Hi all, I am building a supply for a class A amplifier, the bias current is 1.7A per channel at 42Volts. I am hoping to use a CRC Pi filter consisting of 47,000uF R and 47,000uF. per rail. I am thinking of using a 1ohm 25watt resistor, can anyone see a problem with this value, should I try lower or higher values. What are the drawbacks with using this type of filter. Will there be any degradation to sound quality if R is too high.
Thanks for your help.
Alan
Thanks for your help.
Alan
Hi,
Depends on the nature / topology of your class A amplifier,
regarding maximum current demands on the power supply.
BTW a proper Pi filter is CLC, not CRC.
Higher R will reduce the voltage available, but TBH @ 1.7A
around 1R seems about right to be useful but not invasive.
rgds, sreten.
A 5W resistor will do the job, no point complicating it.
Depends on the nature / topology of your class A amplifier,
regarding maximum current demands on the power supply.
BTW a proper Pi filter is CLC, not CRC.
Higher R will reduce the voltage available, but TBH @ 1.7A
around 1R seems about right to be useful but not invasive.
rgds, sreten.
A 5W resistor will do the job, no point complicating it.
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Too large a value for what? Performance in what sense?alibear said:
You need to decide which compromise of voltage drop vs. smoothing best fits your amplifier design. For example, if your amp has very poor PSRR then you need good smoothing and have to tolerate more voltage drop. If your amp has serious second-order distortion then you will have a varying supply current (despite being Class A) and so may need a smaller resistor and have to tolerate more hum.
Hi everyone, thanks for the informative answers. Another question regarding CRC or CLC filters: what is the result if the first C after the rectifier is much smaller than the C after the R or L. For example, first C 10,000uF , final C 47,000uF. What capacitance does the amplifier actually "see"
Thanks
Alan
Thanks
Alan
That depends on the value of the series element. In most cases it will be large enough that the amplifier will 'see' the final capacitor for AC purposes, but will also see that it is fed from a resistance which is a combination of the actual series resistance and the effect of voltage droop on the first (reservoir) capacitor.
Give a man a fish and he eats for a day.
Give a man a GOOD CAD tool and he is too busy to eat for a day.
Duncan power supply designer PSUD2
PSUD2
For ~1.7amps load, 47Ku x 1ohm x 47Ku, 60Hz
42.48 - 42.84V at diode bridge
41.042 - 41.048 at load ripple ~ 6mv, power loss ~ 1.44V
A CLC always performs better than a CRC.
A CRC has "better than calculated" ripple reduction with Class A bias output stages.
I think a 1 ohm resistor is too high and will drop unnecessary voltage, as shown by the 6mv calculated ripple.
Give a man a GOOD CAD tool and he is too busy to eat for a day.
Duncan power supply designer PSUD2
PSUD2
For ~1.7amps load, 47Ku x 1ohm x 47Ku, 60Hz
42.48 - 42.84V at diode bridge
41.042 - 41.048 at load ripple ~ 6mv, power loss ~ 1.44V
A CLC always performs better than a CRC.
A CRC has "better than calculated" ripple reduction with Class A bias output stages.
I think a 1 ohm resistor is too high and will drop unnecessary voltage, as shown by the 6mv calculated ripple.
Agreed, but L must pass high current without saturation, so usually it is expensive.
100 to 200 turns (20 to 30mm OD) of 0.8mm to 1mm diameter enameled wire is not expensive.
ESR does not determine "speed" of a capacitor. ESR limits the current capability.
V/µs in the specification is the maximum speed that the capacitor is capable of.
But the inductance of the circuit to which it is attached generally dominates the HF impedance and determines how fast the capacitor can charge or discharge.
thanks AndrewT, its more clear for me now.
My bad understanding was i believed the first resistor as it is a smoothing one should be "slower" because it is fast pulsed current from the rectifiers and the second as it is the reservoir for modulated current for the load should be "faster" if the final load have only "to see" this last cap.
For sizing : does the L have to support with ampers 8x the current load cusumption for avoid saturation of the ferrite ? for example a self with 8 amper for a circuit maid for 1 amper load ?
My bad understanding was i believed the first resistor as it is a smoothing one should be "slower" because it is fast pulsed current from the rectifiers and the second as it is the reservoir for modulated current for the load should be "faster" if the final load have only "to see" this last cap.
For sizing : does the L have to support with ampers 8x the current load cusumption for avoid saturation of the ferrite ? for example a self with 8 amper for a circuit maid for 1 amper load ?
