Probably leaky paper caps. They tend to absorb moisture.
You can check them with a DC supply more than 80V and a neon light. Connect the neon in series with the capacitor and connect up the power. The neon will blink once as the capacitor charges to somewhere around +V, minus 10 to 60V (depending on the neon light and amount of current pulsed). If it blinks at all within thirty seconds, the cap is leaky and should be discarded. (At 1uF and around +120VDC, this represents 30Mohm. 100V across such a leakage followed by a 220k will put 0.7V on the output, a generous portion of an audio signal.)
Tim
You can check them with a DC supply more than 80V and a neon light. Connect the neon in series with the capacitor and connect up the power. The neon will blink once as the capacitor charges to somewhere around +V, minus 10 to 60V (depending on the neon light and amount of current pulsed). If it blinks at all within thirty seconds, the cap is leaky and should be discarded. (At 1uF and around +120VDC, this represents 30Mohm. 100V across such a leakage followed by a 220k will put 0.7V on the output, a generous portion of an audio signal.)
Tim
FWIW, if you can measure the peak voltages in such a circuit, the exact formula for determining average leakage resistance is:
R = t / [C * ln (Vs-Voff / Vs-Von)]
Where R is the effective leakage resistance, C is the capacitance in farads, t is time between blinks in seconds, Vs is supply voltage, Von is the turn-on (highest peak) voltage of the neon and Voff is the minimum (lowest peak) voltage of the neon, when it turns off.
The natural log comes from taking the inverse of the RC exponential curve.
Typical values I would expect:
Vs = 200V, Von = 80V, Voff = 40V, C = 0.1uF (0.0000001 F), t = 60s, R = 2 x 10^9 ohms (2 gigaohms).
At that rate, some careful watching of a VOM should give accurate enough voltages for high precision with this formula.
Oh, and if you aren't using those caps for much DC, or very slow AC (low frequency oscillator for example), the leakage won't matter much. They'd probably be great for speakers.
Tim
R = t / [C * ln (Vs-Voff / Vs-Von)]
Where R is the effective leakage resistance, C is the capacitance in farads, t is time between blinks in seconds, Vs is supply voltage, Von is the turn-on (highest peak) voltage of the neon and Voff is the minimum (lowest peak) voltage of the neon, when it turns off.
The natural log comes from taking the inverse of the RC exponential curve.
Typical values I would expect:
Vs = 200V, Von = 80V, Voff = 40V, C = 0.1uF (0.0000001 F), t = 60s, R = 2 x 10^9 ohms (2 gigaohms).
At that rate, some careful watching of a VOM should give accurate enough voltages for high precision with this formula.
Oh, and if you aren't using those caps for much DC, or very slow AC (low frequency oscillator for example), the leakage won't matter much. They'd probably be great for speakers.
Tim
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