I just got a new trafo, 20-0-20 650VA.
This will give me 16,25A.....
I was thinking to use the MUR860 as you, Carlos, use on you unregulated PSU.
But, I am affraid that the MUR860 could burn
with such amount of current 
Any advise ?
Thanks.
This will give me 16,25A.....
I was thinking to use the MUR860 as you, Carlos, use on you unregulated PSU.
But, I am affraid that the MUR860 could burn
with such amount of current Any advise ?
Thanks.
I haven't done extensive load testing on it yet, but I used MUR1560's and they seem to be holding up fine on a 650VA, 50,000uf per rail snubbered PSU. I'll let you know hot hot they're getting once I start to push it.
Mike
Mike
Thanks Mike.
I will use the 650VA trafo in a PSU with ~70.000uF p/ rail.
This will feed four LM3886 chips bridged, two for each channel.
If you don't have problems, I think that I will not have 😉
I will use the 650VA trafo in a PSU with ~70.000uF p/ rail.
This will feed four LM3886 chips bridged, two for each channel.
If you don't have problems, I think that I will not have 😉
This surely is the king of faq, but nevertheless I'd like to ask: Does the power rating of the 0R47 and 0R1 snubber resistors really need to be that high (2W and 1W, respectively)?
Estimating the dissipation in these resistors I arrive at values of the order of microwatts. Even with very generous assumptions for the values involved, shouldn't ratings of the order of 1/4 be plenty?
I am asking because I am close to finishing the supply, but I have only access to wirewound resistors with high power rating. Since these (potentially) possess a high inductance, I'd prefer not to use them.
Mick
Estimating the dissipation in these resistors I arrive at values of the order of microwatts. Even with very generous assumptions for the values involved, shouldn't ratings of the order of 1/4 be plenty?
I am asking because I am close to finishing the supply, but I have only access to wirewound resistors with high power rating. Since these (potentially) possess a high inductance, I'd prefer not to use them.
Mick
the amount of energy stored by the capacitor (then dissipated by the resistor) is given by the equation:
P = C * F * (V^2) where V is the ripple voltage.
P = C * F * (V^2) where V is the ripple voltage.
I know, this is how I made my estimate of the power dissipation (and arrived at some milliwatts). 😉
Mick_F said:
that's the correct answer for the energy just looking at the low frequency ripple voltage -- here's a snapshot of what is riding on top of (modulating) the ripple voltage -- if the frequency is high enough you have RFI interference which will couple to the power supply leads. The second part of the image shows the effect of putting a snubber across the diode. This is rather an academic excercise:
An externally hosted image should be here but it was not working when we last tested it.
Hi Carlos,
I am about to build this PSU. It is very close to your latest version, but I have made some minor amendments. I have added two caps before the rectifiers as additional filter stage and I have left the grounds of the pos and neg rails separated. I will use four leads to the amp (in a separate case) and join them there at the power ground star.
You think it looks ok?
Mick
I am about to build this PSU. It is very close to your latest version, but I have made some minor amendments. I have added two caps before the rectifiers as additional filter stage and I have left the grounds of the pos and neg rails separated. I will use four leads to the amp (in a separate case) and join them there at the power ground star.
You think it looks ok?
An externally hosted image should be here but it was not working when we last tested it.
Mick
Yes, that's fine.
Just make sure you place the star ground just in the middle of the two channels, and use very short and thick wires.
This means placing the two amp modules close to each other, not one at each extreme of the case.
Enjoy.
Just make sure you place the star ground just in the middle of the two channels, and use very short and thick wires.
This means placing the two amp modules close to each other, not one at each extreme of the case.
Enjoy.
Mick_F, you have missed something here.
C3+C5 together with C7 alone creates a peak (rather small, typ 6 dB) in the impedance and this can be reduced by inserting a small resistor in series with C7.
I'm aware of that Carlos don't agree on that this resonance exsists but this has been shown both by simulations and from real world mesurement by Joseph_K.
Let me ask: What practical electrical function has R5? (I have the answer)
C3+C5 together with C7 alone creates a peak (rather small, typ 6 dB) in the impedance and this can be reduced by inserting a small resistor in series with C7.
I'm aware of that Carlos don't agree on that this resonance exsists but this has been shown both by simulations and from real world mesurement by Joseph_K.
Let me ask: What practical electrical function has R5? (I have the answer)
peranders said:
Of course there's a peak, so what?
As there is a peak everytime you bypass.
What you are saying the is that you should have the same value caps from the rectifier diodes to the chip's pins? That's ridiculous.
You presumed that my intention was to attenuate the peak created by the bypassing, while I always told it's NOT. I'll leave you in the darkness, anyway.
Say what: build this PSU and listen to the amp.
Then make your changes, with the values you want, and listen again.
You take your conclusions.
Thanks Carlos, for your consultation. I will build up the amp very small and symmetric and take care of proper grounding.
P.A.: I am well aware of the measurements by Joseph K and the simulations by Thorsten (it was him, right?). I know that both show the occurrence of a peak somewhere in the MHz regime. On the other hand, the measurements also clearly show that in the audible regime the total impedance is lowered. This is a fact, and I am not sure whether the presence of a high-frequency peak can fold down and eventually lead to perceptible effects in the audio band.
However, this is not the main argument for me to make it. This PSU is easy to build, and I agree with Carlos that, in the end, you have to listen at it. I will just try. Particularly in the case of this PSU, where it is easy to build and try, I find it not very useful to fight out endless theoretical discussions - it makes much more sense to build and to listen. Thats it.
Mick
P.A.: I am well aware of the measurements by Joseph K and the simulations by Thorsten (it was him, right?). I know that both show the occurrence of a peak somewhere in the MHz regime. On the other hand, the measurements also clearly show that in the audible regime the total impedance is lowered. This is a fact, and I am not sure whether the presence of a high-frequency peak can fold down and eventually lead to perceptible effects in the audio band.
However, this is not the main argument for me to make it. This PSU is easy to build, and I agree with Carlos that, in the end, you have to listen at it. I will just try. Particularly in the case of this PSU, where it is easy to build and try, I find it not very useful to fight out endless theoretical discussions - it makes much more sense to build and to listen. Thats it.
Mick
If this peak isn't so important for you then the snubber at the C9 isn't so particulary important either. 99.99% of all amp designer regard this snubber pretty unimportant, me too, not unexpected.
The snubber for normal people is for removing or reducing the impedance peak, but I'll guess I'll continue wandering in the darkness, Carlos.carlosfm said:
So the real purpose of this snubber is... and the contest starts right now.
- Status
- Not open for further replies.