Catastrophic Current Limiting in HV Supplies using resistor - Why?

[K1JJ]

I wonder if someone can set me straight on the reason for a current limiting resistor in the B+ of a HV supply?

For example:

I currently have a 100W 10 ohm power resistor in series with my 5KV supply.  At first I thought it would help to limit current in case of a direct short to ground. But doing the calculations, it seems senseless...

At 5KV /  10 ohms = 500 amps. (I= E/R)    500 amps wud be the theoretical current of a direct short, assuming the supply could deliver that much current.  (2.5 megawatts!) Since the supply would sag and blow a breaker way before hitting even 2 amps, the resistor seems like a waste of time.  It wud help a tiny bit, but not much.

In addition, it's a power hog -  when it is normally passing 1A of current, this is 10 watts of wasted power. (I2 * R)

Now if I increase this resistor to 100 ohms, this is STILL 50 amps of limited current during a direct short...  still fantasy.  (250 kw) It would eat up 100 watts of power when drawing 1A.

In the case of using a 1K resistor, this  would limit it to 10A, (50kw) but the normal supply regulation would be terrible, as a result.

What am I missing here? It seems like a big waste of power and effort for nothing.  In this situation, I would think the power transformer's internal resistance  and house wiring is about the only MAJOR thing slowing the current flow during a shorted condition.


** Update:  OR, is this resistor meant to be like a fuse, say a few watts rating and blow apart when the shorted current hits a certain point??
If so, I shouldn't be using big power resistors in there like in the primary for a step start circuit, rather small-sized ones..

** Maybe I'd be better off just sticking with my fine wire fuse. This consists of a few strands of the fine braid wire from RG-213 coax - bridged across two stand off insulators in the HV lead. When a short occurs, it vaporizes into a fast arc.  (The stand-offs are spaced about 2" apart)


73,
Tom, K1JJ

[WA1GFZ]

It is called the Rocky Point resistor. The caps in your power supply will deliver a lot of current if shorted. Also a good idea to limit the wire size off the HV supply. It is used to protect the tube. (grids I suspect)

[WB2YGF]

I wonder if someone can set me straight on the reason for a power resistor in the B+ of a HV supply?

For example:

I currently have a 100W 10 ohm power resistor in series with my 5KV supply.  At first I thought it would help to limit current in case of a direct short to ground. But doing the calculations, it seems senseless...

At 5KV /  10 ohms = 500 amps. (I= E/R)    500 amps wud be the theoretical current of a direct short, assuming the supply could deliver that much current.  Since the supply would sag and blow a breaker way before hitting even 2 amps, the resistor seems like a waste of time.  It wud help a tiny bit, but not much.

In addition, it's a power hog -  when it is normally passing 1A of current, this is 10 watts of wasted power. (I2 * R)

Now if I increase this resistor to 100 ohms, this is STILL 50 amps of limited current during a direct short...  still fantasy.  It would eat up 100 watts of power when drawing 1A.

In the case of using a 1K resistor, this  would limit it to 10A, but the normal supply regulation would be terrible, as a result.


What am I missing here? It seems like a big waste of power and effort for nothing.  In this situation, I would think the power transformer's internal resistance  and house wiring is about the only MAJOR thing slowing the current flow during a shorted condition.

73,
Tom, K1JJ

Under short circuit conditions, a capacitor can deliver a nearly infinite amount of current for a short period of time.  Without any limiting to the peak current, a cap can destroy itself or other components, especially if you get into a rat-a-tat repetitive arc condition.  This current has little to do with the average current the supply is capable of delivering under normal loads. (At least this is the reason we put resistors in series with the HV caps for the HV power supplies we make at my work.)

Think about a xenon flash circuit.  The charging circuit is very low current, but once the capacitor is charged, it can deliver a tremendous amount of current to the flash tube for a short period of time.

[W1EUJ]

So why not fuse it with a small fusing wire? Caps will only discharge partially by the time the wire fizzles.

[K1JJ]

Thanks Guys -

I had just updated the post above and then saw the responses. Look like I was partially wrong and partially right... :-)

Right now there's a fine wire from the diode stack to the caps to protect the diodes.  It has worked so far and haven't lost a stack in years. But based on what you've said, I will add another fine wire "fuse" after the HV caps to the rig. This will protect the cap and rig/tubes somewhat.


