Could anyone lay out the math for how you would calculate how much current is being wasted if you used a standard silicone diode to drop a 3.7V source to 3.1V?

I'm trying to evaluate if an LDO is more efficient, than just dropping a voltage using a diode.

Say the base case is the device is sleeping and pulling 1mA of current.

Where do you start on that kind of calculation?

(The real case base case is a module that expects a 3.7V LiPo and is fine from 4.2V to 3.3v, but another IC that wants a 3.6V - 3.3V source. This board is VERY space constrained, so in one sense the diode probably is the smallest way to drop the voltage for the IC in question).

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    \$\begingroup\$ Rather than wasting current, it's the voltage you're "wasting", and that's a weird thing to say... so let's get the correct terminology! Wasting power! \$\endgroup\$ Oct 12, 2017 at 22:19
  • \$\begingroup\$ Sure, power... Where do you start for a measurement like that? \$\endgroup\$
    – Leroy105
    Oct 12, 2017 at 22:23
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    \$\begingroup\$ The battery will not be a constant 3.7V. When it is discharged it might only be 3V or even 2.8V. If you just drop the voltage your circuit will stop operating before the battery runs down. \$\endgroup\$ Oct 12, 2017 at 22:57
  • \$\begingroup\$ Kevin true, in this case the LiPo has got a shutoff ic... But yeah, the diode is not ideal if you can't deal with voltage swings \$\endgroup\$
    – Leroy105
    Oct 12, 2017 at 23:19
  • \$\begingroup\$ the diode, the LDO and a resistor all convert the volts to heat at the same rate (*I) \$\endgroup\$
    – dandavis
    Oct 13, 2017 at 12:25

1 Answer 1


The power wasted with an LDO will be \$P = V×I = (V_{in}-V_{out})×I\$ [W]

The power wasted with a diode will be\$P = V×I = V_f×I\$ [W], (Vf means forward voltage). Depending on the diode you're using it can be everything from 0.2 V to 1.5 V and even higher if you're using a Zener or an LED. I'll use 0.6 V because it's probably something you're having in mind.

So the power wasted with a diode will approximately be \$0.6×I\$ [W]

Let's plug in your numbers and see which one is the most efficient, hint: they are equally bad.

LDO: \$(V_{in}-V_{out})×I = 3.7-3.1 = 0.6×I\$ [W]

Diode: \$0.6×I\$... wait a second, this is the same as for the LDO! That's because you're not using a switching power supply (SPS).

What about using a resistor instead? Maybe that will be better, hint: it's not, it's equally as bad as the above options.

If you know that your system will pull 1 mA, then what is the resistance required to make the voltage across the resistor 0.6 V?

\$U=R×I => R = \frac{U}{I} = \frac{0.6}{0.001} = 600Ω\$

The power wasted will be \$P = V×I = 0.6×I\$, oh no, the \$0.6×I\$ [W] again. Argh!

Size wise a resistor may be smallest, next is probably a diode, third an LDO.
If I were you I'd use an LDO because that is the only one that makes sure that it's 3.1 V at the output, the diode and resistor are just passive, there's no active regulation happening with them.

If you'd want to go for efficiency then a buck converter would make you happy. Sadly they are a little bit larger than resistors/diodes/LDO's because they require an inductor, also the noise from the switching might be too nasty for your load.

Let's assume you'd go for some buck converter (one kind of a switching power supply) that's 90% efficient while supplying 1 mA, it's like a dream come true. Everything just matches and the noise isn't affecting your load at all. I can see a rainbow.

Then the energy wasted would be \$P = P_{waste}×(1-P_{efficiency}) = 0.6×I×(1-0.9) = 0.06×I\$ [W]

Wasting \$0.06×I\$ [W] is much better than wasting \$0.6×I\$ [W], assuming your numbers and rainbows.

  • \$\begingroup\$ This is great thanks. Let me digest this. I understand the forward voltage dilemma in not having a constant voltage output. I am not sure on efficiency. \$\endgroup\$
    – Leroy105
    Oct 12, 2017 at 22:38
  • \$\begingroup\$ Got it with the switching power supply, the IC is pretty sensitive. I think you gotta put in a bunch of passives to filter the noise. No free lunches, right? \$\endgroup\$
    – Leroy105
    Oct 12, 2017 at 22:49
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    \$\begingroup\$ Yes, that would help, but that in turn would make your design even larger, which is something you don't want. -- If you increase the switching frequency (to say 400 kHz) then you can get away with smaller components. The price you need to pay is wasting current in charging and discharging the MOSFET transistor(s) that are handling the switching. And the transitioning from when a MOSFET switches from on to off.... Yeah... there's a lot to think about.... But you're dealing with just 1 mA so I'm fairly certain that you can ignore the efficiency problem and just stick with a LDO. \$\endgroup\$ Oct 12, 2017 at 22:54
  • \$\begingroup\$ @Leroy105 if you plan to mark my answer as correct, don't, let other people get a chance to comment / answer and/or critique. Someone else might come up with a better answer. \$\endgroup\$ Oct 12, 2017 at 22:57
  • \$\begingroup\$ Saw that wrong... It's all good. I understand what now what an ldo is, opamp with a resistor.... \$\endgroup\$
    – Leroy105
    Oct 12, 2017 at 23:11

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