Regulated powersupply, will this work?

[Freddie]

I'm about to build a powersupply for a Borbely Jfet buffer (for my buffered preamp, Buffered Passive Preamp).
I was thinking of using LM317/337, but that's no fun.
So I drew this up. But will this work good? I'm sure that there are other solutions that is much better, but I want something simple.
But it has to be reasonably free from noise.

What do you think?

/Freddie

[Freddie]

Some changes made to the circuit, suggested by UrSv.

/Freddie

[Freddie]

This is the Jfet buffer the powersupply is going to be used with.

/Freddie

Time to draw PCB for these circuits...

[jan.didden]

Freddie,

It will work surely, but I would re-think the 10 Ohms at the output. This isolates the regulator from the last cap for practical purposes. For hum etc, the regulator will be visible, but for load-generated ripple and as a low-impedance supply, in effect the supply consists of just one 4700uF cap. Why do yuo think you need that 10 Ohms?

Jan Didden

[Freddie]

janneman,

Why do yuo think you need that 10 Ohms?

I placed it there to get some filtering.

Perhaps I should remove the 10 ohm resistor?

/Freddie

[sajti]

I use same type of regulator for my valve preamplifier. Output is 240V, the series-pass device is IRF840.
As I see the schematic, the 4700uF is too high at the source of the mosfet. It can results some oscillation, or can damage the mosfet. The 10ohms+4700uF filter looks much better. I think You can simply remove the first 4700uF caps.
To reduce the noise, apply some 100komhs+10uF between the zener strings, and the gates of the mosfets. Increase the C4, and C10 to 10uF, and insert 100kohms resistors between the zener, and the capacitor!. This filter will reduce the niose from the zeners, and gives slow start up.
To protect the series-pass device apply one 10V zener between the filter capacitor, and the source of the mosfet. This will discharge the C4, and C10 during switch off.

Sajti

[peranders]

Freddie,see my answer at http://www.hififorum.nu/topic.asp?TOPIC_ID=15586

Freddie, don't forget that your amp isn't a MC preamp, just a plain buffer. The need for extremely smooth voltage isn't that big.

OK, I'll understand if you want to make an overkill (who want not?) but you must state the minimum need first.

[Werner]

Try putting 1-10 R resistors between the diode bridges and the first capacitors.

[peranders]

20000 uF is very much as smoothing capacitance.

1R+1000uF+1R+1000uF etc a couple of times has very good effect on filtering ripple. The resistor value can be in total up to 10-20 ohms depending on load.

[jan.didden]

Originally posted by Freddie
janneman,
I placed it there to get some filtering.
Perhaps I should remove the 10 ohm resistor?
/Freddie

I guess one wants basically two things from a regulated supply:

One, to supress any chance of hum and noise coming in from the mains side. Your supply, up to the two 10 Ohm resistors, should do a good job as one would want. The 4700 at that point should not be a problem, but 470uF should work Ok as well, and ease the switch-on load on the fets.

Secondly, one would want a low output impedance. Why? As the load current varies, you want the supply voltage to be rockstable. As the fet output is kept at a stable voltage with the zeners, whatever current you draw at the load, it also meets this second requirement (within reason). Now you put in the 10 Ohm. Suppose your load current varies by 100mA. Over that 10 Ohms that is a voltage variation of 1 Volt. So, now you have a perfectly stable voltage at the fets, but a hughe ripple (1V) at the load. You don't want that!

(Yes I know the second 4700uF eases this somewhat, but that is putting the horse behind the cart).

I think you mix up the use of the series resistor BEFORE the regulator, to attenuate some of the noise and hum before it gets to the load. This may be a good idea for unstabilised supplies, but once you decide to use a regulator, it is unnecessary.

Remember: ANYTHING you add after the fets (resistors, coils, badly soldered wiring) will spoil the regulator low output impedance.

Jan Didden