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I'm using an HTRC C150 charger to charge a single-cell LiPo battery for my RC quadcopter.

When I let the device charge the battery, it ends with a reading of 4.2V, however, when I connect a multimeter to my battery it reads 4.05V instead.

While the battery was charging, I took a reading at the output terminals of the device which read around 4.1V. I have two three questions about this:

  1. Is the voltage difference because the cables/device aren't perfect conductors (i.e. they have some resistance), and therefore dissipate some of the DC current's energy over the distance of the wires?
  2. Is this normal for a charger to be off that much? Or did I just get something cheap, and would investing in a higher quality charger fix this?
  3. Is it possible to calibrate the charger somehow to offset for this voltage difference?

Edit: This is the second HTRC charger I've tried, since the first one had a similar issue and I returned it thinking it was the charger, but now I'm questioning my small understanding of electrical engineering haha. Also, to be clear, the 4.05V reading was when the battery was disconnected, and the 4.1V reading was made by pulling out the black/red plugs from the device a little bit and contacting the multimeter to those in parallel to the charging. I also made another reading (that I haven't mentioned yet) on the battery itself while it was charging by pressing the terminals against the exposed metal bits on the JST cables, and that read 4.06V (which makes sense to me why it'd be slightly higher; the charger is sending a slight current into it). I am fairly confident that my multimeter is working correctly, since it reads other batteries of known voltages correctly (like 9V or AA batteries).

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  • \$\begingroup\$ Was that first measurement taken while charging, or after taking the battery off of the charger? \$\endgroup\$
    – hobbs
    Commented May 8, 2023 at 6:02
  • \$\begingroup\$ If the current was close to zero at 4.2 V, it can't be explained by cable or contact resistance. Sounds too high for a decent charger, but better 4.05 than 4.35 V. \$\endgroup\$
    – winny
    Commented May 8, 2023 at 8:05
  • \$\begingroup\$ Are you sure your measuring device is accurate? \$\endgroup\$
    – user263983
    Commented May 8, 2023 at 10:46
  • \$\begingroup\$ At the end of a battery charging process the current should drop (CV mode) so the effect of resistance should be less \$\endgroup\$ Commented May 8, 2023 at 11:26
  • \$\begingroup\$ @winny I'm not at home at the moment, but trying to measure the current is a good idea... I hadn't done that yet heh. I'll make an edit posting the results when I do that. \$\endgroup\$ Commented May 8, 2023 at 20:15

1 Answer 1

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  1. Accuracy. A cheap multimeter to be 2% or 5% off is not something unheard of. The limiting factor for this accuracy are the resistors at the voltage dividers (assuming that at least one range is carefully calibrated at the factory which is not always the case).
  • Chargers are usually (but not always!) somewhat better.
  • Batteries of known voltages are not a calibration tool at any rate. An AA alkaline or zinc-carbon cell can be anything between 1.45 and 1.85 when fresh and brand new (1). 9 V battery is simply 6x AAA cells in series (sometimes they use different form factors inside the 9V package but there are still 6 cells inside) so anything between 6 x 1.45=8.7 and 6 x 1.85=11.1V is to be expected.
  • Even a known good multimeter may develop its own opinion when its battery gets somewhat flat. This may, or may not be immediately indicated by the low battery indicator.
  1. Charging algorithm. A traditional way of charging a Li-Ion cell is CC/CV (Constant Current all the way to the final voltage and then Constant Voltage until the charging current gets below a predetermined value).

Now imagine that the battery is a big-big capacitor and a variable resistor in series. This is quite a good model if one uses the battery in cycle mode (buffer modes are another can of worms).

During the CV phase, the internal resistance of the battery goes up and the "internal" voltage also goes up, this is why the current goes down. A 2000 mAh cell is usually stopped at 50mA - these 50mA mean that the internal voltage is still below 4.2V . Upon disconnect, you can measure the "internal" voltage and it is pretty normal to see something like 4.05 or 4.15 (depending on a lot of factors).

If you wait a few minutes, the voltage will go somewhat lower, which is not obvious from the simple capacitor/resistor model, but it happens anyway.

This is why a fully charged Li-Ion cell is 4.00 +/- 0.05V after some storage.


In short, either improve your measurement equipment or lower your expectations.


(1) Its theoretical voltage is ~2.0V and why it is so much below 2.0 V in order to be called 1.5V is a very complex question.

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  • \$\begingroup\$ Thanks! This is all a new beast to me, so there are a lot of details here I was not aware of before. Thank you for elaborating on my thoughts and pointing out why some of my assumptions aren't safe to make. \$\endgroup\$ Commented May 8, 2023 at 21:24

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