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I'm using this eval board of this battery charger IC

I just started to connect a 3.7V battery, with the programmed charge current of 1A. When I connect the eval board module directly to the battery the red LED glows and the PG LED glows. All is working fine.

Now I want to view the charge current through a series multimeter on the VBAT output, so I inserted a series multimeter on the output and tried to measure the charge current.

When I did this, the charge current displayed on the multimeter is 650mA and then keeps decreasing slowly. I tried to maintain short leads to measure the current. The red LED does not glow, indicating that charging is not happening. The green LED is glowing.

What might be the reason for this behavior? Can someone provide an idea?

To cross-check on what is the problem, I removed the series multimeter. Connected the Eval module directly to the battery and placed a current probe and measured on the oscilloscope. Now, with this setup, the charge current of 1A is clearly visible.

So, when the output battery charge current is measured by inserting a series multimeter (10A current setting), the charging is not happening properly. But when I do the non-invasive measurement using current probe, it is working fine.

Anyone knows the reason on why is this happening?

My suspect thoughts:

  1. In the MCP73831 device (I assume MCP73833 also might be having this feature, but it is not mentioned in the datasheet), section 4.2, page 13 of the datasheet, there is a battery detection feature. It works by sending a current out of Vbat pin and measures the battery voltage and indicates whether it is charging or charge complete through the STAT pins. Maybe, is it because, MCP73833 also has this feature and provides some current which interferes with the current measurement in the series multimeter?

  2. Even with the 10A current measurement setting in the multimeter, maybe there's some voltage drop associated with it. This voltage drop reduces the effective voltage seen at the battery terminals and decreases the charging current? Not sure.

Any thoughts on this?

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    \$\begingroup\$ Try using a high current range on your multimeter; it introduces less series resistance. Try the 10 A range for instance. \$\endgroup\$
    – Andy aka
    Commented Mar 6, 2023 at 11:23
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    \$\begingroup\$ You did not tell that you also connected a power supply. Do you wonder why the charge current is not 1 A? The battery might be "to full" to see the full charge current. Also the DMM's internal resistance might be an additional cause for smaller charge current. Lika Andy told, take the unfused (10 A) or so range. Or what irritates you? What load is connected? \$\endgroup\$
    – datenheim
    Commented Mar 6, 2023 at 11:50
  • \$\begingroup\$ @user263983 the OP said this: a series multimeter on the VBAT output \$\endgroup\$
    – Andy aka
    Commented Mar 6, 2023 at 13:53
  • \$\begingroup\$ @Andyaka, yes I did use the 10A range in the multimeter. \$\endgroup\$
    – user220456
    Commented Mar 6, 2023 at 14:29
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    \$\begingroup\$ What’s the rated series resistance of your multimeter? \$\endgroup\$
    – winny
    Commented Mar 9, 2023 at 10:04

2 Answers 2

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A series ammeter operates by inserting a small, known resistance into the current path, called a burden resistance. According to Ohm's law, \$V=I \times R\$, that resistance develops a voltage across it in proportion to the current through it, and that's what the ammeter uses to calculate and display the current. In fact, it's not wrong to say that it's just a voltmeter measuring the voltage across a hidden burden resistance.

Obviously, with a burden resistor now in the loop between charger and battery, the charger is only able to measure the sum of battery voltage, and the voltage across the burden resistance, between A and C here:

schematic

simulate this circuit – Schematic created using CircuitLab

Without the ammeter inserted, the charger would have direct access to node B, and therefore be aware of the exact voltage across the cell.

Whatever voltage is across the burden resistance is clearly sufficient to influence the charger's idea of the cell's charge state. It is likely to believe the cell is fully charged long before it actually is, since it thinks the voltage across it is higher that it actually is.

A current probe, on the other hand, does not require you to break into the current path in any way, and will not interfere in the same way with the voltages anywhere. It will cause a slightly increased inductance in the loop, but this is not likely to influence the charger's own measurement or response.

If your charger is clever, it might also determine the state of the cell by measuring its internal resistance. As cells age, and their capacity drops, their internal resistance tends to rise. Instead of relying only on measuring absolute cell voltage as a function of absolute current, chargers can get some idea of the health of a cell by measuring its internal resistance, which can be done by measuring change in voltage as a result of change in current.

That's often performed by momentarily increasing current to a higher value, and monitoring the resulting increase in cell voltage, but the presence of a burden resistance in the loop will fool the charger into believing that this resistance is part of the cell's own internal resistance.

To the outside world, a cell appears to be a perfect, ideal cell, in series with a small resistance, just like the setup above. By adding yet another resistor (the ammeter) in series with the cell's own internal resistance, as far as the charger is concerned, the cell's internal resistance is the sum of the two.


