Voltage drop from serial diodes - Page 1
Forum section: electronics / beginners

author: Cricri
author_rank: Contributor
timestamp: May 15, 2024, 01:38:44 pm
content: Hello all,Excuse if this is a dumb question, but I'm a MechEng and not a ElecEng.I use a LiIon battery to power something normally running on 2x1.5V batteries. So I thought, I'm going to drop the voltage a bit, and rather than use a LDO (or god forbid a resistance based divider bridge), I'm just going to add some diodes in serial (not Schottky obviously).So I've got a bunch of 1N4007, decide I wanted to drop the voltage by more than 0.7V, so add 2 in serial.Now that's not the first time I do that, and it works fine, but my question is: each time I did this, I can definitely notice a drop in voltage when adding 1 diode, but the 2nd one always seems to drop the voltage by quite a bit less than the first one. I.e. if I add 1 diode, I may drop say 0.6V, but if I add another one, the drop is always less than twice as much.Can someone explain why that is?Thanks!

author: zilp
author_rank: Frequent Contributor
timestamp: May 15, 2024, 03:08:02 pm
content: Quote from: Cricri on May 15, 2024, 01:38:44 pmThat is because the voltage drop is not a fixed value, but rather depends (logarithmically) on the current through the diode. The higher the current, the higher the drop. Now, if you have a resistive load, say, then the additional diode lowers the voltage across the load, and thus also reduces the current through the load and thus also through the diode ... leading to a lower drop.Note that it's not so much the second diode dropping less voltage, but rather that with lower current either diode drops less voltage, so the total is also less.Which is all the reason why series diodes to change supply voltages aren't great. I mean, there is nothing wrong with it, but it really is only useful for applications where the exact voltage isn't that important.edit: Also, your average diode datasheet will have a graph of Vf vs. If, so you can just look at the datsheet to get a rough idea what you can expect.

author: Cricri
author_rank: Contributor
timestamp: May 15, 2024, 09:54:14 pm
content: Ah ok, thanks, that's very useful!

author: BennoG
author_rank: Frequent Contributor
timestamp: May 16, 2024, 05:28:50 am
content: You know a LDO produces exactly the same heat as a Diode solution. In both solutions the voltage difference will be converted to heat.Only the LDO provide a lot more stable output voltage.If you want reduce heat (power) loss, then it is probably better using some kind of low power buck converter.Benno

author: ArdWar
author_rank: Frequent Contributor
timestamp: May 16, 2024, 06:00:19 am
content: Quote from: zilp on May 15, 2024, 03:08:02 pmAnd of course the reverse is also true.The lower the current, the lower the drop. Down to near-zero drop at near-zero current. That may give you a problem when your load's quiescent current is too low for the diode to drop meaningfully and it gets the whole 4.2 V instead.

author: Cricri
author_rank: Contributor
timestamp: May 16, 2024, 04:45:32 pm
content: Quote from: BennoGThanks. What I like about the diode is it seemless as I just need to make the connecting wire a bit shorter and use the diode as an extension. Also it's cheap and I've got plenty of recycled 1N4007, I've only got a couple of converters that I keep for sensitive applications where the voltage is more critical.Quote from: ArdWarYes, good point, I'll have to be careful. But I already see a drop with just the Fluke, which I don't presume creates much resistive load when measuring voltage?I've got a broken Nintendo Game & Watch that I couldn't fix the buttons, battery voltage is 4.2V measured with the Fluke, after 2 diodes I measure 3.3V, so it actually works pretty well considering that those LCD games consume very little.

author: BrokenYugo
author_rank: Super Contributor
timestamp: May 16, 2024, 06:29:27 pm
content: For such a low power device a low power small signal diode like 1N4148 (another common part you should have a bunch of) will probably give better regulation. More drop at lower currents.

author: Siwastaja
author_rank: Super Contributor
timestamp: May 16, 2024, 06:38:09 pm
content: LDO means Low Dropout, it's a feature, not a component type.

author: Cricri
author_rank: Contributor
timestamp: May 16, 2024, 07:14:39 pm
content: Quote from: BrokenYugoOh yeah, thanks, I'm sure I've got some of those as well, thanks for the tip!Quote from: BrokenYugoAh ok thanks, I've got some LDO voltage regulators so assumed they were just a specific type. I see what you mean though, it's not a name of voltage regulator.

author: kevin.gibbs
author_rank: Regular Contributor
timestamp: August 01, 2024, 03:38:41 pm
content: The voltage drop across a diode always depends on the current flowing through it. When you turn on 2 diodes, the voltage in the circuit becomes smaller, and the current draw is also smaller. Therefore, slightly fewer volts drop across each of the two diodes than when it is a single diode.Look at the diode's datasheet - the forward voltage to the current relationship.

author: mariush
author_rank: Super Contributor
timestamp: August 02, 2024, 06:47:37 pm
content: Quote from: BennoG on May 16, 2024, 05:28:50 amIf we want to be really picky about it, technically a LDO would consume more power because it will also use some power for its internal power consumption. Usually it's below 1mA though so not really important. The benefits (stable output voltage) outweight the power consumption most times.QuoteThere's also charge pump regulators that can reduce by 2 or 3 or double or invert a voltage, without requiring external parts other than a couple capacitors (for low currents, around 10-22uF ceramics would be enough) and maybe a diode.But yeah... step-down (switching) regulators would be most efficient to get a specific voltage.