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!
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?
That 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.
Ah ok, thanks, that's very useful!
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
The higher the current, the higher the drop.
And 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.
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.
Thanks. 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.
And 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.
Yes, 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.
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.
LDO means Low Dropout, it's a feature, not a component type.
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.
Oh yeah, thanks, I'm sure I've got some of those as well, thanks for the tip!
LDO means Low Dropout, it's a feature, not a component type.
Ah 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.