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I have a slight confusion regarding the battery. Suppose I have to charge a battery of initially 12V to 14.2V (full charge). My charger has open circuit voltage of 15V. When I connect it what voltage should it show...the battery voltage or the charger voltage or some intermediate voltage. If I am using a voltage divider than I won't get a Propper battery state if it reads the voltage of the charger or the intermediate voltage between 12 and 15V.

My questions are:

  1. Will I read battery voltage/charger voltage/intermediate voltage between 15V and 12V.
  2. How do we get the exact battery state during charging to show the battery percentage in case it has some display of battery percentage.
  3. I have seen mobile phones charging at 9V....what does that mean. does that mean when I measure the voltage on the charger and battery during charging it will give 9V.

Kindly help me in these simple terms as I have slight confusions in them.

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  • \$\begingroup\$ What chemistry are you using and what is your charging circuit? Vs means volt-seconds; V is the symbol for both singular and plural volts. \$\endgroup\$
    – vir
    Commented Jul 2 at 18:52
  • \$\begingroup\$ What other than the charging voltage do you expect to see? To get current into the battery you need higher potential. A cell phone does not /charge/ at 9V, but the USB power supply connected to the phone supplies 9V to reduce IIR-losses. The battery charger is inside the cell phone. \$\endgroup\$
    – MrGerber
    Commented Jul 2 at 18:52
  • \$\begingroup\$ 9V's means 9 volts \$\endgroup\$
    – kam1212
    Commented Jul 2 at 18:53
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    \$\begingroup\$ No, 9V (or 9 V) means "9 volts". Neither 9V's nor 9Vs mean "9 volts". \$\endgroup\$
    – brhans
    Commented Jul 2 at 19:07

3 Answers 3

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Will I read battery voltage/charger voltage/intermediate voltage between 15V and 12V.

No. If the meter probes are placed across the battery, you will measure the sum of the battery voltage and the voltage across the internal equivalent series resistance (ESR). $$ V_{meas}=V_{bat}+V_{ESR} $$

How do we get the exact battery state during charging to show the battery percentage in case it has some display of battery percentage.

To remove the effect of the internal resistance, the charger must be disconnected. However the charging characteristics for the various battery types are well known so the state of charge can be estimated based on prior knowledge.

The prior knowledge allows the charger display to represent the percentage charge state. As batteries age the prior knowledge may become inaccurate. So without personal experimentation (and even with) "exact" is difficult to obtain. Settle for approximate.

I have seen mobile phones charging at 9V....what does that mean. does that mean when I measure the voltage on the charger and battery during charging it will give 9V.

I can't say for this particular case. Assuming the charger's voltmeter is accurate it is displaying the voltage including the ESR voltage. Having said that, they may compensate for the ESR.

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Most batteries have a high ESR when discharged, and an insignificant (??) one fully charged. So the effect of the ESR may be trivial when all is said and done. If you really want a precise measure then measurement over time of the charging voltage, the charging current and the open circuit voltage is required for full understanding. Voltage measurements do not necessarily indicate the amount of charge transferred to the battery. That can be measured by integrating the current.

Constant voltage charging is only used when the battery is fully charged to compensate for leakage over time. Constant current or current limited charging is always used when charging from discharged. So placing the meter probes across the battery will always indicate the voltage given by the equation above.

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the battery voltage or the charger voltage or some intermediate voltage.

Yes, yes, and kind of yes.

By definition, the battery voltage and the charger voltage are exactly one and the same because they are connected directly in parallel. So, the voltage you measure is both the charger voltage and the battery voltage.

The voltage out of then the charger when disconnected is different from its voltage when connected to the battery. The moment you connect the charger to the battery, the charger voltage drops to the battery voltage. As the charger send power into the battery, that voltage increases, not because the charger increased it, but because the battery's electrochemistry and physics set a higher voltage as power flows into it.

  • During most of the charge, the battery sets the voltage, not the charger.
  • At the end of charge, the charger can limit the voltage it puts out.

But regardless, the battery voltage and the charger voltage are exactly one and the same.

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    \$\begingroup\$ So actually if we have a battery to charge and I need to monitor the voltage accurately then can I disconnnect it for some short interval or make the current flow minimum to measure the voltage accurately or I can simply use a VDR to measure voltage of battery directly during charging??? \$\endgroup\$
    – kam1212
    Commented Jul 2 at 19:42
  • \$\begingroup\$ > ???. Please don't do that: it's quite annoying. I believe I already answered your question. Please reread that part that start with : "By definition, the battery voltage and the charger voltage are exactly one and the same..." \$\endgroup\$ Commented Jul 2 at 20:16
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    \$\begingroup\$ Sorry I donot mean to annoy you. \$\endgroup\$
    – kam1212
    Commented Jul 2 at 20:21
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  1. If your battery is 12V and charger is 15V then connecting the charger to the battery should show 12V or something very close to 12V as the battery voltage cannot jump instantaneously as it needs charge in order for the voltage to rise. Resistances like ESR of the battery or wire will show up in your measurements as the ESR of the battery and resistance of the wires is not 0.

  2. By counting the charge. Or with some lookup table. Most batteries do not have a linear relationship with voltage and their state of charge, and if you are talking about lithium batteries, they are charged up to 4.2V per cell with constant current but when 4.2V is reached at full current the voltage is held constant until the battery draws only something like 10% of the original constant current.

Your mobile phone takes in 9V from a USB power supply. Your phone has a lithium battery charger inside it that takes care of the charging in any way it wants. The 9V has nothing to do with it. The 9V is not applied to battery. It is only a supply voltage for the battery charger. The phone knows how much charge has been moved into the battery and out of the battery to show some kind of percentage, because the phone knows how large capacity battery is installed and how empty it can be discharged and how full it can be charged.

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