Power Supply Drain

[J Rindt]

I guess there is a law that states a power supply must drain down to 25 Volts in 30 seconds...or something along those lines.
So we often see something like a 5 Watt 100k resistor across the reservoir capacitor (from + to -).
When you shut the amp off, the resistor will start to discharge the power supply caps.
I have two questions:
1. Is there a "simple"  formula to arrive at an Ohm Value that will be small enough to do the drain job, but big enough so it does not apply undue load to the power supply.?

2.In this Vox schem, is that the purpose of the 18k resistor on the screen supply cap.? It kind of seems like an awfully low value...but I guess THAT relates to my question #1.
That 18k would not be a very good way to drop some B+ from the screens would it.?
Thank You

http://www.drtube.com/schematics/vox/ac100jp.gif

[John]

I think the screens are fed directly from the 100uf cap. That resistor appears to drop the voltage going to the preamp. I hope I'm reading it right anyway, it's drawn a little differently that what I'm used to.


Oh, and for the drain resistor, I think a lot of guys use a 220k to 470 across one of the filter caps. Higher value means slower drain time, but also I would think a 1-watter would be plenty.

[eleventeen]

18K offhand *does* seem rather low, implying high current through it, thus implying big watts. Assuming that the choke drops zero volts, node "B" sees 466 volts. E = IR.


466 / 18K = I = .0258 amp = 26 mils. Yeah, that's more current than I'd like to see going through there, and since Power (watts) = I * I * R, then that resistor needs to be good for .026 * .026 * 18K = 12.02 watts. Should be a big fat 25 watt resistor. 

try this:


http://mustcalculate.com/electronics/capacitorchargeanddischarge.php



I would expect that resistor would drain the power supply in RC time, and I probably have this wrong, but 200 uf (both caps, essentially in parallel) * .018 megs (ohms * farads or megohms * microfarads) =
1 TC = 3.6 seconds drains by 63%, to 37% = 172.4 VDC, and
2 TC =7.2 secs to 37% of 177 = 63.8 volts and
3 TC =10.8 secs to 37% of 63.8 = 23.6 volts and
4 TC =14.4 secs to 37% of 25.2 = 8.7 volts and
5 TC = 18 secs to 37% of 8.7 = 3.2 volts

I could have this backwards, that each 3.6 second period drops 37% of the cap voltage, to 63% of what was there...thus





3.6 seconds drains by 37%, to 63% = 293.5 VDC, and
7.2 secs to 63% of 293.5 = 185 volts and
10.8 secs to 63% of 185 = 116.5 volts and
14.4 secs to 63% of 116.5 = 73.4 volts and
18 seconds to 63% of 73.4 volts = 46.25 volts and
21.6 secs to 63% of 46.25 = 29.1 volts and
25.2 secs to 63% of 29.1 volts = 18.35 volts


I think the first table is right, generally 5 time constants is considered adequate to "fully discharge" a cap. The discharge happens MUCH faster during the first 2-3 periods. 

If the power transformer has 12 watts of power to throw out, then it does, but you have to have that fattish 25 W "bleeder" resistor in there. Offhand, I would triple that 18K to about 50K, to get bleeder current down to the 8-10 mil range. 

Where you see 220K resistors, often in the higher-power Fender supplies, that resistor serves a different function, it "balances" currents across TWO e-caps in series. This allows one to use lower voltage rating caps. All you need is a mil or so going through such a cap, so it can be a high value = low current = low watts. Yes, those resistors will in fact bleed off B+ after power is shut off, but verrrrry slowly.

[J Rindt]

I think the screens are fed directly from the 100uf cap. That resistor appears to drop the voltage going to the preamp. I hope I'm reading it right anyway, it's drawn a little differently that what I'm used to.


Oh, and for the drain resistor, I think a lot of guys use a 220k to 470 across one of the filter caps. Higher value means slower drain time, but also I would think a 1-watter would be plenty.

