resevoir cap value

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Q = CV

Differentiating:
dQ = C.dV

For an f Hz, V Volts mains supply:

dQ =~ (1/(2f)).I ..... where I is current

I.1/(2f) =~ C.dV

dV =~ I/[2fC]

dV is the approximate ripple. How much one allows depends on PSRR and CMRR of the amplifier and the OWNER.
 
tks, there's a lot to read i'll have a look later

these are from the schematic for my amp, its a big old mosfet, which I love the sound of

it barely get warm at listening levels

and blows the rcd with inrush!!!

its rated 450w x2 = 900 watt into 4 ohms

yet it says it can drive 2 ohms

280w x2 into 8 ohms

on the psu

its a 1.5 kw(kva? ) transformer

so that' makes a 6 amp fuse or therabouts (230 volts)

yet the fuse is 15 amps (for 110 volt use?)

its quite old and will need re-capping

is the fuse correct

and what size resevoirs do i need, stock is 15,000uf

I was thinking of upping to 22k or even 47k, but is it warranted or desirable, please, its not that much extra cost

if i do, its it likely to blow the bridge, or anything else internal?

100 volts mandatory, 80v rails, I could go 100,00uf at 80v!!!!!!!!!!??????????

thanks
 
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PRR

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Joined 2003
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...If the amp trips your RCD, then there is leakage of current to earth.

Older GFIs (and RCDs) could trip for lack of exact balance in the difference coil. A 50mA threshold with a short 50A surge demands 0.1% balance, a tough figure. "Ordinary" loads do not show the huge peak of a hi-fi capacitor input.

Resistance IS the answer but people resist it.
 
It gets very complicated with amplifiers as most of the time they aren't on full power or if they are its music signal not sine or square waves.
This is why some manufacturers start getting into PMP etc.
Some of the cheap Chinese stuff is terrible for rating many hundreds of watts but only use a small transformer and caps.

Q=CV=IT Q=charge, C=capacitance, V=voltage, I=current, T=time.
I tend to use worst case which is a square wave.
Then you need to decide on max ripple allowed and use formula back to work out capacitance.
 
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Joined 2002
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Rule of thumb for the math-challenged:

A 1A current discharges a 1F cap in 1 sec by 1V, giving you 1V ripple.
So it discharges a 1000uF (1/1000F) cap by 1000V; but the discharge time in a 50Hz mains is only 10ms (1/100 sec) so you are back to 10V ripple.
Up your cap from 1000uF to 10,000uF, ripple down to 1V.
But if your amp draws 5A current, you're back to 5V ripple.

Roughly.

Jan
 
Animosity? Let's see! :D

Applying the 'basic Physics', C = It/V, couldn't be simpler - I even gave a worked example - what more is needed?

I don't agree that Jan's 'rule of thumb' helps the mathematically challenged at all - sorry Jan!

I could go on endlessly... ;)
 
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