On Mon, Jan 11, 2016, at 02:00 PM, Gene Heskett wrote:
> On Monday 11 January 2016 13:10:10 John Kasunich wrote:
> 
> > SSRs can fail shorted, I don't think this buys you much in terms of
> > reliability and safety.
> >
> > I think there is a little over-kill going on.
> >
> > Do you really need a fast bleeder (<1 minute to safe)?  Or would a
> > five minute bleeder be OK?
> >
> > My understanding is that you have two 68000uF caps in series, charged
> > to about 130V. (If this is incorrect let me know - the thread is long
> > and rambling and refers to other thread.  I'm basing this on something
> > written in the first message of this particular thread.)
> 
> Actually 10 6800 u-f in parallel, working at their labeled 63 volts 
> ratings.
> 
> But yes, you have the right idea. :)
> >
> > Series caps divide, so you have 34000uF.  Let's say that "safe" is
> > 20V.  So you need to discharge from 130V to 20V, that is to 15% of the
> > original voltage.  Ln(0.15) is -1.9, so it takes 1.9 RC time constants
> > to get to a safe voltage.  Call it 2.0 for easy math.  If you want to
> > get there in 5 minutes, then one time constant is 2.5 minutes = 150
> > seconds. T = RC, solve for R, gives you R = T/C = 4411 ohms.  At 130V,
> > that would draw 29mA and burn 3.8 watts.  Next lower 10% value is
> > 3.9K, that would draw and burn 4.3 watts. I like to derate power
> > resistors by 50%, so you want a 3.9K 10W resistor.
> 
> I actually have, atm, 6 ea 5k 10 watters in parallel clipped onto it 
> right now that I was going to use once the relays get here.  Thats 833  
> ohms which would drain it even faster than just the one. 

The 5K's are across the entire bus, or across each cap?

If across the entire bus, 130V would result in 3.38 watts per resistor. 
Less than the 5W rating, but more than half the rating, which is the derate
I like to use to keep things from getting too hot and ensure long life.

I'd suggest 3 of the 5K across each side of the bus.  At 65V across 5K
each resistor would be running at 0.85W, nice and cool.  Total resistance
of the 3-parallel 2-series arrangement would be 3.33K, RC time constant
113 seconds, bleed-down time from 130V to 20V = 212 seconds = 3.5
minutes.  And free voltage balancing between the two sets of caps.

> And, somewhere 
> on an undermanned galley in the pacific, is a 250 ohm 250 watt resistor 
> which would do the bleed-down even faster, and the current peak would be 
> nominally .5 amps in that case. I can't see that as being capable of 
> welding a relay contact.

Making 0.5A probably won't weld any contacts.  Breaking 0.5A DC on
a relay designed for AC is another story.  And keep in mind that contact
bounce = breaking and making.

I'm just not a big believer in switched bleeders.  Significant hassle and
risk.  If fixed bleeders can meet your discharge time requirements, call
it good and move on.

Context - I design large motor drives for a living.  The vast majority 
used permanent bleeders only.  One recent design does augment them
with a switched bleeder.  But that drive has 90 caps, each rated 6800uF
400V, in a 30-parallel by 3-series array, total of 68000uF.  Nominal bus
voltage is 975V DC.  Stored energy is 32 kilo-joules.  UL requires that
it discharge below 50V in five minutes.  The fixed bleeders would do it
in 10.5 minutes, so we added a switched bleeder to help it along.

Your unit has 34000uF at 130V, for a stored energy of 287 joules.
 
> > If you do indeed have two caps in series, they ought to have balancing
> > resistors anyway, so you could split the bleeder into two sections. 
> > Make each section 2K or 2.2K and 5 watts.
> >
> That too, I was contemplating, but the 51 ohmmer would need to be 
> dismounted to gain access to the pcb bus connecting them all together.  
> Since I have them already 2 each of the 5k's across each bank would seem 
> to be about right.  They are well discharged atm, and I could do that 
> yet this afternoon.  If my math is correct, at 5k and 63.5 volts=0.80645 
> watts per, so if I spread them out on the busses, the heat shouldn't 
> hurt the caps.  