Yes...I believe you are right...I made a mistake. The screens say Node-B...and then After that we see the 18k.
My Goof.
But anyway...so they are using that resistor to drain the PS on shut-down...And to drop voltage to the preamp.?
Doesn't that resistor drop all 466 Volts (as eleventeen says below).?
To my knowledge, I have never seen this before...
Thank You
Actually...looking at this schem again, I would say that 18k is in  parallel with the 100uF-500 cap for the screens.
The drawing is kind of ambiguous...do not know why they separate the cap and resistor so much on the schem.....

[J Rindt]

eleventeen -
Thanks for the info.
Yeah...I understand about the "balance" resistors that are used on Series Caps.
I was talking about...like on a (modern day) Marshall Plexi ...some guys will put a 100k across one of the cans to make a "drain".
Thanks Again

[PRR]

> I guess there is a law that states a power supply must drain down to 25 Volts in 30 seconds...

No.

It may be a good idea. But it is often overlooked.

Also in my experience there is (in a healthy amp) considerable bleed-down in the few seconds while the tube heaters are coming off red-heat.

If desired, bleeders are a compromise between rapid bleed-down and waste and cost. A nice quick bleed on a 100uFd cap-array may be 30K, which in a 400V amp is 5.3 Watts of excess heat and 13mA of excess power supply load.

It is more usual to budget for 2W resistors, figure for 1 Watt dissipation, and live with whatever bleed-rate that turns out to be. At 400V this will be 160K, giving around 16 seconds to 63% (48 seconds to 25%).
___________________________________

> In this Vox schem, is that the purpose of the 18k resistor

I suspect that is mis-copied. It has around 466V across it. 466V in 18K is *12* Watts. Bleeders must be very conservative. A 20 Watt part is not really enough. Even in a 100W amp, this is excessive.

If a zero has been lost, 180K is 1.2 Watts and "could" work for some years with a 2W resistor.

[J Rindt]

Thanks for the reply.....but nope, I have seen the originals. They are indeed 18k. They are the tubular wire-wound I guess. They look to be 10 Watts.
So at that location, it IS just a drain right.?
With one end at ground, and the other end at screen potential, it is dropping all the voltage there is.
If they wanted to drop volts to the preamp supply, they would need to use a series dropping resistor, wouldn't they.?
But that makes me wonder why VOX chose that location for a drain. Perhaps it was an easier application than the reservoir cap.?
Thank You

[jazbo8]

The 18K is wired from the can cap to ground, not sure of its power rating though, but certainly no 100W...

[sluckey]

The schematic below shows the plate node at 470V and the screen node at 450. That's a 20v drop across the choke. Without the 25mA load of that 18K resistor, the screen node would be much closer to 470V. Maybe the lower screen voltage was a factor in using that 18K resistor?

The 18K will dissipate 11.25W if there is 450V across it. The resistor in the pic seems to be underrated to me. I'd expect to see a 20 or 25 watter there.

     http://el34world.com/charts/Schematics/files/Vox/Vox_ac100.pdf

[J Rindt]

So there is the answer...Thank You.
I see there are no Screen Resistors.
I guess you could add those to get the voltage down instead.
But anyway...thanks for clearing up the mystery. I have not seen as many amps as some of you guys...but I have seen quite a few. I had never noticed  this before, and was curious. It seems like an "unorthodox" way to get things done, but I am not an engineer.
Thanks Again Sluckey

[HotBluePlates]

I'm thinking the purpose of the 18kΩ is to be an "old-school bleeder" resistor, rather than simply to discharge cap voltage on turn-off.

A class AB amp draw significantly less current at idle from the power supply than at full-output. If you were designing such an amp in the very old days, you might be concerned with maintaining a stable supply voltage in the face of the changing current demands (improving voltage regulation). This would be especially true if you used a choke-input power supply, as the choke's inductance drops when only a trickle current is pulled through it, where its rated inductance is manifested when a current closer to its rated maximum passes through it.

