A variable voltage regulator that acts as a passthrough if voltage is too low - Page 1
Forum section: electronics / projects

author: WarFreak131
author_rank: Contributor
timestamp: March 29, 2021, 01:56:20 am
content: Hello all, I am making project where I want to build a variable voltage regulator using an LM317 or something similar. The caveat is that I want the circuit to act as a passthrough for the input voltage if it is too low to operate the regulator. So for example, if the LM317 is set to regulate to 12 V, and needs a minimum input/output differential of 1.5V to operate (meaning a minimum of 13.5V input), then the table of input/output voltages would look something like this:Input voltage Output voltage5 59 912 1213 1313.5 12... 12... 12This might be possible using a Zener diode, but I specifically want the ability to change the regulated voltage to any arbitrary value. I tried sketching something out using diodes and transistors but I couldn't figure it out. Does anyone know of a way to do this?Thanks

author: Buriedcode
author_rank: Super Contributor
timestamp: March 29, 2021, 02:10:45 am
content: Sounds like you need a Low Drop-out regulator - LDO. Every linear regulator needs to have an input voltage higher than the output (regulated) voltage, but the minimum difference can be down to tens of milivolts depending on the current draw and pass element used.If you're talking amps being passed, and a requirement of <50mW then it becomes difficult. But that also creates problems if the input/output differential is high eg: 3A 16V in 10V out = 18W power dissipation = large heatsink.Edit: typos.

author: bdunham7
author_rank: Super Contributor
timestamp: March 29, 2021, 05:01:49 am
content: A typical LDO regulator does basically that, except that there is a very low minimum voltage (1-2 volts, look at datasheets) before you get anything and then there is a small (again--datasheets!) difference between input and output until the input goes high enough that it is operating in regulation again.

author: Berni
author_rank: Super Contributor
timestamp: March 29, 2021, 05:51:18 am
content: As others have said just get a low dropout linear regulator chip.It's fairly common to find dropout voltages of about 300mV, for the better ones 100mV and under. Tho keep in mind that this voltage is also typically dependent on the load current since the pass transistor inside is still a transistor, so more current the higher its saturation voltage.I suppose is you really wanted the lowest possible dropout at high currents you could add a comparator to the feedback pin of such a regulator so that when the output voltage drops out of regulation a large beefy mosfet is turned on to connect the input and output together with a few miliohms, so it should get just a few milivolts of dropout even at 1ADropout voltage of exactly 0V is of course not possible. Even the wire you use to connect the regulator has more than 0 voltage drop.

author: Gyro
author_rank: Super Contributor
timestamp: March 29, 2021, 09:29:48 am
content: You didn't mention required output current, but take a look at the LM2940 datasheet just as an example of a well behaved LDO. This one has some nice graphs showing its low input voltage behaviour - hopefully other LDOs are similarly well behaved. Note from the figures that you really need to be pulling a fraction of the rated current for best results.

author: WarFreak131
author_rank: Contributor
timestamp: March 29, 2021, 01:38:31 pm
content: Thank you all for the replies. Do LDO regulators operate that way? I thought that if the voltage isn't high enough, the chip doesn't even operate. Does an LDO act like a passthrough?Gyro, I will take a look at that datasheet. It will be very little current, just enough to operate an atmega uC and a 16x2 LCD screen. But I have some other plans as well, which is why I'll need the adjustable function for other projects.EDIT: Well I'll be damned, the outputdoestrack the input for voltages less than the set point...

author: ejeffrey
author_rank: Super Contributor
timestamp: March 29, 2021, 07:12:07 pm
content: Quote from: WarFreak131 on March 29, 2021, 01:38:31 pmOnce you see how they operate it is pretty clear why. The internal error amplifier and feedback network can operate down to a fairly low voltage, probably in the 1-2 V range. Whatever that voltage, it will be less than the lowest input voltage at minimum output. If your input is above that the regulator is basically working. In that case, dropout just means that the feedback loop can't drive the pass transistor hard enough to raise the output voltage any farther so the amplifier will saturate in the "fully on" state and the output will track the input with only a couple hundred mV voltage drop.

author: tunk
author_rank: Super Contributor
timestamp: March 29, 2021, 09:10:17 pm
content: I'm a bit confused, you say you are going to use an atmega+display (3.3-5V?).Why do you need up to 13V - or is that just an example?And what's your input, is it e.g. a solar panel?

author: drussell
author_rank: Super Contributor
timestamp: March 29, 2021, 10:38:51 pm
content: Quote from: WarFreak131 on March 29, 2021, 01:38:31 pmWhy wouldn't it?A linear regulator is going totryto keep raising the output voltage if it thinks it is low, it just won't be able to actuallydoit if the input voltage falls much below what it is trying to output. Older, simpler regulators "drop out" of regulation and just run "full-on" with perhaps a volt or two input-output differential. Newer LDO (Low Drop Out) voltage regulators will still stay in regulation to alowerdifferential (often only 100-300 mV or so of drop, as pointed out in posts above) which may be negligible (enough) in some applications. YMMV, depending on your application, of course...In order to actually output a voltage higher than your input, you would need to move to some sort of switching converter that includes boost mode operation.

author: WarFreak131
author_rank: Contributor
timestamp: March 30, 2021, 04:48:57 pm
content: Quote from: tunk on March 29, 2021, 09:10:17 pmThere will be other components in the circuit that can operate on a variable input voltage. The input is a simple power jack.

author: WarFreak131
author_rank: Contributor
timestamp: March 30, 2021, 04:50:35 pm
content: Quote from: drussell on March 29, 2021, 10:38:51 pmI always just assumed that if the input/output differential is not high enough, that the device simply doesn't operate at all, meaning output voltage = 0. But this is good to know, thank you!