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19th April 2018, 02:45 PM  #1 
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Join Date: Sep 2009
Location: Catalonia  Europe

Electrolytic capacitor for mains PSU
What are the best specs have to have the electrolytic capacitor for the best SQ, the same for tubes or SS, the same for preamp or poweramp?
TIA Felipe 
19th April 2018, 03:58 PM  #2 
diyAudio Member
Join Date: May 2007

Correct capacitance value. Lowish ESR. Adequate ripple curent rating (if used as a reservoir cap). Good toleration of temperature.

19th April 2018, 04:00 PM  #3 
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Join Date: Sep 2009
Location: Catalonia  Europe

Thank you, resuming low ESR, high ripple & high temperature.
There is a relation between the current draws & the amount of capacitance? I ask because normally for tube preamps drawing 60mA I used two caps 100uF each & was enough but when used the same PSU for an OTL headphone amp drawing 140mA I had to increase to 660uf (3 caps 220uF). 
19th April 2018, 04:15 PM  #4  
diyAudio Member
Join Date: Apr 2011

Quote:
For DC current, I = C x delta(V)/delta(t). Decide what % ripple is needed for your given supply voltage, and then you know delta(V). Also, the supply refresh rate delta(t) is 1/120 seconds (for 60Hz). 

19th April 2018, 05:26 PM  #5 
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Location: Catalonia  Europe

Thanks for support rayma.
C = I / (delta (V) / delta (t)) ??? 
19th April 2018, 05:38 PM  #6 
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Lifetime is a good thing to look at too.
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19th April 2018, 05:46 PM  #7  
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Join Date: Apr 2011

for a
Quote:
Then rearranging, we have Q = C x V. Differentiate both sides wrt time: dQ/dt = (C x dv/dt + v x dC/dt) We know that dQ/dt = i, the current. Normally C is considered to be a constant, so dC/dt = 0. Then we have i = C x dv/dt For DC we can use for dv = delta(v), the pp amount of ripple voltage between rectifier recharge pulses. For a FWR this is 1/120 second for 60Hz, or 1/100 second for 50Hz. The i is now I, the amount of constant DC current drawn. Then C = I / (ripple voltage/recharge time), or C = I / (delta(v) / delta(t) ) So we have C = I x delta(t) / delta (v), which is the capacitance needed for the desired amount of ripple delta(v), neglecting the ESR. Last edited by rayma; 19th April 2018 at 06:10 PM. 

19th April 2018, 06:10 PM  #8 
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Join Date: Sep 2009
Location: Catalonia  Europe

Please could you do the maths for a known current, for example 100mA?

19th April 2018, 06:13 PM  #9  
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Join Date: Apr 2011

Quote:
This would typically be less than 5% of the DC supply voltage. Also, what's the rectified DC supply voltage? For 8.5V ripple, 100mA DC current, and 60Hz operation, we have: C = I X delta(t) / delta(v) and then C (in farads) = 0.1A x (1/120 Seconds) / (8.5V), so C = 98uF. Use a 100uF, 400V capacitor. This will be a problem for a tube rectifier. Last edited by rayma; 19th April 2018 at 06:41 PM. 

19th April 2018, 06:26 PM  #10 
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Join Date: Sep 2009
Location: Catalonia  Europe

B+ 170V =8.5V ripple
Rectified voltage 338V 100mA 60Hz Mine calculations: 0.14A x (1/120 seconds) / 8.5V = 16.1476uF Now the question: This is for the 1st input mains capacitor or the total capacitance of the mains PSU? 
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