Popping sag resistor, too much inrush current?

[spacelabstudio]

Hey, so I'm testing a PSU build on the bench.  It uses a bridge rectifier and a "sag" resistor before the first reservoir cap.  See attachment.

What I've found is that I've burned through two 100R/3W resistors right in a row.  The don't smoke or light up or give any other visible sign of distress but they both wind up dead open in the circuit.  I didn't have the wherewithal to measure their resistance before trying them in the circuit, but I figure the odds of getting two bad resistors from the factory are pretty slim.  I know that at the instant of turn on the current and power through the resistor will be much higher than rated, but I figured you're just counting it on that time period being short enough that nothing blows up while that first cap charges up.  But maybe it's too much too fast?  Any other theories?

Chris
 

[sluckey]

Use a higher wattage resistor.

[jjasilli]

25W is common.  Ohms Law (do you know the Pie Chart?):  Power (Watts) = V²/R.  V= voltage drop which we don't have.  Let's say it's 25V.  W = 25²/100Ω; W = 625/100= 6.25.  Your 3W R's don't stand a chance.  And inrush current worsens the situation, maybe by a factor of 2, rounding UP = 13W.  Double that to run cool gets you to 25W as the closest commercially available number.  25W metal housed power resisitor heat-sinked to the chassis is best.  See this discussion:  http://www.justinholton.com/hotrod/sag.html

[spacelabstudio]

Yeah, once caps are charged up expected steady state is ~ 100mA, P = I^2 * R = .1 * .1 * 100 = ~1 watt. 

But at power on, there's inrush current.  Immediately, current is stupidly high and then calms down.  The time constant is 100R * 100uF = 10ms.  The charge in the cap is q = C * V = 100uF * 450V = 45 millicoulombs.  Let's say that the current decreases linearly (it doesn't) and that it takes roughly two time constants to charge the cap.  The average current during power on is then 45 millicoulombs / 20ms = 2.25A.  P = I^2 * R = 506.25W!!  Which probably isn't any more helpful a number than just "a lot". 

A wirewound resistor might be a boon since the inductance will slow down the rise time of the current at power on.  Apparently something significantly larger than what you need in the steady state.  It seems to just come down to trial and error, how big a resistor do you need to absorb the initial shock?  I haven't seen published any specs for resistors on instantaneous power.  Maybe if you knew the internal temperature you weren't allowed to exceed, and how much material you were working with, and it's specific heat, you could just say that an amount of energy equal to what gets dumped into the reservoir cap will also get dumped into the resistor as a pulse, so you just calculate delta T and see where you land.  And given that that's not terribly practical, it probably just means trial and error, order some big resistors and try them out. 

Chris

[sluckey]

You are way overthinking this. I don't think the inrush current is cooking that resistor. I think the steady state current is cooking it. Either way, just use a higher wattage resistor.

[spacelabstudio]

You are way overthinking this. I don't think the inrush current is cooking that resistor. I think the steady state current is cooking it. Either way, just use a higher wattage resistor.

I know it's not the steady state, because I've been testing with no load.  The inrush is the only current.  Assuming I get the current with load that I'm after, the dissipation will be ~1 watt.

Chris

[jjasilli]

If you were right, your resistors wouldn't be blowing.  Try a 25W resistor; it won't blow.  Measure the voltage drop across it.  Then use Watts = V²/R.  

[sluckey]

I know it's not the steady state, because I've been testing with no load.  The inrush is the only current.

No way is that resistor burning up just because of the inrush current to charge those filter caps. You must have a circuit error or there is something you are not telling us.

[tubeswell]

I know it's not the steady state, because I've been testing with no load.  The inrush is the only current. 

Incorrect. Anywhere there is a voltage drop across a resistance there is current flowing through the resistance. Current x Voltage = Power. If the resistor is rated at 3W and is trying to carry 6W worth of VA, it will burn.

[spacelabstudio]

I know it's not the steady state, because I've been testing with no load.  The inrush is the only current. 

Incorrect. Anywhere there is a voltage drop across a resistance there is current flowing through the resistance. Current x Voltage = Power. If the resistor is rated at 3W and is trying to carry 6W worth of VA, it will burn.

And there is no voltage drop, because there are no tubes.  I was just doing a basic check of the PSU before building further.  The only current is whatever gets dumped in that cap on turn on.

Anyway, it's good to hear what values have worked for other people.  Thanks guys.

Chris

[jjasilli]

No tubes!     Then you need to do the light bulb limiter test.  But I think      we already know the result.  Must be a short somewhere in the PS.   

[PRR]

> P = I^2 * R = 506.25W!!

Correct.

Some resistors will take 10X rated Power for a few milliSeconds. A few will take 100X. Here you have 253X rated power.

What is your point? 100 ohms is not much sag. And with 100uFd after it, sag will be slow. 100u is a LOT of capacitance. I'd be thinking 500 ohms and 20uFd.

You might hunt-down an OFF-ON-ON switch. The first ON brings in <10 Ohms 10W to bang that 100uFd up near 450V. The second ON inserts the few-hundred sagger, _after_ the main cap is charged.

(You could automate this with a relay and a transistor to sense >400V to trip the relay and un-short the sagger.)