If your ammeter has a 10A range, you could try that, since the burden resistance will be much lower. You may have better results with that, at the cost of a less precise reading. The truth, though, is that any additional resistance in that loop will influence the charger's opinion of the state of the battery.

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  • \$\begingroup\$ thank you very much for this answer. But if I look into the datasheet of the Battery charger device, I'm not able to find or know whether this IC actually trying to understand the SoC of the battery. However, if you look into the datasheet of another battery charger IC, MCP73833, it has this battery detection mechanism as mentioned in the first point in my question. What do you think about this? Does this battery detection, helps to detect the SoC of the battery. I think it is just to understand the presence/absence of battery. Please let me know what you think? \$\endgroup\$
    – user220456
    Commented Mar 15, 2023 at 8:53
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    \$\begingroup\$ I think, with a resistance between the battery and charger, that all bets are off. The charger cannot possibly know with any certainty, what the SoC is, or what its internal resistance is. NO charger can know, whether it's a MCP73831 or any other, what the SoC is unless it's connected directly to the battery. However, it can know if a battery is connected, because suddenly there's a potential difference there, and current flows when it tries to push current through it. SoC is different from "detection". \$\endgroup\$ Commented Mar 15, 2023 at 9:26
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    \$\begingroup\$ Don't forget, all these ICs do is measure voltage, compare it with internally set voltage thresholds, and react according to when those thresholds are reached. If the information it receives is wrong, because you've inserted something that makes it wrong, how can it possibly react appropriately? I can't say for sure what the symptoms will be, like high-impedance outputs or LEDs lighting up, or wotnot, but I'm certain that behaviour will not be what the datasheet says it will be. \$\endgroup\$ Commented Mar 15, 2023 at 9:30
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    \$\begingroup\$ From what I read in that datasheet, it's just "presence" detection, yes. It compares battery voltage to pre-programmed thresholds to determine what "phase" of charging to employ, like fast charge, trickle, constant current etc, and one of the those thresholds is the "under-voltage lockout" UVLO threshold. If a battery isn't there (no voltage), or the battery voltage is too low, UVLO is in force, which will inhibit charging completely. \$\endgroup\$ Commented Mar 15, 2023 at 9:44
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    \$\begingroup\$ You're welcome, thanks for the reputation bump! \$\endgroup\$ Commented Mar 15, 2023 at 9:56
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From the Dev board data sheet you've linked to, the red LED being off, and the green LED being on, indicates the controller thinks it's finished charging. So adding the ammeter in series does something to make the controller think the battery is charged.
[edit] This is actually wrong, I misread the question. Only the power-good LED is on, meaning the charger is either in standby, timer fault, or temperature fault. I'm leaving the rest of my answer as is [/edit]
Also, as you've said you've seen the current decrease from 650mA steadily to zero, this also looks like the end of a charge cycle.
When charging these batteries, initially the charger will charge with a constant current, in your case 1 A, increasing the voltage as required. Eventually, the voltage will reach 4.2 V per cell, at which point the charger will decrease it's current output until a set value, after which it will stop charging. Figure 2-17 in the battery charger datasheet you've linked shows this.

The key here I think is the relationship between charging current and voltage. As I'm not the first to point out, adding the ammeter in series will introduce an amount of extra resistance into the circuit. This could convince the controller that the battery was at a higher level of charge, as an extra amount of voltage to drop across the ammeter would be required. It becomes much more likely if you're using budget-friendly equipment, and of course it's not just the ammeter, but also the resistance of the wiring.
I'd recommend you do the following.

  1. completely discharge the battery normally, to 3.7V
  2. Attempt to start charging with the ammeter. If you see 1 A charging current from this stage then it's probably the ammeter causing the charging to finish prematurely as per my explanation
  3. Measure the voltage dropped across the ammeter. You can use this to calculate the resistance of the ammeter, be sure to measure from the ends of the leads to include the wire resistance.
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  • \$\begingroup\$ thank you very much for the answer. There are two Green LEDs and one Red LED. When placing the ammeter in series, only the PG LED was glowing and not the other green LED that indicates that the charging of the battery is completed. \$\endgroup\$
    – user220456
    Commented Mar 15, 2023 at 8:56
  • \$\begingroup\$ Sure, I will try to do the recommendations that you suggested. When I measured the voltage drop across the ammeter in the 10A range, while the setup was on, I measured somewhere around 20mV. \$\endgroup\$
    – user220456
    Commented Mar 15, 2023 at 8:58
  • \$\begingroup\$ Oh well then I've misread your question, that's my mistake. Still, at least now you know it's 20mV, and you can calculate the series resistance from that and hopefully, that helps! \$\endgroup\$
    – LordTeddy
    Commented Mar 15, 2023 at 10:53

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