Quick DC-DC Question - 35V to 12V?

Can anybody let me know the easiest way to drop 35V DC to 12V DC? All I need is 85mA. I could easily drop that load to 55mA by removing three status indicator LEDs if that will make the circuit more efficient. Continuous load is only 18mA which could be dropped to about 10 by removing an indicator LED.

Would a resistor be more efficient or a 78L12 regulator? It's 5:50am and I've been up all night so my brain isn't working as it should.

Thanks, Dave

Reply to
Dave
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Either way, you'll have to dissipate (35 - 12) * .085 = almost 2 watts; Drop a watt or so of that in a resistor of P = I^2 R,

1 = (.085 * .085) * R, therefore R = 1/.007225 = 138.4, so a 120 - 150 ohm, 2 watt resistor will keep your 7812 from getting way hot. I'd check the max. input voltage on the 7812, 35 sounds kind of marginal. You could also use a series zener, or even a shunt preregulator.

Good Luck! Rich

Reply to
Rich Grise

"Simple" means not-switcher, I think. And that means you have to dissipate the 2W that Rich already mentioned -- that is, to supply the

85mA at 12V you have to drop away 23V also at 85mA. Whatever you can cut current requirement that down to, goes in your favor in terms of dissipation, of course. The continuous load of 18mA, if that's the bulk of it, means that 18/85 of 2W or about 400mW is all you'd need to burn off. More tolerable.

The 78L12 usually has an absolute maximum specification of 35V, so your supply is worrisome. Normally, these are used with voltages above 14.5V or so, but well below 35V. And if your 35V can be 36V, then you'd be above the absolute max. Not so good.

Resistors aren't more efficient than a 78L12, or less efficient. It doesn't matter how you throw away the voltage -- it's all equally bad, unless you use switching. But resistors don't regulate well and with the load you mentioned varying its current requirements all over the map, I don't think a resistor would be the regulator you want it to be.

Could consider a 12.6V zener, a current setting resistor for it, and a BJT emitter follower. Three parts. I think you are in the right ballpark, load-wise, for something like that. Still need to dump the wattage, though, so a BJT in a TO-220 on a small heatsink.

Jon

Reply to
Jonathan Kirwan

Do people still say, "Duh"? ;-)

Thanks! Rich (that's me saying, "duh", by the way)

Reply to
,

Thanks everybody for your help, glad I didn't try hooking something up at

6am when I was working on it. I kept thinking 'there must be a simpler way of doing this!' but due to the voltage being 35V (higher than standard regulator maximum input voltage) and the need to dissipate 2W of heat, plus the varying current load it wasn't as simple as a dropper resistor.

I like the zener idea, very elegant. If I have a spare 12.6V zener sitting around I'll hook something up. Otherwise, I just found a spare 100mA 12V wall-wart, probably easier to just use that instead! It's only going to be used for a couple of weeks anyway. It is being used to control a door opener, and the door motor operates at 35VDC.

Thanks, Dave

Reply to
Dave

no. it'd be about equal. if the load is constant, or precision of the 12V not important use a resistor, else use

the 7812 with a resistor in the input picked to drop the imput voltage into the allowable range.

Bye. Jasen

Reply to
Jasen Betts

Exactly. However resistors can easily be use as a preregulator for a 7812. Create a resistor divider with 5W resistors that cuts the

35V down to 16-17V, then feed that into the 7812. Then of the 23V that needs to be dropped, the resistors handle nearly 18V of it and the regulator only needs to handle the last 5V or so, of which 3V are required in headroom.

The 7812 has thermal protection and the like. I think it's better to use a regulator as opposed to trying to emulate one.

BAJ

Reply to
Byron A Jeff

removing

What about a string of diodes in series each dropping a bit of voltage?

R

Reply to
Roger Dewhurst

Aside from the fact that it's not likely to have good resolution (roughly .7V per step), the voltage of each varies on the order of

60mV per decade of current. To drop some 23V, he'd need some 33 diodes. With current varying from say 18mA to 85mA depending on LEDs and perhaps other things, this would be a rough factor of 5 -- about 42mV per diode. Times 33. Almost 1.4V variation. Not so good.

