0
\$\begingroup\$

I'm an undergraduate student and new to the real-world design of electronic devices. I'd like to make some smart home devices (like Mijia).

I use 5V pin on USB programmer to power my microcontroller, so the voltage may be not a problem at that time. However, when it comes to purchasing components like li-ion battery, fans, or simply some LEDs, I cannot decide which version of voltage should I choose. I suppose that the higher voltage will cause the lower current (and lower heat cost) with the same power, and lead longer battery life with the same capacity. Higher voltage battery also cause safety considerations, but I think it's not a problem if under safety voltage 36V.

So it comes my questions in three points briefly:

  1. Does the higher supply voltage have significant effect on saving battery life and reducing the heat (in real-world practice)?
  2. Any safety considerations when choosing battery, except under safety voltage 36V?
  3. If higher-voltage is better, why there still exists lower-voltage batteries? I don't know if it's in the case that the higher-voltage version is series-wired lower-voltage version.
\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

when it comes to purchasing components like li-ion battery, fans, or simply some LEDs, I cannot decide which version of voltage should I choose

It's not that you have any choice in most of these cases:

  • The chemistry of a battery defines its voltage. You get no say in that. It's confusingly so that different vendors tend to specify different aspects (open loop voltage at maximum charge, nominal load voltage at full charge, something else); but no matter whether your LiIon battery says 3.7 V or 4.2 V: same battery. so, not really a choice.
  • The color of an LED defines its forward voltage. That's a direct result of how light is made in an LED (photons falling from a higher voltage potential to a lower one, both being separated by a well-defined "band gap"). So, unless you're free to choose the color, no choice here

I suppose that the higher voltage will cause the lower current (and lower heat cost) with the same power

Nope. Think about this: you connect a simple 10Ω resistor to a power source, say 5V. The current through the resistor is V/R = 5 V / 10 Ω = 0.5 A then; its power is P = V·I = 5V · 0.5 A = 2.5 W. If you double the voltage supply, that resistor then also carries twice as much current, 1A; the power, P = V·I = 10 V · 1 A = 10 W. The power consumption got quadrupeled!

Not reduced.

For LEDs, you always need a series resistor to set the right current through the LED. So, when you double your supply voltage, you need to significantly increase your series resistor, too, and that series resistor will simply "burn" the additional voltage. You just waste more power.

So:

  1. Does the higher supply voltage have significant effect on saving battery life and reducing the heat (in real-world practice)?

No. Usually, the opposite is the case.

  1. Any safety considerations when choosing battery, except under safety voltage 36V?

no, not really any more for higher than for lower voltages.

  1. If higher-voltage is better,

it's not, in general

why there still exists lower-voltage batteries?

You choose the battery that fulfills your application needs. If you need 12 V to drive something, you don't get to use a single 4.2 V battery, or you need to add a step-up converter, which itself is inefficient and burns energy and thus battery life. Same is true in the opposite direction: if your microcontroller needs 2.5 V, and you add a linear regulator to go from a 20V battery pack down to that, you're burning 87.5% of your battery capacity in that regulator and not what you want to use it for.


looking at this from the "getting as much runtime from the batteries I buy" perspective:

As said, there's not arbitrary battery voltages. Say, for the LiIon chemistry, we just say (to make the math a bit easier) the fully charged battery voltage is 4V. Say, the most cost-effective batteries you can buy have a capacity of 3000 mAh = 3 Ah (== 3 amperes delivered for an hour).

Now, if you want 8V, you need to buy two batteries, and put them in series. That makes 8V, and you can still draw 3 A for an hour, so that's a total of (8·3) Watthours in energy, 24 Wh.

If you want 4 V, but longer live time, you buy two batteries, and put them in parallel, so it's still 4 V. But now you can draw 3 A for two hours, or 6 A for one hour, so it's (4·6) Watthours, 24 Wh.

You essentially pay for the energy stored in your batteries, not for the voltage or your battery pack. Each of our example batteries contains 12 Wh of energy. Whether you stack them serially to get more voltage or parallel to get more current makes no difference in terms of that.

\$\endgroup\$
4
  • \$\begingroup\$ Thanks for your answer! So how about a motor or other inductive components, if I'm free to choose the components' voltage. In that case, I think higher voltage with constant power will cause lower heat cost (like 10W 5V Motor vs 10W 12V Motor), which is my intended meaning in the post. \$\endgroup\$ Mar 4, 2023 at 11:25
  • \$\begingroup\$ hm, no? That's not correct in this generality. Both motors use the same energy in a given amount of time: that's what the 10W say. Which one is more efficient can't be directly deduced. \$\endgroup\$ Mar 4, 2023 at 11:34
  • \$\begingroup\$ But a motor have inner resistance. May lower current reduces heat power (P=R*I^2). That's what been taught in textbook. May be different voltage versions of motor have different inner resistor? \$\endgroup\$ Mar 4, 2023 at 11:43
  • \$\begingroup\$ yes, but how do you know your 5V motor would not have a lower inner resistance than your 12V motor? That's simply not a given. You're assuming you can directly read that from your voltage specs, but that's not the case for small motors. \$\endgroup\$ Mar 4, 2023 at 11:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.