Agreed.  But why not use all 6 of the 5K?

> >
> > Inrush limiting is a separate issue.  Separate resistor, bypassed by a
> > contactor or relay for normal operation.  Put it on the AC side so you
> > don't have any issues with DC contact ratings.
> 
> That is what I am doing right now with the SSR's, one to switch on the 
> power thru that 51 ohm resistor, and a second SSR about 5 seconds later 
> to short out the resistor.
> 
> > The main risk is that 
> > if you try to run with the bypass relay open you can fry your
> > resistor.  One way to avoid that is to make the "resistor" a 100W
> > incandescent light bulb.
> 
> Which rigid tapping would quickly make history.  The spindle turnaround, 
> even if I have it rate restricted, still pumps the supply high enough to 
> blow a C7 bulb in 4 or 5 pecks.  Jon's pwm servo amp is effectively a 
> full 4 quadrant controller, so it dumps motor energy back into the 
> supply when it reverses the motor. I haven't actually measured the 
> voltage as my digital meters aren't near fast enough to catch the true 
> peak. I have about a 1 second rate ramp in the hal file to ease that to 
> about 1 second from 2500 to -2500, without that its positively brutal.
> But thats yet another problem that I should stretch the allowed time on.  
> One setp in the .hal.

We must not be talking about the same thing.  I'm not talking about 
putting a light bulb across the bus as a bleeder, in fact a couple 
paragraphs down I specifically say that is a bad idea.  I'm talking 
about using a light bulb as a _charging_ resistor - in your case, it
would replace the 51 ohm part.

Using a light bulb as a pre-charge resistor does NOT subject it to the
DC bus voltage.  The bulb is in series with the incoming AC line, 
before the rectifier, and can never see more than the AC line voltage.
In addition, during normal operation the bulb is shorted out by the
bypass relay.

> > Using bulbs for bleeders is risky because they run all the time, and
> > when they burn out they leave the caps charged with no indication of
> > the risk.  But using a bulb for charging is different.  The bulb only
> > lights for a few seconds when you first apply power.  If it is burned
> > out the caps won't charge.  If you try to run with the bypass open the
> > light will glow to let you know.
> >
> > A 100W 120V bulb has a hot resistance of 144 ohms.  With 34000uF the
> > charging time constant is 4.9 seconds.  As the bus charges the voltage
> > drop across the bulb goes down, it cools down, and its resistance
> > decreased - that is good, it make the "long tail" of the charging
> > curve go faster.  The inrush during the first cycle or so will be
> > based on the cold resistance of the bulb, but if your switch and
> > diodes can handle a 100W light bulb you know it can handle any size
> > cap bank fed thru that same 100W light bulb.
> 
> Some how, given the quality of 100 watt light bulbs today, I trust this 
> 50 yo 51 ohm 200 watt resister a lot better. And that gives me a T=RC=
> 1.734 seconds, so at 5 seconds, the current delay in then .hal file, they 
> should be well charged.

There are two risks with that approach:

1) suppose Jon's drive develops a shorted IGBT or a cap fails.  Your 51
ohm resistor will dissipate 331 watts with 130V across it.  It is only rated
for 200 watts.  Fine for the 5 seconds or less that a normal charge takes,
but going to get very very hot if something goes wrong and the bus can't
actually charge.  The light bulb is designed to take rated voltage indefinitely.

2) closing the bypass relay 5 seconds after power-on without actually
knowing that the bus is charged is risky.  Again, what happens if there
is a short somewhere and the bus voltage didn't come up?  Huge inrush
when you close the bypass device.  Hopefully all it will do is blow a fuse.
Might toast your SSR and/or diodes.

> I'll go put the 5k's, 2 in parallel per bank of caps, in yet this 
> afternoon & report back.

Good plan.  But while you're up there, why not put three in parallel?

-- 
  John Kasunich
  jmkasun...@fastmail.fm

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