what should the value be per 100 watts
and how do designers come to a value
does it really matter to oversize them
it just seems what other amps do and a rough rule of thumb, tks
and how do designers come to a value
does it really matter to oversize them
it just seems what other amps do and a rough rule of thumb, tks
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.
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.
C = Q/V = (I t)/V
Where:
I = load current
t = discharge time 1/50Hz (20ms) for half wave rectifier and 1/100Hz (10ms) full wave rectifier, or 1/60Hz and 1/120Hz for countries with 60Hz mains.
V = capacitor ripple voltage
Where:
I = load current
t = discharge time 1/50Hz (20ms) for half wave rectifier and 1/100Hz (10ms) full wave rectifier, or 1/60Hz and 1/120Hz for countries with 60Hz mains.
V = capacitor ripple voltage
Example calculation for a load current of 10A, a ripple voltage of 5V, a full wave rectifier and a 50Hz supply.
Firstly: t = 10ms = 10/1000s = 0.01s
Then: C = (I*t)/V = (10*0.01)/5 = 0.02F
Lastly: 0.02F = (0.02*1000000)uF = 20,000uF
Firstly: t = 10ms = 10/1000s = 0.01s
Then: C = (I*t)/V = (10*0.01)/5 = 0.02F
Lastly: 0.02F = (0.02*1000000)uF = 20,000uF
Down load "Motorola AN1308" by Andy Hefley, it's about amp design, a 100W and a 200W. Yours questions should be answered.
Craig
Craig
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
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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Please post the schematic or, at least, tell us the make and model of the amp.
If the amp trips your RCD, then there is leakage of current to earth.
If the amp trips your RCD, then there is leakage of current to earth.
The short answer is to replace the reservoir capacitors with ones of the same capacitance, DC voltage and ripple voltage ratings.
Increasing the capacitance can put undue strain on the transformer and rectifiers.
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.
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.
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
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
^^^^^^^ this back to basic Physics approach is all that´s needed and should be a sticky.
As a side benefitial effect it would avoid endless threads and member to member animosity. 😛
As a side benefitial effect it would avoid endless threads and member to member animosity. 😛