I'm gonna install the new fuse now. (2" spaced stand-offs with 2 strands of fine wire from coax braid as the fuse in the HV lead to the rig)



BTW, Frank, is the, "Rocky Point Resistor"  supposed to blow up like a fuse?

T

[WA1GFZ]

Tom,
I would leave the resistor in and add the fuse. AH yes the time the cat shorted the PDM supply....Fuses blow slower than grid wires

[K1JJ]

OK will leave the 10 ohm resistor in for the cap protection and add a fine wire fuse to the rig.

TNX lots!!

T

[WB2YGF]

Tom,
I would leave the resistor in and add the fuse. AH yes the time the cat shorted the PDM supply....Fuses blow slower than grid wires

But based on my calculations in the first post above, why bother using it? Should this be a small resistor to blow up OR a big one like I have now?  A big one doesn't make sense based on the currents I calculated under shorted condix.

Well, it needs to be large enough to handle the steady state dissipation so it can't be too small.  Also I would think it would be preferable to replace a fuse than a resistor, so you would want it to limit the peak current but not be so small that it would be the first thing to fail under a short.  Also 1A may be the normal steady state current, but what is the worst case steady state current it might see?

[K1JJ]

OK, John -

So please clarify...  is the 10 ohm, 100w doing anything if I keep it in series in addition to adding a fine wire fuse, based on the peak currents I calculated during shorts? Or does the fuse pretty much take it's place and do a better job on the output?

Or as Frank suggested, use both the resistor and fuse on the output?  The resistor hogs power, so I just assume to get rid of it if I can.

T

[WA1GFZ]

a fuse is very slow to open.....

[WB2YGF]

OK, John -

So please clarify...  is the 10 ohm, 100w doing anything if I keep it in series in addition to adding a fine wire fuse, based on the peak currents I calculated during shorts? Or does the fuse pretty much take it's place and do a better job on the output?

Or as Frank suggested, use both the resistor and fuse on the output?  The resistor hogs power, so I just assume to get rid of it if I can.

T

It's all about timing.  A resistor is infinitely fast at limiting the current, while a fuse does not limit current at all, but opens after some time delay that is a function of the amount of overload and the characteristics of the fuse element.  I think that's why Frank suggested to use both.

[K1JJ]

Got it.

Well, I guess because a capacitor could deliver almost infinite current, maybe the resistor limiting it to even 500 A is a good thing.  I'll add both.

Just wanted to understand the reasons why, before proceeding.

Thanks again for the fast answers, guys!

T

[N3DRB The Derb]

my reading of it was that you want a limit to the absolute peak joule limit that the filters can deliver. very small slice of time.

[K1JJ]

my reading of it was that you want a limit to the absolute peak joule limit that the filters can deliver. very small slice of time.

Yep, you got it concisely there Derb. Those caps turned out to be the bad MoFo to watch out for.  I just rewired the supply to give a cap rating of 140 uf at 10KV for lots of MaxHeadroom.  That's a lot of jingling jewels.

That's why I'm running scared with these precautions... :-)

T

[K1ETP]

Jingling joules fer sure there, Vu.  A mushroom cloud from Marlboro when those caps let go.

Rich

[N3DRB The Derb]

I put in some NOS Sprague "blue devils" in the Gonsets. wirewounds.  Never use those "cement block" resistors - if you get a discharge they will blow sandy looking crap into every nook and cranny in your chassis.

I wonder if there is modeling software out there to take a given voltage, current, and supply joule capacity that would tell you the optimum resistor size and wattage to limit the timeframe of the peak discharge the best? 

[W1EUJ]

Isn't there a resistance and inductance associated with filter caps that would have a small-but-limiting effect on current upon initiation of the dead short? Wouldn't this keep the current down until the fuse blew?

