Hi Jim,
They both appear to be reasonable values for this purpose. The 10 uF/100K combination will charge up to the final value about twice as fast as the 22 uF/100K combination, which isn't important, but it won't be down quite as far at 60 Hz as the latter. That difference probably isn't important either, although I can't say that with certainty as I'm not familiar with the internal design of modern digital multimeters.
Specifically, the 3 db bandwidth of the low pass filter (i.e., the frequency at which 3 db of rolloff will have occurred) = 1/(2 x pi x R x C). Let's refer to that as "bw" (bandwidth).
For 10 uF/100K, bw = 0.16 Hz
For 22 uF/100K, bw = 0.07 Hz
The rolloff at 60 Hz will be 1/(square root(1 + ((60/bw)squared))), converted into db based on 20 x logarithm of the resulting numerical value.
(That looks more complicated than it really is; it could be illustrated more simply if it didn't have to be shown as text. Some people depict "squared" as "^2" and "square root" as "^(1/2)", but I'm not sure if for most people that would make the equation more clear or less clear).
For 10 uF/100K that calculates to -52 db at 60 Hz, which would reduce 120 volts to about 0.3 volts.
For 22 uF/100K that calculates to -59 db at 60 Hz, which would reduce 120 volts to about 0.14 volts.
Also, the time to charge to very close to the full value of the DC that is present will be approximately RC x 5, which is how I derived the 10 or 15 second figure for the 22 uF/100K combination.
Best regards,
-- Al
They both appear to be reasonable values for this purpose. The 10 uF/100K combination will charge up to the final value about twice as fast as the 22 uF/100K combination, which isn't important, but it won't be down quite as far at 60 Hz as the latter. That difference probably isn't important either, although I can't say that with certainty as I'm not familiar with the internal design of modern digital multimeters.
Specifically, the 3 db bandwidth of the low pass filter (i.e., the frequency at which 3 db of rolloff will have occurred) = 1/(2 x pi x R x C). Let's refer to that as "bw" (bandwidth).
For 10 uF/100K, bw = 0.16 Hz
For 22 uF/100K, bw = 0.07 Hz
The rolloff at 60 Hz will be 1/(square root(1 + ((60/bw)squared))), converted into db based on 20 x logarithm of the resulting numerical value.
(That looks more complicated than it really is; it could be illustrated more simply if it didn't have to be shown as text. Some people depict "squared" as "^2" and "square root" as "^(1/2)", but I'm not sure if for most people that would make the equation more clear or less clear).
For 10 uF/100K that calculates to -52 db at 60 Hz, which would reduce 120 volts to about 0.3 volts.
For 22 uF/100K that calculates to -59 db at 60 Hz, which would reduce 120 volts to about 0.14 volts.
Also, the time to charge to very close to the full value of the DC that is present will be approximately RC x 5, which is how I derived the 10 or 15 second figure for the 22 uF/100K combination.
Best regards,
-- Al