Hoffman Amplifiers Tube Amplifier Forum

Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: tubenit on August 18, 2011, 07:23:48 pm

Title: capacitance in series
Post by: tubenit on August 18, 2011, 07:23:48 pm

Is this correct?

(C1 x C2)  divided by (C1 + C2) = C

(.002 x .00039)  divided by  (.00239) =  .0002928  or  293pf ?


With respect, Tubenit

Title: Re: capacitance in series
Post by: Rich on August 18, 2011, 08:08:40 pm

Just cranking the numbers I get 3.263598

Title: Re: capacitance in series
Post by: Fresh_Start on August 18, 2011, 08:09:39 pm

AFAICT formula is correct, calculation is not.

Chip

Title: Re: capacitance in series
Post by: tubeswell on August 18, 2011, 08:12:58 pm

Quote from: tubenit on August 18, 2011, 07:23:48 pm

Is this correct?

(C1 x C2)  divided by (C1 + C2) = C

(.002 x .00039)  divided by  (.00239) =  .0002928  or  293pf ?


With respect, Tubenit

Yes the formula is correct (except you made a mistake in your calculation)

CT =  (C1 × C2)/(C1 + C2)

I prefer to do it as:

1/CT = 1/C1 + 1/C2 ...

which in your example is 1/.002 + 1/.00039 = 1/.00032 (which if you think about it is how the answer will always be, the sum will be smaller than the smallest contributing value, same as with resistors in parallel)

Incidentally, for the purpose of popping these into a freq roll-off formula, a microfarad is 1 millionth of a farad, so 1uF would be 0.000001F (and  0.002F would be 2,000uF and 0.00039F would be 390uF). But me wonders whether you meant to mean 0.000002F (2uF) in series with 0.00000039F (390nF)?, or 0.0000002F (200nF) in series with 0.0000000039F (3n9F)?, or 0.00000002F (20nF) in series with 0.00000000039F (390pF)?