110v bulb to bleed 24v DC cap?

Here's the LTspice simulation of a 100W, 120V incandescent light bulb empirically derived model that reasonably matches OBW0549's resistance measurements in post #13.
The model does not include any thermal lag that the real bulb would have.
For that reason It also only works for DC voltages, not AC.




Below is the simulation of the 100,000uF cap discharging from 50V through the 100W bulb compared to discharging through a resistor dissipating the same power at 50V.
Note the faster discharge with the bulb due to its non-linear resistance with voltage.

a resistor with the value calculated as 1k per volt (1 milliamp current) will discharge the cap to 37% of the original voltage in 100 seconds, and to about 1% in 500 seconds. a similar thing is done with high voltage power supplies, the 10M resistor built in on microwave oven capacitors discharges the 2500 volts to less than 50V in the time that it takes to remove the topcover of the oven (actually the cap is at less than 50 volts within 5-10 seconds (assuming the resistor hasn't gone open circuit, which is why the cap should always be discharged with a grounded screwdriver).

for a permanent bleeder resistor, always calculate the amount of power the resistor will dissipate, and pick a resistor wattage at least twice the result. a 100,000uf cap charged to 25V can ruin a good screwdriver, so a bleeder resistor will save you money in the future.

so, using the example of 100,000uF@25V, using a bleeder of 2500 ohms, would drain at an initial rate of 10mA, and would discharge the cap to about 1% in 50 seconds. 10mA @ 25V is 0.25W, so a half watt resistor would work here. 10mA is almost nothing to a power supply using such a large capacitor (the charging current for the cap is likely to be several amperes at least), so the bleeder resistor will not have an adverse effect on the power supply (in power supplies for vacuum tubes, a bleeder can draw enough current that it adds some ripple to the output).

as an indicator, you could always parallel an LED and current limiting resistor across the cap to indicate whether the cap is charged or not (this could be another branch that helps discharge the cap, as many LEDs are happy between 10 and 50mA).

I have a mig set that has an annoying feature if you touch the wire to ground you get a spark even without the trigger depressed & long after its been welding, its not big enough to flash you but annoying non the less.
I keep meaning to pull the lids off & see if it can be modded.

dr pepper said:

I have a mig set that has an annoying feature if you touch the wire to ground you get a spark even without the trigger depressed & long after its been welding, its not big enough to flash you but annoying non the less.
I keep meaning to pull the lids off & see if it can be modded.

Click to expand...

Every one I have ever used does the same thing. My thoughts were it is the "stinger/torch" liner, since it is a coil, it becomes an inductor. If you leave it set for a small mount of time it doesn't do this spark.

A wacky power supply I used to troubleshoot after everyone else kept replacing parts yearly, I finally fixed it for good. It had like a 10 M 200 Watt power resistor for the bleeder. I could look up the actual value because the schematic is online. It was a part I replaced.
It was a 30 kW ~15 kV variable power supply at 1.5 AMPs. Most of the other parts were part of a series of 625K resistors that formed the voltage divider for the tube regulator. Some 625K resistors were replaced, but all of the resistors got Locktite 222. The rest of the repair was tightening loose connections and replacing a power transistor in the HV regulator. My predecessor replaced that transistor yearky,
Input was 208 3 phase 60 AMPS.

A unit from a different manufacturer had 90 A, 208 V, 3 phase input as a power source.

I fixed a 1940's 100kV, 0.1 A variable supply basically the same way. It used a Variac to set the voltage.

One system I never had to work on had a 3000 A 6.3 V power supply in it.

Other times I was chasing electrons and measuring currents about 2 pA at +-100 V.

You sure about that shortbus.
The liner is in contact with the wire so is 'live', It usually is insulated at one end so no current flows through it, the spring steel its made of is not that conductive and would melt if it carried the weld current, the latter is carried in a heavy conductor in the torch lead.
Some sets have a burn back timer, if you dont give the set time to burn back this can causeit, but mine doesnt have that function.
The set here at work doesnt do it, but it wouldnt theres no capacitors in it, being 3 phase its doesnt desperately need them.

dr pepper said:

You sure about that shortbus.

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Not at all sure, it's just the only idea I could come up with. But even when I was working and used them feed by 3phase, they still did it. I don't think, again may be wrong, that my Millermatic has caps on the output.

An inductor only stores energy when either its subject to a magnetic field or when it has voltage on it (same thing), when voltage is removed it 'flies back' & looses its energy.
Or something like that.
So I think what your getting is capacitance, or perhaps leakage across a contact.

dr pepper said:

An inductor only stores energy when either its subject to a magnetic field or when it has voltage on it (same thing), when voltage is removed it 'flies back' & looses its energy.
Or something like that.

Click to expand...

an inductor stores energy when there is a current flowing through it. when the current is removed, the magnetic field collapses (which acts like a flywheel tending to keep current flowing) and a) a large voltage is developed across the inductor if the terminals are open circuit, or b) current is sourced into the connected circuits, or c) a combination of a and b. this is the reason for putting a reverse biased diode across a relay coil, to keep the inductive "kick" from blowing up the transistor that drives the relay.

so, with inductors and capacitors you can readily see the difference in how they behave just using an ohmmeter. connecting to a capacitor, the capacitor looks like a short at first, and as the capacitor charges up the resistance reading climbs until it reads open circuit. with a sufficiently large inductor, it acts like an open circuit at first, and then the resistance reading steadily drops until it settles at the DC resistance of the coil

Agreed, I was trying to say something like that in as few words as possible.