sink to soak up the joules
[4] wire it to a bypass relay which can be switched in when the caps are topped up.
This does ALL these things:
* minimizes time-to-charge
* ... at safely limited current flow
* is simple to implement
* doesn't expose the relay to persistent high voltage
* is cheap
The 100 ma CC regulator will charge the caps in 5 seconds to 500 volts. This can NOT be shorted any further without increasing the current flow. If "100 ma" is safe, then this is definitionally safe. The current regulator will still need to dissipate those 125 joules though, in 5 seconds. That's an average of 25 watts. Heat sink!
Further, since the regulator is in the ground-side leg, it can produce a low-voltage (through 2-resistor voltage divider) signal to run to a comparator, to drive the relay. i.e. "when delta-V across regulator drops below 10 volts... engage relay".
You get the best of all the design ideas - fastest charging, (actually "near-constant charge time!"), easy implementation, trivial bypass (so the current regulator doesn't get in the way of the operating point of the amplifier). Moreover, one might choose to use the same switching signal to drive an additional high-voltage side reed relay to engage the B+ to the amplifier's tube finals.
This gives a nice A/B operating point: "A" is "disconnect finals, charge capacitors" and "B" is "bypass safety/charge-limiting capacitor charging circuitry, and engage amplifier finals". Worth considering!
GoatGuy
Here's my "final (and best)" solution.
[1] lose the 22K resistor in the 500 B+ leg.
[2] insert a 2-terminal high voltage 100 ma regulator in the ground leg.
[3] give it a mid size heat sink to soak up the joules
[4] wire it to a bypass relay which can be switched in when the caps are topped up.
This does ALL these things:
* minimizes time-to-charge
* ... at safely limited current flow
* is simple to implement
* doesn't expose the relay to persistent high voltage
* is cheap
The 100 ma CC regulator will charge the caps in 5 seconds to 500 volts. This can NOT be shorted any further without increasing the current flow. If "100 ma" is safe, then this is definitionally safe. The current regulator will still need to dissipate those 125 joules though, in 5 seconds. That's an average of 25 watts. Heat sink!
Further, since the regulator is in the ground-side leg, it can produce a low-voltage (through 2-resistor voltage divider) signal to run to a comparator, to drive the relay. i.e. "when delta-V across regulator drops below 10 volts... engage relay".
You get the best of all the design ideas - fastest charging, (actually "near-constant charge time!"), easy implementation, trivial bypass (so the current regulator doesn't get in the way of the operating point of the amplifier). Moreover, one might choose to use the same switching signal to drive an additional high-voltage side reed relay to engage the B+ to the amplifier's tube finals.
This gives a nice A/B operating point: "A" is "disconnect finals, charge capacitors" and "B" is "bypass safety/charge-limiting capacitor charging circuitry, and engage amplifier finals". Worth considering!
GoatGuy
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It still exposes the relay to transient high voltage. If it can stand that, I can't see any reason why it won't work.
One suggestion though... Don't switch the B+ on the final, just bias the tube off by upping the grid bias.
I don't want to spoil the soup either, but:
a) As already said by FoMoCo, the relay still sees the full HT voltage at the moment the mains is switched on. Choose a relay that is capable of withstanding that voltage with the contacts open.
b) The regulator sees the full voltage at switch-on as well. It must be capable of coping with that.
c) Suddenly connecting the preheated finals to a B+ line with full charged caps will lead to an awful thump that probably costs you a set of speakers at each switch-on. Rethink that idea.
Rundmaus
a) As already said by FoMoCo, the relay still sees the full HT voltage at the moment the mains is switched on. Choose a relay that is capable of withstanding that voltage with the contacts open.
b) The regulator sees the full voltage at switch-on as well. It must be capable of coping with that.
c) Suddenly connecting the preheated finals to a B+ line with full charged caps will lead to an awful thump that probably costs you a set of speakers at each switch-on. Rethink that idea.
Rundmaus
[1] ... reed relays fit the ticket well.
[2] ... see earlier post (IXYS p/n IXCP10M90S)
[3] ... right you are: don't switch the B+, but instead turn on the final tubes' filaments with the same relay-closing decision-point signal. Soft turn-on.
GoatGuy
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