Reducing rectified output voltage

[JBourke]

Hi all,

I am building a bench power supply from scratch based on a step-down transformer, rectifier bridge, smoothing caps and a constant current/voltage DC step down buck converter module I bought off the Ebay.

Without any load connected I get about ~42V output as the caps are charged to peak value. However the converter module has a maximum input voltage of 40V.

Am I right in thinking I could add a suitable 40V Zener diode to reduce the rectified output to a safer level? Or are there any other ways of reducing the output voltage (preferably without generating lots of heat)?

Thanks!

John

[tautech]

A few tricks come to mind.....
Has your transformer any primary side taps, say a 250V one when our mains is ~235V will reduce the secondary voltage a little.

Or try 2 x 24V auto bulbs as a load before the regulation to see how much a small load pulls the secondaries down.
If you can get the supply below the regulator max, then measure what the regulator's quiescent current is at various settings, that alone might be enough.

Failing that a wirewound resistor before the regulator might be needed as a permanent addition.

Maybe even a light dimmer before the transformer to control the input voltage but that's something I'd want to check carefully with a scope before connecting the regulator.

[crazyguy]

add a simple pre-regulator stage

[crazyguy]

Maybe even a light dimmer before the transformer to control the input voltage but that's something I'd want to check carefully with a scope before connecting the regulator.

I think a light dimmer is not a good idea for this application, the output of the dimmer to the transformer will be no longer a pure sinewave

[rstofer]

You didn't say how many milliamps or amps the PS is supposed to provide.  Any way you do it, when you drop voltage, you give off heat.  Some schemes are more efficient (less heat) than others but, basically, you wind up building a heater.

A silicon diode (1N4001 - 1 Amp) drops about 1.1V so 3 or 4 in series will drop quite a bit.  But it's still heat!  Each diode is generating about 0.7V * <whatever current in amps> -> power in Watts.

http://www.vishay.com/docs/88503/1n4001.pdf

Pay attention to the peak repetitive current.  When you put a LOT of capacitance at the output of the rectifiers, the cap voltage stays pretty high.  This means that the diodes only conduct when they have a voltage higher than the capacitors and that conduction angle (in degrees, converted to time) might be pretty short.  Infinite capacitance would result in a 0 degree conduction angle.

But again, without knowing how much current is desired, it's hard to recommend a solution.

[jeroen79]

Why not get a more suitable transformer?
Or a switching AC-DC converter.

[JBourke]

Really appreciate the feedback everyone. Potential amps capability is about 6A before the buck module. Module itself is max 3A rated but it wouldn't last long with the puny heat sink it comes with. Given the component layout fitting a bigger heat sink will be a interesting problem in itself :-). Although the transformer etc may be a bit overkill at this stage I am planning to add a second buck module later to get to a dual power supply.

I just hooked it up and including the fan and the power indicator output voltage hovers around 38.5V but it's evening time here in New Zealand and mains are usually a bit lower than during the day.

Any takers for a zener diode approach or is that not advisable?

Cheers,

John

[Zero999]

add a simple pre-regulator stage

I would use a Darlington pair, omit the zener and make the capacitor quite large. This will drop a few voltage and improve the ripple rejection.

[Arjan Emm]

Does your transformer allow for winding a few extra turns on it? If you wind them the "wrong" way around, each turn should reduce the voltage by a fixed amount. This is with very little losses, just the losses in the extra length of wire.
If the secundary is on the outside, you could also remove a couple of turns ofcourse.

[Brumby]

Any takers for a zener diode approach or is that not advisable?

Your mention of a 40V zener indicates a very basic arrangement.  Without a diagram to show us how you would use it, it's difficult to give a definite answer,  but I suspect it is not likely to be practical assuming it would work at all.


My first suggestion would be to throw two 1N5404's in series, which will drop you the couple of volts you are worried about and they will take 3A.  It's not sophisticated, but it should work.

[jeroen79]

Any takers for a zener diode approach or is that not advisable?

It would not be energy efficient.
A zener would clamp the voltage by shunting away current from the buck converter.
It would then either draw enough current to reduce the transformer's output or draw current through a resistor to drop the voltage.
In the first case the current would heat up the zener, in the second case the resistor would also drop voltage when the buck converter draws current.

[Brumby]

Let's say you use a 40V zener as a simple shunt regulator.  If it needed to shunt 1A to bring the voltage down, then it would dissipate 40W.  If 2A, then 80W.

I don't know of any single zener that could do that - and if there was, it wouldn't be cheap.

Irrespective - that's a really poor idea.

[Arjan Emm]

As Brumby pointed out, all the solutions dissipate possibly a lot of power. Since this a one off, just alter the transformer. Not a suitable solution for mass production, but easy and lossles in this case, unless its a potted transformer, wich makes changes more or less impossible.

[KerryW]

You could add a small transformer that opposes the primary or secondary of the main transformer.  That would generate almost no heat.

[K5HJ]

Adding a choke between the rectifier and filter capacitor will reduce the voltage.  Look up "choke input filter."
Might be kinda large for your current requirements.

Randy