On 28 February 2012 21:49, Jim Barber <audioguy@q.com> wrote:
> [forgot to CC this back to the list]
>
> Thanks, Jim.
>
> I believe your comments about heavier bleeder current being a carryover
> are probably true.
>
> I'm going to pencil in four 100k, 20watt resistors with a working
> voltage of something greater than 1KV. I don't know if that would bleed
> 4KV in 32uF off in 2 seconds or less, but it would get there. If I trip
> over any energy storage equations in the next few days I may try to work
> it out just for fun.
>
> Thanks and 73,
> Jim, N7CXI
>
>
The equation is very simple. If the capacitors are initally charged to a
voltage Vo (says 4 kV), then the power removed from the transformer, then
the voltage 'V' after a time 't' will be:
V=Vo E^( t/(C*R))
Where E = 2.718, t is in seconds, C in Farads and R in Ohms. (Or more
conveniently, C in uF and R in M Ohms).
If you use 32 uF and 400 kOhm, the timeconstant is 32*0.4=12.8 seconds.
After 2 seconds, the voltage V will be:
V = 4000 E^(2/12.8)
= 3421 V.
So no, it wont be safe after 2 seconds, but to do so, using a simple
bleeder, will require an excessive amount of dissipation.
BTW, try this on WolframAlpha
http://www.wolframalpha.com/input/?i=NSolve[+50+%3D%3D+4000+Exp[t%2F%2832+10
^6+400000%29]%2Ct]
I've found the time 't' for which the voltage has dropped to 50 V from 4000
V using 400 k Ohms and 32 uF. Just change the numbers to suite your needs.
IMPORTANT. THE IMPLICATIONS OF GETTING THESE CALCULATIONS WRONG CAN BE LIFE
THREATENING. I'VE ONLY WORKED THESE OUT QUICKLY AND MIGHT HAVE MADE A
MISTAKE. USE WITH CARE.
Dave, G8WRB
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