The old:

: +35 : | : | : \\ R1 : / 2.2k : \\ 1/2W +35 : / | : | | : | |/c Q1 : +-----| TO-220 : | |>e : ---/ | : / ^ '----->

: / \\ 12.6V : --- zener : | 1/4W ,---->

: | | : gnd gnd

Would be better. Base would vary less than 10mV. Vbe would also vary no more than 40mV, I think. So about 50mV over the load variations. Double all that and call it 100mV. Still a lot better than a diode string. And you can dissipate 2W in a TO-220 easily.

Jon

Reply to
Jonathan Kirwan

Thanks Jon, that's definitely the best answer yet!

Unfortunately this is all now a moot point, but I'll keep that schematic of yours for if I need it next time for something else. I hooked up the remote control box (the thing that needed the 12V supply) up to the door control box, but unfortunately the receiver causes a lot of interference and stops the box from working! Very strange but I've seen it happen a couple of times before. There isn't enough shielding on the motor control circuit or the receiver box (or both) so I have to now mount the two in different corners of the room with an interconnecting control wire. It'll be much easier to just use a separate plug-pack now, as there is a powerpoint mounted in a convenient position. Running the wire along the ceiling will now be the tricky part.

Thanks again for your help, I'm sure I'll use this for something else soon, it won't go to waste!

David

Reply to
Dave

That could work, but you're dissipating more power (in your divider) with that scheme than with the regulator alone. The extra couple of watts may not matter much for a mains-powered device though.

A single resistor in series with the regulator can also work, with the same power dissipation as the regulator alone, but only over a limited range of currents. Specifically, you need to ensure that you have sufficiently high minimum current for the voltage drop across the resistor to keep the input to the regulator below its maximum Vin, and sufficiently low maximum current that the drop across the resistor keeps the input to the regulator above 15V or so.

Oh, and a 78L12 hasn't got a hope in hell of dissipating the required power, even though the required 80-odd mA is below the 100mA current limit of the 79Lxx series. I find it's quite rare in practice that I get to design for the nominal 100mA current limit of 78Lxx regulators, rather than the thermal spec. With a thermal resistance for the TO-92 junction-ambient of 230C/W and a max junction temperature of 125C, you can only dissipate (125-25)/230 = 0.43W and that's with the unrealistic assumption of a 25C ambient and no safety margin on the junction temp. 0.43W at 0.1A means a maximum voltage drop of just

4.3V, even with that unrealistic ambient temp and zero safety margin. Bump the ambient temp a bit and try to keep the junction below, say, 100C and you're struggling to dissipate enough power to pass the rated 100mA at the minimum voltage required for regulation.

A heatsink helps, of course, but you're not going to find anything very substantial to fit a TO-92 or any of the other packages you can find a 78Lxx in.

Tim

--
Shares are your votes in a pigologocracy.
Reply to
Tim Auton

how many diodes?

that's why not.

Bye. Jasen

Reply to
Jasen Betts

Quite a lot, but they are cheap.

R
Reply to
Roger Dewhurst

I know of someone who uses a 7.5 volt drill pretty continuously in the field. He has modified the battery pack to hold a bunch of diodes and to connect to the 12 v car battery. It works apparently.

R
Reply to
Roger Dewhurst

I'm surprised no one drew a schematic of this idea that someone mentioned.

------

35V + ---[165R]---+-------+---In|LM7812|Out--+---> +12 | | ------ | [1K] [.22uF] | [.1uF] | | | | Gnd --------------+-------+---------+--------+--->Gnd

The 165 ohm is 5 watt, the 1K is 3 watt, both values based on availablity at from Allelectronics at 3 for $1.00. (The 165 will dissipate a bit over 2 watts, the 1K a bit under 1 watt). The LM7812 in a T0220 package can handle

1 amp; in the circuit at .085 amps it will dissipate about .34 watt. Allelectronics can also supply the LM7812 at 50 cents, .1uF cap at 17 cents each, and a .22 uF cap at 5 for $1.00.

Ed

Reply to
ehsjr

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