discharge 440v AC capacitor with a light bulb?

Thread Starter

tbb2

Joined Jun 27, 2014
14
Picked up from:
discharge capacitor with a light bulb
https://forum.allaboutcircuits.com/threads/discharge-capacitor-with-a-light-bulb.163670/

Am I correct in understanding I can connect a 120v 60w bulb to a 440VAC capacitor and a the bulb voltage rating does not apply? Is the only consequence that a 120v bulb will burn out sooner or would the filament actually blow quickly not allowing the capacitor to discharge or somewhere in-between?
 

Hymie

Joined Mar 30, 2018
976
At switch off the maximum voltage across the capacitor will be 600V (440V x √2)

You can then calculate the stored energy from E=1/2 C.V.V

If we assume a 10µF capacitor then the stored energy will be:-

½ x 0.00001 x 600 x 600 = 1.8 Joules

Given that a 60W lamp can dissipate 60 J/s – you would need quite a large capacitor to shorten the expected life of the lamp.
 

Thread Starter

tbb2

Joined Jun 27, 2014
14
Hymie,
Trying to educate myself and not understanding how 600v comes into the mix?
"600V (440V x √2)" ... I imagine (440v times SqRt of 2) but that is 662.
If that is even close ... why?
 
Last edited:

Hymie

Joined Mar 30, 2018
976
My calculator says 440 x √2 = 622V (but I rounded down).

Using 622V in the calculation results in a stored energy of 1.9J rather than 1.8J.
 

Thread Starter

tbb2

Joined Jun 27, 2014
14
My calculator says 440 x √2 = 622V (but I rounded down).
Using 622V in the calculation results in a stored energy of 1.9J rather than 1.8J.
Okay. I am finding E = ½ CV2,
but not having luck finding (440 x √2) ... (capacitor Voltage rating x √2=maxV).
What is this equation?
 

Hymie

Joined Mar 30, 2018
976
The peak voltage of a sine wave is √2 times the rms value – so 440Vac (rms) will have a peak voltage of 622V.
 

Thread Starter

tbb2

Joined Jun 27, 2014
14
The peak voltage of a sine wave is √2 times the rms value – so 440Vac (rms) will have a peak voltage of 622V.
That shows some hits.
If I am getting it, RMS is specific to AC and does not apply to DC.
- Which was the info about capacitors I was hitting.
Vm=√2V≈1.414 ... which is what you said to begin with.
I do like the why.
Thank you
 

neonstrobe

Joined May 15, 2009
153
At switch off the maximum voltage across the capacitor will be 600V (440V x √2)

You can then calculate the stored energy from E=1/2 C.V.V

If we assume a 10µF capacitor then the stored energy will be:-

½ x 0.00001 x 600 x 600 = 1.8 Joules

Given that a 60W lamp can dissipate 60 J/s – you would need quite a large capacitor to shorten the expected life of the lamp.
There are some issues here.
First the danger of 622V - make sure you don't touch both terminals at the same time under any circumstances!
Second - a cold filament has a very low resistance (about 1/10 of normal operating). Surge current will be high.
Fusing depends more on i^2.t and may be a better way of thinking about the discharge.

I'm also amazed that the original thread user was not alerted to the danger of discharging a flash capacitor - its capacitance is typically hundreds of uF and is designed to deliver a high current pulse. Discharging that is serious stuff not to be taken lightly.

In summary - use a suitable resistor rather than a lamp!

e.g. a 1W resistor at 622 v is 390k. Use a resistor rated at 750 or 1000V.
Allow 10 time constants to discharge. e.g. 10uF RC=390k x 1e-5 = 4s therefore with this resistor should allow 40s.
Can also connect a neon lamp with a suitable dropper resistor (1M for standard neon) in series, in parallel with the discharge resistor. That should show whether there is high voltage on the capacitor. Also, can check the cap voltage with a voltmeter?
 

shortbus

Joined Sep 30, 2009
8,586
Probably going to show my ignorance again, but how does a capactior store any voltage of the AC type? DC yes but AC? Don't caps on AC pass the energy through them, not store it?
 

shortbus

Joined Sep 30, 2009
8,586
Depends on the point of the cycle when the lead is 'whipped' off! ;)
I thought of that later. But isn't that kind of a hit or miss? If it was easy to get the *whipping off* right, why is a zero crossing detector so hard?

And wouldn't what ever the cap was working with have drained it, in the time it took to disconnect it from the circuit?
 
Reminds me of the stories my highschool electronics teacher told of playing "capacitor roulette" during his days in the army. Touch an AC capacitor to a 120VAC source and then toss it to one of your buddies.:p

[Don't do this, it's dangerous.]
 
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