Hi its quite hard for me to find capasitors rated for tube voltages.
The only way i can get em is by taking apart old tv.
I was wondering cap is series lower the capasitance
but do they increase voltage rating
i have ----160V 33uf ----160V 100uf ---
what will my resulting capasitor be
The only way i can get em is by taking apart old tv.
I was wondering cap is series lower the capasitance
but do they increase voltage rating
i have ----160V 33uf ----160V 100uf ---
what will my resulting capasitor be
yngwie said:
1/c = 1/33uF + 1/100uF
I believe the formula is the same like the parallel resistors formula.
If the voltage ratings are equal, you can double your applied voltage. Some people rely on the increasing leakage of the caps with increasing voltage to equalize them, but IMO that's risky and bad design. You need divider resistors across the caps to insure that the voltage divides evenly, lest you overvoltage one or the other. You can also use different voltage ratings, and divide accordingly, but that's a bit too weird 
well i plan to rectify 240VAC ( not sure how high it will result as DC) . So theoretically these should be fine with 320V which i doubt it'll reach. Could you eleborate on this resistor.
Assume i have very little knowledge in such a thing.
Cheers
Assume i have very little knowledge in such a thing.
Cheers
ynqwie,
Rectifying 240V Ac will give you 240 x root 2 = 340V. I hope thats from a transformer and not from the mains - rectifying the mains and using that directly is not only dangerous its actually illegal in this country.
When series connecting capacitors we usually use capacitors of the same value and the same voltage rating. Thats because they will share voltage according to their capacitance value.
Charge q=CV so V = q/C and q =I(current) x time(seconds)
If for example you connected the 33uF and the 100uF from your post in series the charging current through the caps will cause the 33uF to charge to 3 times the voltage that the 100uF will charge to.
Capacitors of EQUAL value will share voltage according to their leakage current.
In order to MAKE them share voltage equally you see balancing resistors across each cap. The balancing resistors MUST have a value such that they will conduct 3 to 5 times the expected leakage current.
If you don't know what the leakage current is there is a "Rule of Thumb" to use which is:
Leakage Current = 0.006CV +10uA
CAUTION: Some caps (cheaper ones and old ones) will exceed that value and you will want to use lower value balancing resistors.
Example:
Your 100uF/160V cap leakage current = 100 x 10^-6 x 160 x 0.006 = 100uA (round figures).
So if (for example) you were going to put 2 of those 100uF/160V caps in series and put say 280V across them you would want bleeder resistors of 140/ 5 x 100uA = 270K across each cap (rounded to nearest value).
Hope this helps.
Cheers,
Ian
Rectifying 240V Ac will give you 240 x root 2 = 340V. I hope thats from a transformer and not from the mains - rectifying the mains and using that directly is not only dangerous its actually illegal in this country.
When series connecting capacitors we usually use capacitors of the same value and the same voltage rating. Thats because they will share voltage according to their capacitance value.
Charge q=CV so V = q/C and q =I(current) x time(seconds)
If for example you connected the 33uF and the 100uF from your post in series the charging current through the caps will cause the 33uF to charge to 3 times the voltage that the 100uF will charge to.
Capacitors of EQUAL value will share voltage according to their leakage current.
In order to MAKE them share voltage equally you see balancing resistors across each cap. The balancing resistors MUST have a value such that they will conduct 3 to 5 times the expected leakage current.
If you don't know what the leakage current is there is a "Rule of Thumb" to use which is:
Leakage Current = 0.006CV +10uA
CAUTION: Some caps (cheaper ones and old ones) will exceed that value and you will want to use lower value balancing resistors.
Example:
Your 100uF/160V cap leakage current = 100 x 10^-6 x 160 x 0.006 = 100uA (round figures).
So if (for example) you were going to put 2 of those 100uF/160V caps in series and put say 280V across them you would want bleeder resistors of 140/ 5 x 100uA = 270K across each cap (rounded to nearest value).
Hope this helps.
Cheers,
Ian
That's not true I'm afriad. It's only valid when the caps have the same capacitance. If not, you must calculate how much voltage each cap gets. Smaller cap gets more voltage than a larger one.Conrad Hoffman said:
Example:
100 uF + 200 uF and 100 VDC => 67 volts across 100 uF and the rest over 200 uF.
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