Right, but that inductor has only a few mH (assuming no ferrite core) and that is not enough for a solid state amp power supply. For a really low ripple, choke with more than 30 mH is needed.
DC current may saturate choke's core, high DC current chokes are not cheap.
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Eldam,
use air cored.
Just wind it your self on a bobbin with side cheeks. Then slide off the cheeks and bobbin for the next one.
The air core will have a bit of resistance creating an rCRC filter. This is a 2pole filter and rolls off the harmonics at 40dB/decade.
The inductance has L creating an rCLC filter. This is a 3pole filter and rolls off the harmonics at 60dB/decade.
The combined filter is a 3pole, still 60dB/decade, rC(L+R)C
and costs a few tens of pence for the wire.
You are not creating a choke input filter which is what Sonce is referring to. A choke input filter inductor, for high current at lowish voltage, turns out to be enormous, possibly as big as the mains transformer.
The transformer secondary circuit includes all the resistors and rectifier diodes as well as the winding in series. All must pass that same current. For a capacitor input filter the charging is always done at a low duty cycle and thus must have high peak currents. The low duty cycle helps reduce the total heating effect, unfortunately not as much heat as the I²R is producing, which heats even more.
The resistance after the first C sees a much smoother long duration current. The heating effect of assuming it is constant DC is probably close enough for sizing the resistor dissipation capability.
use air cored.
Just wind it your self on a bobbin with side cheeks. Then slide off the cheeks and bobbin for the next one.
The air core will have a bit of resistance creating an rCRC filter. This is a 2pole filter and rolls off the harmonics at 40dB/decade.
The inductance has L creating an rCLC filter. This is a 3pole filter and rolls off the harmonics at 60dB/decade.
The combined filter is a 3pole, still 60dB/decade, rC(L+R)C
and costs a few tens of pence for the wire.
You are not creating a choke input filter which is what Sonce is referring to. A choke input filter inductor, for high current at lowish voltage, turns out to be enormous, possibly as big as the mains transformer.
The transformer secondary circuit includes all the resistors and rectifier diodes as well as the winding in series. All must pass that same current. For a capacitor input filter the charging is always done at a low duty cycle and thus must have high peak currents. The low duty cycle helps reduce the total heating effect, unfortunately not as much heat as the I²R is producing, which heats even more.
The resistance after the first C sees a much smoother long duration current. The heating effect of assuming it is constant DC is probably close enough for sizing the resistor dissipation capability.
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I am sorry, my explanation was not clear enough, thanks to AndrewT for pointing it out. I didn't refer to a LC choke input filter but to a regular CLC pi filter, so it is the inductor L not choke L.
Simple CRC filter with R=2,2 ohms has almost the same ripple as CLC filter with L=4.7 mH and internal R=2.2 ohms (air core). L has to be at least 10 mH (internal R=2.2 ohm) for a low ripple in the alibear's case, as simulated by the Duncan's power supply designer PSUD (with L=30mH voltage is super smooth). 10 mH air core inductor with internal R=2.2 ohm would be big and expensive.
Sonce,
you are simplifying your reply to the extent that important lessons are being omitted.
The rCRC and the rC(L+R)C pi filters with the same r and R will have nearly the same ripple when the DC current load is the same.
BUT!!!!!!!!
the shape of the ripple is very different.
The L creates a 3pole filter. It becomes very much more effective at attenuating the high harmonics.
The second harmonic (100Hz or 120Hz) is virtually identical.
But, start comparing the 3rd, 5th, 7th, 9th, 11th .... etc.
The Small value L of the air cored inductor makes a big difference to the high harmonics. It's the reduction of the high harmonic content that changes the "shape" of the sawtooth ripple to a curved ripple.
you are simplifying your reply to the extent that important lessons are being omitted.
The rCRC and the rC(L+R)C pi filters with the same r and R will have nearly the same ripple when the DC current load is the same.
BUT!!!!!!!!
the shape of the ripple is very different.
The L creates a 3pole filter. It becomes very much more effective at attenuating the high harmonics.
The second harmonic (100Hz or 120Hz) is virtually identical.
But, start comparing the 3rd, 5th, 7th, 9th, 11th .... etc.
The Small value L of the air cored inductor makes a big difference to the high harmonics. It's the reduction of the high harmonic content that changes the "shape" of the sawtooth ripple to a curved ripple.
I agree with you, CLC filter has an advantage over CRC at higher harmonics. Amplitudes of higher harmonics are attenuated in both cases, but more effectively with CLC filter, especially with harmonics 5th, 7th and higher up.
But, to be truly effective at 3rd harmonic, inductor L must be big enough.
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