Is it possible in a CRC to have a filter resistor with such low resistance that it has a negative effect on the power supply? I've a 0.1 Ohm 3W resistor with 630 mA running through it as an external single rail 5V PSU powering a DAC. So it's only dropping 0.06V and the power is way more than required. I find that with the mids and bass lack body and weight with the resistor in place, but are normal when the resistor is replaced by a straight wire. (Vishay RS02B)
Having R = 0 means there is no R, just all the supply Cs are in parallel.
Then there is no filtering of the ripple voltage, which may or may not matter to you.
If you like it better without an R, that's fine.
Then there is no filtering of the ripple voltage, which may or may not matter to you.
If you like it better without an R, that's fine.
I can appreciate that rayma, yet my R is not zero but 0.1. Are you indicating that the R value has direct bearing on how much ripple is reduced? Also since this is not a class A amplifier but a DAC, does that mean CRC is not likely to be useful and a decently rated Cap should handle that lower ripple? Obviously speaking in general terms since I haven't actually measured the ripple (no facility).
The DAC should have internal voltage regulators to isolate the circuits from the 5V supply.
If there are no internal voltage regulators, then it just relies on the op amp supply rejection.
Since the DAC is intended to be used with a simple 5V supply, an added series R might
have some unintended effect. Better to improve the supply itself, than to follow it with RC filtering.
A good supply should provide a low source impedance, which the RC filter will disrupt.
If there are no internal voltage regulators, then it just relies on the op amp supply rejection.
Since the DAC is intended to be used with a simple 5V supply, an added series R might
have some unintended effect. Better to improve the supply itself, than to follow it with RC filtering.
A good supply should provide a low source impedance, which the RC filter will disrupt.
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In what sense do you mean "follow" the supply with RC filtering, since the CRC comes between the bridge and the voltage regs? I guess if you usually design unregulated power supplies, the CRC would follow. Is that hwo you are refrring to it?
I also guess that the amount of ripple voltage in the first cap is the amount of voltage you want the resistor to drop, which if correct answers my questions in post 3.
I also guess that the amount of ripple voltage in the first cap is the amount of voltage you want the resistor to drop, which if correct answers my questions in post 3.
Maybe you should post the schematic of the power supply that you are using.
If the R is before the regulator, it can cause drop outs when the AC line is low.
The DC voltage drop across the resistor is different from the ripple voltage.
Most of the ripple voltage will be dropped across the R (as long as the capacitors are large enough),
regardless of the actual value of R. But too large R will drop too much DC voltage and waste power.
If the R is before the regulator, it can cause drop outs when the AC line is low.
The DC voltage drop across the resistor is different from the ripple voltage.
Most of the ripple voltage will be dropped across the R (as long as the capacitors are large enough),
regardless of the actual value of R. But too large R will drop too much DC voltage and waste power.
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The regulation circuit needs 9V input to operate properly and I was only shaving of 0.065V so assume it should not cause a problem that way. No worries, I don't have an easy answer and was just experimenting to see if CRC before the regs improved the subjective output rather than just a simple C. I think I'll just pull the resistors and be happy with what I had before.
Your description of the filter resistor's relation to ripple is what I had assumed before, but suddenly thought I'd gotten it all wrong. Thanks for getting me back on track. And thank you for your time and patience in addressing my questions.
Your description of the filter resistor's relation to ripple is what I had assumed before, but suddenly thought I'd gotten it all wrong. Thanks for getting me back on track. And thank you for your time and patience in addressing my questions.
Power supplies are rather more complex than might be expected, and it's not so easy to describe them
in general terms. Best to pick an approach, and work out the details from there.
If the input is 9VDC, then the regulator drops 4VDC. With the 0.63A current, that's dissipating 2.5W in the regulator.
A low dropout regulator would let you cut that in half, by using a lower input DC voltage.
in general terms. Best to pick an approach, and work out the details from there.
If the input is 9VDC, then the regulator drops 4VDC. With the 0.63A current, that's dissipating 2.5W in the regulator.
A low dropout regulator would let you cut that in half, by using a lower input DC voltage.
In the end, the CRC topology did not seem to work well for the particular circuit it was feeding. When I removed the pretty good resistor, the sound imporved noticeably. Ah well, I'm happy with the results.
Hey - thanks for returning to the thread after a period and updating it with your findings; thats a really helpful thing to anyone who might search in future - and often doesn't happen.
Folks,
I'm going down a somewhat similar route for my DAC supply: The transformer's nominal output is 7V, rectifier is an LM4320 based NMOSFET rectifier, followed by a 4x 3900µF C-Bank which I intend to break in half and make it a CRC configuration to reduce Ripple. This would then feed a 5V Low Drop regulator board in the DAC box. Total max. current draw on the CRC supply is ~1.5Amps.
My question: How do I select or calculate the correct Resistor for the CRC circuit??
Thanks and Regards,
Winfried
I'm going down a somewhat similar route for my DAC supply: The transformer's nominal output is 7V, rectifier is an LM4320 based NMOSFET rectifier, followed by a 4x 3900µF C-Bank which I intend to break in half and make it a CRC configuration to reduce Ripple. This would then feed a 5V Low Drop regulator board in the DAC box. Total max. current draw on the CRC supply is ~1.5Amps.
My question: How do I select or calculate the correct Resistor for the CRC circuit??
Thanks and Regards,
Winfried
Seems like a pretty slim margin. Maybe there's a few tenths of a volt left, so possibly 0.22 ohms would work.
Suggest you breadboard the circuit with full loading, and see how it behaves with varying AC input.
Suggest you breadboard the circuit with full loading, and see how it behaves with varying AC input.
@rayma
Thanks a lot for your breadboarding suggestion with full loading which I will follow. I don't have the ability to vary AC supply, but will see the "nominal" margin this way.
I'm missing the dimensional rationale behind the .22Ohms suggestion. How would I calculate a value?
Thanls and Regards,
Winfried
Thanks a lot for your breadboarding suggestion with full loading which I will follow. I don't have the ability to vary AC supply, but will see the "nominal" margin this way.
I'm missing the dimensional rationale behind the .22Ohms suggestion. How would I calculate a value?
Thanls and Regards,
Winfried
I think that's the most resistance that you could insert, without dropping out of regulation with normal variations.
Thanks for the feedback!
The 5V Supply is LT3042 + Transistor based.
The rectifier drives 4x 3.9mF or 2x3.9mF + R + 2x 3.9mF .
Regards,
Winfried
The 5V Supply is LT3042 + Transistor based.
The rectifier drives 4x 3.9mF or 2x3.9mF + R + 2x 3.9mF .
Regards,
Winfried
Ok, then at full loading you should have at least 1.5VDC across the regulator, to allow for AC line sag, and for rising capacitor ESR.
More, if possible.
Sincere Thanks! I'll first build the supply without the resistor and look at the DC Margin. At some point (when the target system runs) I may get back to the CRC configuration and have a listen if there's an audible improvement anyway 😉 I had originally thought that CRC would be technically advantagous in any supply config, incl. regulated supplies.
Regards,
Winfried
Regards,
Winfried
A RC noise filter before the regulator is a great idea as a normal regulator runs out of gain at high frequencies.