Assuming you couldn't spare the space and expense of using a true regulated power supply, you might install a bleeder circuit which draws a significant percentage of full-load current even at idle. If you have a voltage divider and you use such a bleeder, the added change in load current from no-load to full-load is now a smaller percentage of the total current pulled through the divider, and the resulting output voltage stays more-consistent. If you were using a choke, you have now set a minimum current draw below which the load current never drops, which keeps the inductor functioning properly all the time. The downside of this approach (though cheap in terms of circuit components) is that it is wasteful of energy and creates waste heat.

What I think I see (now that we can confirm the 18kΩ isn't a drawing error by viewing the original Vox schematic and a chassis picture) is a 100w amp in which the designer wanted to ensure a steady screen voltage, and who was versed in old techniques. An EL34 might idle at 5-10mA of screen current per tube, rising to 20-25mA per tube at full output. No screen dropping resistors are used to prevent the increased screen current from creating a voltage drop across the screen resistor, which would drop screen voltage somewhat (which tends to clamp plate current, right as you're trying to make maximum output power).

Assuming the screens idle at 5mA per screen, the 18kΩ bleeder draws another 25mA at 450v, for a total of 45mA through the choke. The 100mA choke probably is operating at full inductance under these conditions, and the bleeder resistor accounts for ~56% of the choke current. If each EL34 draws ~20mA of screen current at full power, the circuit pulls 105mA through the 100mA choke (which may be limiting that current rise somewhat), and the bleeder accounts for ~24% of the total load current through the choke. The preamp current draw is insignificant in the face of the screen and bleeder current.

This approach may have made some sense when the AC100 was first designed, and every company was fighting to make the first true 100w guitar amp prior to the advent of good PA systems. It is also an approach built to a price point. I wouldn't recommend emulating this approach in any modern build (unless you're designing a choke-input power supply amp, in which case you should be building a class A amp which doesn't need a bleeder to draw idle current to keep the supply voltage from spiking to a high value).

[J Rindt]

Wow...thanks for the engineering info.
Amazing.....whether it be War, Civil Rights, or Guitar Amps...I guess you have to consider the "tone of the era" when you look back on stuff.
Correct me if I am wrong...This is a Pi Filter...so it "should be" OK to ditch that 18k, and use some (1k5 perhaps) screen resistors in its place.?
And if safety is a concern ...use (100k perhaps) drain resistor across the reservoir cap(s).
Thanks

[HotBluePlates]

This is a Pi Filter...so it "should be" OK to ditch that 18k, and use some (1k5 perhaps) screen resistors in its place.?

Yes. The only change should be a slight reduction in maximum clean power output, and maybe a touch more audible compression as you approach full power.

If you are building your own copy, I'd do as you're describing. If you have an original AC100, I'd leave it as-is. If it hasn't burned up by now, it probably won't.

And if safety is a concern ...use (100k perhaps) drain resistor across the reservoir cap(s).

Fender typically has 2x 220kΩ resistors across series filter caps for 440kΩ from B+ to ground (though these perform 2 functions at once). I'd probably use a 270kΩ-470kΩ from B+ to ground, calculate the current draw for the B+ you'll have, at least double the resistor dissipation to select the proper resistor wattage rating, and call it a day. It doesn't matter to me if it takes 60 seconds before it's safe to grab B+ after unplugging, as opposed to 16-20 seconds (you shouldn't be in a hurry to grab B+ anyway, and should have a eter connected to know the voltage has dropped to a safe value).

[J Rindt]

10-4
Thank You

[sluckey]

Removing the 18K resistor will cause all the downstream voltages to rise. Probably not enough to mess up the magic mojo sound.

[J Rindt]

Removing the 18K resistor will cause all the downstream voltages to rise. Probably not enough to mess up the magic mojo sound.

Good Point.
Probably better to // a couple of 40k 10-15 Watt resistors.
Or look for a  20-25 watt 18k-20k.