> A wirewound resistor might be a boon since the inductance will slow down the rise time

No. Not enough to measure.

Wirewound *may* take the transient Power a bit better than some sand resistors which are just 1/2W parts in ceramic shells.

[sluckey]

I guess I never considered inrush to be that much. Never too old to learn something.

[tubeswell]

I know it's not the steady state, because I've been testing with no load.  The inrush is the only current. 

Incorrect. Anywhere there is a voltage drop across a resistance there is current flowing through the resistance. Current x Voltage = Power. If the resistor is rated at 3W and is trying to carry 6W worth of VA, it will burn.

And there is no voltage drop, because there are no tubes.  I was just doing a basic check of the PSU before building further.  The only current is whatever gets dumped in that cap on turn on.

Anyway, it's good to hear what values have worked for other people.  Thanks guys.

Chris

Resistors don't burn for no reason, those resistors are passing current and forming (at least part of, if not all of) the load on the B+, otherwise they wouldn't even heat up. Not even for startup surge. Therefore there must be a short to ground in the B+ somewhere after those resistors

[spacelabstudio]

I know it's not the steady state, because I've been testing with no load.  The inrush is the only current. 

Incorrect. Anywhere there is a voltage drop across a resistance there is current flowing through the resistance. Current x Voltage = Power. If the resistor is rated at 3W and is trying to carry 6W worth of VA, it will burn.

And there is no voltage drop, because there are no tubes.  I was just doing a basic check of the PSU before building further.  The only current is whatever gets dumped in that cap on turn on.

Anyway, it's good to hear what values have worked for other people.  Thanks guys.

Chris

Resistors don't burn for no reason,

True.

those resistors are passing current and forming (at least part of, if not all of) the load on the B+, otherwise they wouldn't even heat up. Not even for startup surge. Therefore there must be a short to ground in the B+ somewhere after those resistors

Well, if I just put a jumper there, it all looks good.  There are no shorts.  Inrush current is the only reasonable explanation.

Chris

[spacelabstudio]

> P = I^2 * R = 506.25W!!

Correct.

Some resistors will take 10X rated Power for a few milliSeconds. A few will take 100X. Here you have 253X rated power.

What is your point? 100 ohms is not much sag. And with 100uFd after it, sag will be slow. 100u is a LOT of capacitance. I'd be thinking 500 ohms and 20uFd.

That's a reasonable point.  I don't know yet what amount of sag is going to make my ears happy.  I'll proceed without a sag resistor for the time being and then start playing with different ideas after the rest of the amp is working.  I'll probably also compare the sound of a sag resistor to just the sound of putting some resistance in series with the screens.

Thanks!

Chris

[tubeswell]

I know it's not the steady state, because I've been testing with no load.  The inrush is the only current. 

Incorrect. Anywhere there is a voltage drop across a resistance there is current flowing through the resistance. Current x Voltage = Power. If the resistor is rated at 3W and is trying to carry 6W worth of VA, it will burn.

And there is no voltage drop, because there are no tubes.  I was just doing a basic check of the PSU before building further.  The only current is whatever gets dumped in that cap on turn on.

Anyway, it's good to hear what values have worked for other people.  Thanks guys.

Chris

Resistors don't burn for no reason,

True.

those resistors are passing current and forming (at least part of, if not all of) the load on the B+, otherwise they wouldn't even heat up. Not even for startup surge. Therefore there must be a short to ground in the B+ somewhere after those resistors

Well, if I just put a jumper there, it all looks good.  There are no shorts.  Inrush current is the only reasonable explanation.

Chris

How are the resistors wired in? - in series after the rectifier?, or one on each side of the secondary?

[spacelabstudio]

How are the resistors wired in? - in series after the rectifier?, or one on each side of the secondary?

There's a schematic attached to the top post.  The resistor in question is R36.  It is in series with the rectifier, just before the first filter cap.  I have replaced it with a jumper for now.

[tubeswell]

If the 100R was frying, there had to have been a circuit for the startup surge to 'rush' through. So there must've been a short to ground somewhere after the 100R (Maybe one of the filter caps?). If there was no connection to ground, then no current would rush into the resistor, not even startup current. (Both ends of) the resistor would just sit at whatever voltage the rectifier was putting through.

Edit:  Oops I see!  'Tis the initial charge up current for the filter caps. My apologies for not reading the thread properly. 

[sluckey]

If the 100R was frying, there had to have been a circuit for the startup surge to 'rush' through. So there must've been a short to ground somewhere after the 100R (Maybe one of the filter caps?).

At t₀ (the moment of power on) the filter cap is a dead short and charging current is maximum, limited only by the series resistance. As it begins charging at an exponential rate, the current decays to zero (or leakage current). This high charging current is popping the resistor. Using a higher wattage resistor is the cheap solution.

If you have an analog meter you can see this action quite clearly, especially with a 100µF cap. Just set the meter to some high resistance range and connect the probes across the cap(Observe polarity and know that the red probe ain't necessarily connected to the positive voltage inside your meter). The needle will quickly jump to zero (which is max current thru the meter), hover there for some time, then slowly increase to some high value resistance as the charging current decreases, and finally stabilize (unless you have a lot of leakage). This was a quick check of big caps back in the day when every shop had an analog meter.