Seeking advice on low-current power supply - Page 1
Forum section: electronics / projects

author: Involute
author_rank: Contributor
timestamp: March 17, 2013, 09:29:54 pm
content: 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.

author: Thor-Arne
author_rank: Unknown
timestamp: March 17, 2013, 10:10:03 pm
content: 7805?At 10mA it shouldn't dissipate too much heat.

author: Harvs
author_rank: Super Contributor
timestamp: March 18, 2013, 01:01:18 am
content: Is it just a single 10mA load you need to supply? I'm a little confused at it being implemented on 25-75 PCBs?If it's just a single load as Thor-Arne said a 7805 would be fine. Or you could probably get away with a 78L05 if space was a problem.However if it's actually 75*10mA, I'd be inclined to do something a little more elaborate. But possibly not, depends on what this is being used for.

author: ivan747
author_rank: Super Contributor
timestamp: March 18, 2013, 01:06:11 am
content: A 78L05 will do fine as others suggested. It comes in a small TO-92 package just like signal transistors so it is very compact. A zener diode with a resistor is a very cheap way to do it, but that cost difference is not worth the sacrifice in performance, specially when you are hand assembling 75 boards.

author: Involute
author_rank: Contributor
timestamp: March 18, 2013, 03:13:57 pm
content: My concern about linear regulators has been the heat from the 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. I guess the power is going to get dissipated somewhere, though, 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.As for the 78L05, I thought those max'd out at 20V input, but I see there's a smattering that go to 35V. I'll give one of those look.As an aside, is there some way to get an e-mail notification of a post here? I'm a newb and couldn't find how to set that up. Thanks.

author: SeanB
author_rank: Super Contributor
timestamp: March 18, 2013, 03:30:54 pm
content: Series resistor to dissipate power so the 78L05 only has around 10V input at highest current draw, which will reduce the dissipation in it so you can use the e line version if needed. You will need a resistor of around 1k 0.5W to do this, easy enough as this will reduce dissipation in the regulator to a low level which will make it more reliable. Still use a 35V part though, it will have the full 33V from the wall wart applied at start up.

author: mariush
author_rank: Super Contributor
timestamp: March 18, 2013, 03:33:00 pm
content: There are regulators that work up to 30v:http://www.digikey.com/product-detail/en/LP2950-50LPRE3/296-20933-1-ND/1216905http://www.digikey.com/product-detail/en/LM78L05ACZ/LM78L05ACZFS-ND/458699etc etchttp://www.digikey.com/scripts/dksearch/dksus.dll?FV=fff40027%2Cfff80182%2Cc00001%2C1140050&mnonly=0&newproducts=0&ColumnSort=1000011&page=1&stock=0&pbfree=0&rohs=0&quantity=100&ptm=0&fid=0For 10mA, you'd have (24v-5v)x0.01 = 0.19 watts dissipated as heat. If you make a bit of copper pour around the regulator holes I think you will be fine.

author: SeanB
author_rank: Super Contributor
timestamp: March 18, 2013, 07:11:21 pm
content: If you want a power on LED then feed the power to the regulator through it. At 10mA most LED's will be bright and will not be overrun. As a plus it will double as a activity indicator, brightening slightly as the mosfet is turning on.

author: Involute
author_rank: Contributor
timestamp: March 18, 2013, 10:02:11 pm
content: Quote from: SeanB on March 18, 2013, 03:30:54 pmSeanB, can you walk me through how you arrived at .5W? I calculate .01A x 10V = .1W. I like the idea of a dropping resistor, though.

author: mariush
author_rank: Super Contributor
timestamp: March 18, 2013, 10:56:29 pm
content: V = I x RR = V / I = 10 / 0.01 = 1 kOhmPower dissipated in resistor P = I^2 x R = 0For 10mA, it is indeed 100mW.But consider the efficiency of the regulator, and the risk of having some sudden extra current... with a 20mA current, the power dissipated already jumps to 400mW.You also want to consider how much the resistance drifts with heat, and how hot it would be with such power dissipation. A 0.5w will heat much less with 100mW going though it, compared to a 0.125w or a 0.25w resistor.

author: KeithBrown
author_rank: Contributor
timestamp: March 19, 2013, 04:10:37 pm
content: Quote from: mariush on March 18, 2013, 10:56:29 pmI suppose every body has noticed that the calculation should have 0.1 on the RHS?Anyway, the "what if" case of 20 mA actually shows a different problem: if 20 mA is being drawn then 20 V "wants" to be dropped in the 1 k resistor: this will force the regulator out of regulation (24-20=4) so the circuit may not work. Even if it is believed that this will not happen, it might happen at start-up when any electrolytics are charging, which might cause the processor to not reset properly. An alternative is to use a zener in series, at least that would never drop a lot more than its nominal voltage.Always a trade-off!

author: SeanB
author_rank: Super Contributor
timestamp: March 19, 2013, 04:21:29 pm
content: Worst case will be the power being connected with the regulator shorted, so the resistor has to do the full supply. This gives a little over a half watt dissipation, which a 0.5W resistor will handle without becoming very unhappy too fast. As a bonus in normal operation it will not get much above ambient temperature as well, a smaller resistor will run hotter, and with a higher load will go crispy pretty fast.