Seeking advice on low-current power supply

17 posts / 0 new
Author
Message

I'm building a project that will be powered from a 24VDC wall wart which will power a solenoid. I need to step this down to 5VDC at 10 mA for my logic. The circuit will be implemented on 25 - 75 simple PCBs I'll be hand-assembling, probably using through-hole components. I'm looking for a power supply circuit that's simple, reliable, and cheap. I've been leaning toward a voltage divider or Zener diode, but thought I'd see if the community has some better ideas, or suggestions/caveats regarding these two approaches. Thanks.

My first thought is this: Is there any chance of switching to a 12 V solenoid, and hence a 12 V power supply?

Doing so cuts the power dissipated as heat in the resistor feeding the zener in half.

That said, 10 mA is pretty low, so the resistor only needs to dissipate ~ 250 mW, (continuously). What is the ambient temperature of the device going to be? Room temperature, inside a larger box with lots of heat, under the hood of a vehicle, etc?

You can do lots of thermal calculations, but if the device will be operating at room temperature, then pick a 2 W or 4 W resistor, air mounted, and see how warm it gets while under your test load.

Obviously you could set up an LED to draw 10 mA as your test load.

Recall that zeners don't have a sharp cutoff voltage.

Note that a voltage divider is a poor option if the load current changes. If the load is truely constant then for low current draws it is an option. If the load changes then so does the voltage.

I still recall a little bipolar power supply PCB I built, 24 V in, one of the outputs was 5 V via a linear regulator. Poor choice... Perhaps if I bolted a car's radiator on the regulator it would stay cool enough not to go into thermal shutdown. :?

JC

Unfortunately, has to be 24V. Room temp, enclosed in a box, but the current sink (an ATTiny85) is the only power dissipator. I still need to do some tests to see how much the current swings, as that was my concern, too, regarding the divider. Since the CPU will be going in and out of sleep, it will be a lot on a percentage basis, but it may not matter given the small absolute numbers involved.

I thought of using a regulator, and feeding it from a Zener or divider to reduce the dissipation in the regulator, but it seemed like I was just moving the power dissipation around while adding complexity (regulator and a couple of caps).

Sounds like a good place to use a 'chopper'... a sort of primitive switcher.

Basically, pulse width modulation of the 24 VDC supply at ~21% duty cycle will provide5 volts (into a sufficiently large capacitor).

There are a few such regulators available as packaged ICs, but a 555 timer and a MOSFET will be cheaper and just as effective. Probably your best bet is to 'chop' the 24 volts down to around 8 volts, and use a standard 5 volt linear regulator from there.

Use SMPS chips of the like MIC4680?? its only a 8 pin smd switch mode power supply chip and requires a few components to go with it. I have used it. They come in 5V or 3.3V or even adjustable options.

That's a good part to know about, luvocean1, but it's a lot of parts compared to a voltage divider or Zener if I can get either of those to work.

Well what is your 5V voltage tolerance? Standard 5%? What logic do you power with it?

Voltage divider won't clearly work. If the logic can take 0 to 10mA and you need 5V voltage to be within 5% or change of 0.25 per 10mA is allowed then you would need the voltage divider to pass 760mA when idle. It will heat up at power of over 18 watts.

Remember that for the zener diode to stabilize the voltage at 5V, you need a certain current through it or the voltage is higher or lower. Depending on the zener the current might be 10mA or 20mA, so that easily doubles your 5V power consumption.

So if max current your logic takes is 10mA, you need at least 10mA available at 5.0V (plus tolerance). So a resistor should limit 24V to 5V at 10mA, around 1k8 then. And in case the logic consumes 0mA sometimes, the zener needs to limit to 5V or 5V1 at 10mA that is available. Much more reasonable, the resistor needs to be able to dissipate only 0.19W at all times. Zener only dissipates 50mW max when no load connected.

But seriously, no matter what, it will always turn into heat in one way or another with linear regulators, perhaps the best thing is to just put a linear regulator there? Something like LP2950? It only needs input and output capacitors and the output capacitor you must anyway have for the logic, and the input capacitor depends on how long wires you have for the supply. Though at 10mA it will heat 32C above ambient

Seriously, I cannot see the problem with using a bog-standard 7805 ... educate me!

Ross McKenzie ValuSoft Melbourne Australia

If he needs 10mA max, a 7805 could easily hog 5-10mA of quiecent current max, doubling the consumption and heating.

But, it seems that at 65C/W rise means at 20mA the temp rise is only 25C above ambient.

Yes, why not a standard 7805 and a 330nF input and 100nF output cap.

According to a few maker's datasheets, the 7805 quiescent is 4.2mA or 5mA typically and 8mA max. A bit of pcb copper will help any temp rise. And it has over-temp and short circuit protection; let's see you do that with your voltage divider or zener kludge.

Ross McKenzie ValuSoft Melbourne Australia

My concern about linear regulators has been the heat from that 25C rise, as well as the size; this is only driving an ATTiny85 and a small MOSFET, so a TO-220 would make the board bigger than a couple of resistors or a resistor and a Zener, which I thought I might need even with a regulator to bring down the voltage differential. As I mentioned earlier, though, the power gets dissipated somewhere, so maybe the regulator's the way to go. The board space isn't a huge deal, and this is getting mounted inside a square aluminum tube, so I might even be able to mount the regulator to the tube, getting it off the board and eliminating any power dissipation issues.

The 2-transistor Black regulator (by Roman Black) is one way.
Another way is to use a switcher module that's in a 7805-like form factor; RECOM is one of several that make these.

"Dare to be naïve." - Buckminster Fuller

You might consider using this stop-down module or a similar one :
http://www.pololu.com/catalog/pr...

Costs \$2.99/100, depending on what your defitition of expensive is.

Price isn't a big deal in this application, so that's a possibility. Thanks.

With a 24v power source, and an output of 5V@10mA, your power dissipation any design of linear regulator will be 190mW. A 78L05 regulator in a TO-92 package would do the job. The junction temperature would be 43 degrees above ambient, or about 69 degrees. This is well below the maximum 125 degrees the device can handle.

Prices for both are down in the 30 cent range from Digikey.
If you're willing to go surface mount, then the SOIC-8 package would have a junction temperature of about 60 degrees.

Chris-Mouse wrote:
With a 24v power source, and an output of 5V@10mA, your power dissipation any design of linear regulator will be 190mW. A 78L05 regulator in a TO-92 package would do the job. The junction temperature would be 43 degrees above ambient, or about 69 degrees. This is well below the maximum 125 degrees the device can handle.

Plus quiescent current of max 5mA for 78L05 adds another 120mW at 24V supply, so total dissipation could be 0.31W.

At thermal resistance of 230C/W my datasheet says for TO-92, temperature rise is about 71C, and the 125C max means the ambient max is only about 53C.

If the box is at room temperature it could work.