10-05-2013, 09:45 AM
Don’t know where I got this info from, but when rummaging through my various bits of paper I’ve collected over the years, I came across this formula for working out typical acceptable leakage values to be expected from a cap that has been reformed and is in good condition:
Plain foil: 0.15CV in uA.
Etched foil, 0.05CV in uA
Where C = cap in uF and V = Working Voltage of cap.
Hence, for example:
8 uF cap, 350 VW: 0.15 x 8 x 350 = 420 uA = 0.42 mA
8 uF cap., 450VW: 0.15 x 8 x 450 = 540 uA = 0.54 mA
32 uF cap, 450VW: 0.15 x 32 x 450 = 2160 uA = 2.16 mA
I can’t verify whether this formula and the values it suggests are realistic or acceptable, but it seems to be more or less in line with the notes on the capacitor tester/reformer project that appeared in Radio Bygones, and later in the BVWS Bulletin, which I built and have posted details on the forum state as follows:
Quote:
What leakage current is acceptable? At the end of a successful reforming process, with the voltage at the maximum working voltage for that capacitor, the leakage current should be less than 0.4mA, with current as low as 100uA flowing through it (0.1mA).
End quote.
As to the length of time to reform a cap, the formula suggested in that article was:
T (in minutes) = M +5 where M = total months that the capacitor has remained unused, either in un-powered equipment or left unused ‘on the shelf’. So for example, if a radio is thought to have been unused for 40 years that would be minutes = 40 years x 12 months + 5 = 485 minutes. (A capacitor can’t be ‘over-reformed’ so long as the rated voltage isn’t exceeded and that the leakage current is low).
My thread I posted back in May 2011 on the Radio Bygones Capacitor Reformer project that I built and have found to be very useful can be found here:
http://golbornevintageradio.org/forum/sh...hp?tid=620&highlight=Capacitor+Reformer
Hope that’s of interest, but I'm amenable to any other suggestions as to calculating acceptable leakage values, not that it's keeping me awake at night!
Plain foil: 0.15CV in uA.
Etched foil, 0.05CV in uA
Where C = cap in uF and V = Working Voltage of cap.
Hence, for example:
8 uF cap, 350 VW: 0.15 x 8 x 350 = 420 uA = 0.42 mA
8 uF cap., 450VW: 0.15 x 8 x 450 = 540 uA = 0.54 mA
32 uF cap, 450VW: 0.15 x 32 x 450 = 2160 uA = 2.16 mA
I can’t verify whether this formula and the values it suggests are realistic or acceptable, but it seems to be more or less in line with the notes on the capacitor tester/reformer project that appeared in Radio Bygones, and later in the BVWS Bulletin, which I built and have posted details on the forum state as follows:
Quote:
What leakage current is acceptable? At the end of a successful reforming process, with the voltage at the maximum working voltage for that capacitor, the leakage current should be less than 0.4mA, with current as low as 100uA flowing through it (0.1mA).
End quote.
As to the length of time to reform a cap, the formula suggested in that article was:
T (in minutes) = M +5 where M = total months that the capacitor has remained unused, either in un-powered equipment or left unused ‘on the shelf’. So for example, if a radio is thought to have been unused for 40 years that would be minutes = 40 years x 12 months + 5 = 485 minutes. (A capacitor can’t be ‘over-reformed’ so long as the rated voltage isn’t exceeded and that the leakage current is low).
My thread I posted back in May 2011 on the Radio Bygones Capacitor Reformer project that I built and have found to be very useful can be found here:
http://golbornevintageradio.org/forum/sh...hp?tid=620&highlight=Capacitor+Reformer
Hope that’s of interest, but I'm amenable to any other suggestions as to calculating acceptable leakage values, not that it's keeping me awake at night!
Regards, David.
BVWS Member.
G-QRP Club Member 1339.
'I'm in my own little world, but I'm happy, and they know me here'
BVWS Member.
G-QRP Club Member 1339.
'I'm in my own little world, but I'm happy, and they know me here'
