Need advice on 5v regulator for Arduino, need to take 10-20v - Page 1
Forum section: electronics / beginners

author: Jackster
author_rank: Frequent Contributor
timestamp: January 22, 2017, 11:39:31 am
content: Hi all.So I need some advice on how to regulate a voltage range from 10v to 20v for my Arduino project.I am using the Nano source and adapting it for my use.The stock regulator on the Nano takes up to 9v.So I will need to replace this with something else.I have been playing with the L7805 but it is a bit to large for my project and it is not very efficient.Does anyone have any recommendations on something a bit smaller?Need around 500mA to power everything (this includes overhead).ThanksJakc,

author: danadak
author_rank: Super Contributor
timestamp: January 22, 2017, 12:14:25 pm
content: 768L05 also comes in some pretty small packages -http://www.ti.com/lit/ds/symlink/lm78l05.pdfRegards, Dana.

author: orolo
author_rank: Frequent Contributor
timestamp: January 22, 2017, 12:30:25 pm
content: Worst case, your power supply will be drawing 0.5 amps from a 20V source. That means 0.5·15 = 7.5 Watts dissipated by the regulator, a lot of power for a small package. There are some very good SMD parts that work to 20V and give up to 1 amp, but you must check their SOA. A quick search at mouser gives, among others,this one. If you check page 2 of the datasheet, this chip should give you up to 0.5A at 20V input, just at the limit of the SOA. On page 7 they specify that this limit is for 2 oz copper and 5mm x 5mm pad, no air flow, and quote a thermal resistance of 73ºC/W --> for 5 Watts that means a 365ºC temp increase, maybe their SOA chart is optimistic.Anyway, with that input voltage range and high output current, you should consider a SMPS, if you find one small enough.

author: Ian.M
author_rank: Super Contributor
timestamp: January 22, 2017, 12:44:25 pm
content: 500mA @5V out and 20V in = 7.5W of losses in a linear regulator. That's a 7805 and a *chunky* heatsink if you want a linear regulator solution.Your best bet would be a switching regulator, and if you don't want to have to either make a PCB or do some rather fiddly construction on matrix board, you'll need a pre-built module.I suggest:https://www.pololu.com/product/28315.5V-36V in, 5V, 1A out, 0.7"x0.5". They also do an even tinier 500mA versionhttps://www.pololu.com/product/2843but if you've underestimated any startup surges or transient loads, that spike over 500mA that might go into current limiting, dipping the 5V rail, and repeatedly rebooting your Nano.

author: Seekonk
author_rank: Super Contributor
timestamp: January 22, 2017, 01:15:08 pm
content: You can't beat those little 2596 or XL boards on ebay for a little over a buck. I assume anything external to the micro will have the buck converter 9-12V as a source and the nano will see no more than 100ma. use a resistor of a couple diodes to drop the voltage to the nano.

author: mariush
author_rank: Super Contributor
timestamp: January 22, 2017, 01:42:54 pm
content: Here's a whole bunch of them :http://www.digikey.com/short/3rnhz9I put some filters 5v to 5.2v (your arduino will work with up to 5.5v), at least 400mA, at least 16v input voltage. All shapes and sizes, some look like to-220 devices, meaning they can be used as 7805 replacement basically, some have long perpendicular pins which means they could be soldered on top of your board.. some are like DIP packages etc..If you switch from fixed output voltage to adjustable you may get even smaller circuits but you'd need at least one resistor (or two most often) between pins of device to configure output voltage.

author: Zero999
author_rank: Super Contributor
timestamp: January 22, 2017, 02:42:17 pm
content: One thing you could do is to put a resistor in series with the regulator to drop some of the voltage before the regulator. The downside is doesn't work over a wide range of voltages and 10V to 20V could be too wider range for this to reliably work.

author: Ian.M
author_rank: Super Contributor
timestamp: January 22, 2017, 03:08:17 pm
content: Not much help. A 7805 needs 2V headroom, so you can only throw away 3V from the 10V in the dropper resistor at the max current of 500mA. That moves 1.5W from the regulator to the resistor, reducing the max regulator dissipation to 6W - still a big heatsink. Even if you use an ideal LDO, you can only move 2.5W to the resistor. A better linear approach would be to split the dissipation between two cascaded regulators, a 7812 and a 7805. Both will need heatsinking but the worst case dissipation is only 4W for the 7812 and 3.5W for the 7805 so you can use two cheaper heatsinks. Below 14V, the 7812 will drop out of regulation, but it should still be capable of delivering 500mA with no more than 2V drop, leaving enough headroom for the 7805 at 10V.However, a switched mode buck module is a *vastly* preferable solution, far more compact, energy efficient and cheaper unless you've got a free *big* heatsink.

author: Jackster
author_rank: Frequent Contributor
timestamp: January 22, 2017, 04:22:25 pm
content: Oh wow completely forgot about heat.Just tested it with the L7805 and it burnt my finger so yep...Thinking I might hand off the design to someone else. This stuff is complicated xDThanks for the support guys.Much appreciated.

author: Ian.M
author_rank: Super Contributor
timestamp: January 22, 2017, 04:27:12 pm
content: ... which is why I recommended the Pololu module.You could also use a generic Chinese buck converter off EBAY, but if you do that get the adjustable current limit type (with two trimpots, voltage and current) so you can set the limit down to something safe for your circuit.

author: Jackster
author_rank: Frequent Contributor
timestamp: January 22, 2017, 04:37:59 pm
content: Quote from: Ian.M on January 22, 2017, 04:27:12 pmI have had a go with the little tiny eBay ones and they do get a tad hot. Though nothing like this. Main thing is I don't want to heat the case because there is a sensor in there that will offset if the case inside is getting a few degrees warmer :/

author: Ian.M
author_rank: Super Contributor
timestamp: January 22, 2017, 04:44:09 pm
content: *EVERYTHING* that uses electricity at more than a few mA or V, gets hot. Your design needs proper thermal management - maybe a miniature 5V fan so that air is drawn from the outside over the sensor then over the parts that get hot with the hottest nearest the air outlet, or maybe a close-fitting polystyrene block between the PSU and the sensor as a thermal barrier (though that doesn't work well in a metal case).

author: james_s
author_rank: Super Contributor
timestamp: January 24, 2017, 06:30:04 am
content: I've used a bunch of those 99C ebay buck regulator modules in stuff. I normally would build my own but I can't buy the regulator IC alone for twice the price it costs to have a whole module shipped to my door. Linear regulators only make sense up to a few Volts drop and/or at very low currents. With a linear regulator the input current will *always* equal the output current and all that extra voltage gets burned up as heat. With a switching regulator the input current can be lower than the output current and so little energy is wasted that in many cases you need no heatsink at all beyond the PCB.

author: hosko
author_rank: Newbie
timestamp: January 29, 2017, 06:09:56 am
content: Not a good answer but it might be easier to use a Nano v3.0 which has Input Voltage (recommended) 7-12 V and Input Voltage (limits) 6-20 V. Should be trivial to reduce the input voltage down to work within its window. Its the arduino board I use when I don't need ethernet.