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I know it as half the capacitance times the square of the voltage, but don't chokes affect this?
WarmestTimbo in Oz
The Skyptical Mensurer and Audio ScroungerAnd gladly would he learn and gladly teach - Chaucer. ;-)!
'Still not saluting.'
Follow Ups:
You have the eq correct.Chokes will affect how fast that energy can be obtained. You can store all kinds of energy in the capacitor bank, but a choke cannot change it's current instantly.
Cheers, John
Hi.Joule in electricity defines as the energy expended by a one watt load draining power for one second. Put it in perspective, 1 watt-hour is 3,600 joules. This is a electrical energy unit with time factor always.
So how is it applied to a PSU? I want to see an example.
c-J
But maybe you want to talk coulombs which = farads(voltage). One amp is one coulomb per second. Average power in a cap (in watts) is [farads(volts^2)]/2(seconds).Does this answer your question?
Russ
Hi.This is what the poster asked. I asked for a calculation example for its joule value of a given PSU.
c-J
Actually the original post asked for the formula for energy storage in an inductor. That question has been answered, incorrectly by me and then correctly by several other people. What is your problem?
So I wouldn't say you were "incorrect". Truth be told, it was the first time I ever saw you make a "slip". I just felt duty bound to make the small correction.As to what CJ wants....Jeez I gave him any number of ways to relate joules, watts, and time. Now just how joules of storage one needs is a matter of opinion (to some degree).
Russ
Energy in a CapacitorWe know that:
1. Energy = charge × voltage
2. Q = CV.
This second relationship tells us that the charge – voltage graph is a straight line.
The capacitor is charged with charge Q to a voltage V. Suppose we discharged the capacitor by a tiny amount of charge, dQ. The resulting tiny energy loss (dW) can be worked out from the first equation:
dW = V × dQ
If we discharge the capacitor completely, we can see that:
Energy loss = ½ QV
By substitution of Q = CV, we can go on to write:
E = ½ CV2
I'll leave you to do the units ........
Hi.Assuming your dW=V.dQ is correct (relevant to the electrical defination of a joule), how many joules are you talking about a given PSU say 350V with its final cap being 8uF ?
There are tons of similar formula around, my question is how to apply them in the joule calculation to assess a PSU.
c-J
E joules = 1/2 C V 2350 volts, 8 μF
8 μF = 8 * 10 -6 F
E = 1/2 * 8 * 10 -6 * 350 * 350E = (1/2 * 8 * 350 * 350) * 10 -6
E = 490,000 * 10 -6
E = 4.9 * 10 +5 * 10 -6
E = 4.9 * 10 -1
E = .49 joules
Cheers, John
Hi.This makes the didfference btween actually knowing it or just hearsay about it.
c-J
a clue:1 joule = 1 watt.sec
The formula is the same:Joules = 1/2C(E*E) for capacitance and 1/2L(E*E) for inductance. Inductors store energy but the voltage and current phase relationship is opposite compared to capacitors. I thinks this means that the energy storage is not additive or subtractive, it is shared. In a resonant circuit the energy is passed back and forth between the two. Only parasitic losses or intentional loading remove the energy and dampen the oscillation.
but that is not relevant for amplifier filters.What Timbo in Oz wants to know is the energy available instantaneously to the amplifier circuit. This would be the energy stored in the capacitor in direct connection with the amplifier. Any capacitors on the other sides of chokes don't count, as the chokes will prevent rapid change in the current through them.
Choke-input filters require higher transformer voltages for the same output but have the large benefit of preventing rectifier ringing if designed properly. This is more effective than use of HexFRED or similar fast-recovery silicon rectifiers in conjunction with conventional capacitor-input filters.
Kinda reminds you of the old 1/2at^2. When you look at in that light you see inductance is akin to inertia. One resists a change in velocity and one a change in current.IMHO the real reason to use chokes are ripple reduction and increased diode conduction angle. A choke input filter has many important advantages.
Russ
Yep, when one uses a filter cap, one has a one pole R C low pass filter where the r is the sum of the equivalent resistance in the rectifiers, transformer etc and the C.
Adding an L makes it a two pole filter and an LCLC makes it a four pole low pass filter.
The lower you make the corner of that filter or the higher the order, the less AC component can come through.
Best,Tom
and, the resonant behavious helps, too!?Goodoh!
WarmestTimbo in Oz
The Skyptical Mensurer and Audio ScroungerAnd gladly would he learn and gladly teach - Chaucer. ;-)!
'Still not saluting.'
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