adding some diodes to the input of a 3 terminal regulator

you can, it will just move some of the dissipation from the regulator to the diodes

-Lasse

On a sunny day (Sun, 4 Aug 2013 09:59:25 -0700 (PDT)) it happened snipped-for-privacy@gmail.com wrote in :

That thing has a dropout voltage of about .5V it seems. So stay above 6 on the imput, do not forget bottom ripple... But it would ruin line regulation....

Maybe get the TO-220 version and / or a small heatsink?

Else use a switcher.

How come you can't use a heat sink?

** To avoid heat, you could put a few NiCd or NiMH cells in series with the supply and reg IC and charge them.

.... Phil

I was trying to find a clever way to cheat. It sounds like the heat has to be dissipated, if not in the 3 terminal regulator than in the diodes (so I might as well stick with just the regulator). Next time I'll design a switching regulator into the board!

Can you get to the AC input? Probably not, but if...

Use a nonpolarized capacitor in series between the secondary and the bridge. Calculation is beyond my ability, but experiment until you have enough overhead to feel comfortable. Speaker crossover caps work fine, and can be paralleled for low ESR and proper value. My starting point; I'd start with 40 uf and see which way to go. Mikek

Maybe I missed something. How would adding diodes to the input of the regulator ruin line regulation? A resistor might make some impact since it would produce a change in voltage with a change in current and some of that would appear at the output, although not so much really. But a diode is essentially a constant voltage device once biased on with very small changes in voltage drop with large changes in current. So how would this mess up line regulation as long as you keep the voltage above the dropout requirement?

--

Rick

If you are using a TO-220 type device with leads, you can replace it with a switching regulator in a similar package. CUI makes them, sold by Digikey, for most output voltages including +5v of course. I use them in a test fixture and they seem to work well, even with a lower dropout voltage than the spec requires. I'm running 5 and 12 volt outputs from a 15 volt input, saves a lot of power.

--

Rick

Somewhat, this depends on the source: is that a regulated, or just a filtered 9-12V? To shed heat outside the regulator, you could either put a resistor between the source and regulator input, or a resistor/capacitor parallel pair, if surges are expected. A 1W diode and its mounting/heatsink is more expensive than a 1W resistor and its mounting. The voltage drop on the resistor has a higher dV/dI than an equivalent string of diodes, though.

Thinking outside the box, you could go to a lower-current regulator and a PNP transistor, with some current-sharing resistors, and move most of the heat dissipation into the transistor (assuming the transistor can take higher temperatures than the IC regulator). There's also the possibility of adding a choke to the AC into your source rectifiers, which will drop V without dissipating power.

Some three terminal regulators are quite sensitive to any input resistors, since they start to oscillate. To eliminate this, a big capacitor directly at regulator input to ground is required.

Put the resistor/diode/choke between the rectifier bridge and the main reservoir capacity, so the capacitor is next to the regulator input.

Choke input power supplies (or C-L-C) were popular in the tube rectifier era, since tube rectifier could not handle huge peak currents into a simple big capacitor. The choke in L-C or C-L-C configuration extended the conduction angle and the average DC voltage was close to the secondary RMS voltage.

Unfortunately the choke at 100/120 Hz needed to be big and heavy and needed to also handle DC current (air gap), so it is understandable that chokes are seldom used these days.

Total heat dissipation will be the same; all that would do is distribute the heat sources. Want more efficiency and less heat generation? Use a switcher.