[John K5PRO]

The resistor is recommended by Eimac, RCA, and ? if I remember correctly. A fuse wire is too slow to save a grid wire inside a tube in case of an arcover. Remember that you have to limit the energy deposited to 5-10 J if you want the small wire grids to remain intact. So, to do this, the arc will have maybe 50 volts across it, then the remaining instantaneous voltage is your power supply - 50, for example, 2350 volts. This is across 10 ohms, leaving 235 instantaneous amps,  until the cap is drained and the breaker kicks or blows a fuse. Besides, if you try and fuse 235 amps at that voltage, you get a plasma across the fuse, and it will continue to burn there until your grid wire is toast first. Without the resistor, take 2350 and divide by .1 or whatever the internal wiring impedance is. You will see that you might get thousands of amps for a some microseconds. 30 AWG wire melts with about 1500 amps through it, for a few hundred uS.
I have been testing this very thing at work few weeks ago, with a big crowbar circuit. My tube requires that a 29 AWG wire be dropped across the 225 uF power of capacitors, charged to 32kV DC, and not melt. I divert the fault current to a parallel device, in this case a large mercury ignitron. I have about 1.5 ohms in series, with lots of cooling air. The wire survives beautifully. So my grid wires will survive if the tube decides to burp, outgas, from Rocky Point effect. In smaller plate supplies, a series R is used without the shunt crowbar device sometimes, as ham equipment usually does.

[W3RSW]

Interesting thread.

Say, a WW resistor also acts as an inductor, even if only a few micro-Henries.  That will affect the leading and trailing edges of a squarewave, in effect, soften your very sudden short's high freq. components in addition to instantaneously limiting the peak surge.

The old designers were not only very technically smart, they were intuitively very canny.

[WA1GFZ]

Yup E=L di/dt

[N4LTA]

The WW resistor acts as an inductor and limits the di/dt. To an infinite step in current, the resistor looks like an infinite impedance.

A fuse is not nearly as safe as you think- Look at the curve on a small standard fuse. Also, look at the interupting current rating. A fuse that can interupt a large current pulse at 5 KV is very expensive.

A 5 amp fuse does not "blow" at 5 amps. Most simple fuses can carry twice the rating for many  seconds.

A small 250 volt fuse will do nothing but explode into an arc at 5 KV.

Proper fusing is a complicated process and involves knowing the time damage curve on what you are protecting, the available fault current, and the voltage across the fuse at the time of short circuit.

[K1JJ]

The WW resistor acts as an inductor and limits the di/dt. To an infinite step in current, the resistor looks like an infinite impedance.

A fuse is not nearly as safe as you think- Look at the curve on a small standard fuse. Also, look at the interupting current rating. A fuse that can interupt a large current pulse at 5 KV is very expensive.

A 5 amp fuse does not "blow" at 5 amps. Most simple fuses can carry twice the rating for many  seconds.

A small 250 volt fuse will do nothing but explode into an arc at 5 KV.

Proper fusing is a complicated process and involves knowing the time damage curve on what you are protecting, the available fault current, and the voltage across the fuse at the time of short circuit.

Good info.

I like the idea of the wire wound and the di/dt  opposing effect. I'm using two large 100w 5 ohm wire wounds in series, so that will help.

For "fuses,"  I use two stand-off ceramic insulators spaced about 2" apart. I bridge them with 2 fine wires from the braid of RG-213. I've used this for year between the diode stack and the filter capacitors to protect the diodes.   I used to lose a diode stack about once a year due to foul ups on my part.  But since adding the wire fuse, I've not lost a stack in ten years.

The fuse disintegrates and I see some black residue on the standoffs showing their was an arc. But the diodes survive every time.  This may not apply for the capacitor bank when discharging into the rig, as has been discussed.  But for now, I have an additional 10 ohms of wire wound resistance and another 2" wire fuse in line to the rig.    At 140 uf / 10KV rating, (at 5kv) those caps can hold some nasty potential.

Later -

T

[K6JEK]

Yup E=L di/dt

I always wondered what YupE meant.  Now it's clear, although I've been spelling it wrong all these years.

[WB2YGF]

  At 140 uf / 10KV rating, (at 5kv) those caps can hold some nasty potential.

Not just potential (V), but more importantly, a stored energy of 1750 J (1/2 CV^2)

By contrast, a 1.2 KW, 5kV supply using high frequency inverter technology we would manufacture at my work, has a stored energy at the output of about 1.2 J.

[WU2D]

That's some high voltage and you have some Joules.

Mike WU2D