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Battery resistance while charging

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Newbie level 3
Jun 19, 2011
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Hi, While i was trying to design a battery charging circuit using dc-dc converters i got this doubt, does batteries offer any resistance during
charging , if so how to calculate that ,how will they vary and when it will be maximum......plz help...?

i think you are talking about the battery internal resistance.

a battery sinks high current when it has not charged.
when it is being charged and goes to full charge state , the current starts rapidly reducing and stays at a low value.

Go here for some good reading on batteries and battery charging techniques.
Battery information from Battery University

The simplest way to charge a Pb-acid battery with a DC wall-wart is to set up the circuit like you would an LED. You know the float voltage of the battery (VBattFloat), and can determine the float current (Ifloat). See the articles mentioned above to determine these values.

Once you know these two values, you can look at the voltage from the power supply when it's putting out Ifloat amps (can use trial and error with a few resistors, measuring the output voltage... I = V/R). Using Ohms law (V=I*R), you can calculate the value of the series charging resistor.... [Vsupply(at Ifloat amps) - VBattFloat] / Ifloat = R (ohms). Then size the resistor such that it can handle the power dissipation when the battery is fully discharged (~1.75V/cell... or 10.8V on a 12V battery). Pdissipated = (voltage across resistor)^2 / R.
Why don't you use a battery charging IC from Microchip, TI, Linear Technology or STMicroelectronics instead?
DC-DC converter is intended for load regulation and battery charging IC is usually placed on the same regulated supply rail in parallel to the load. In some battery charging IC, the load is placed behind/after the IC, so the DC-DC converter only "sees" the load impedance of the battery charging IC.

You are right, the battery 'resists' the charger. How?
The basic electrical model of a battery is formed by 2 elements:

A voltage source (V_bat)
An eternal resistance (R_bat)

Both (V_bat) and (R_bat) may change during charge and discharge.
Also (R_bat) usually increases with time.

But how can we measure (V_bat) and (R_bat)?
It will be easy if one has a voltmeter [DC volt] and Ammeter [DC ampere].

During the charge, we read the charging current (I_chg) and at the same time we read the voltage (V_chg) on the battery terminals.
We cut the charge [that is (I_chg) becomes 0 Amp] and at the same time we re-read the battery voltage which is now (V_bat) itself at this moment (since its current is zero).

We will notice that V_chg > V_bat
So now we have (V_bat), let us calculate (R_bat) at the moment of measurement.

We know that during the charge:
V_chg = V_bat + I_chg * R_bat
This gives:

R_bat = (V_chg - V_bat) / I_chg

The same method can be followed during the battery discharge:

R_bat = (V_bat - V_dis ) / I_dis

Now if you have the meters you can have fun in measuring V_bat and R_bat... both hidden inside the battery :wink:


In case of discharge the measurement is easier since we need a voltmeter only.
As we saw (V_bat) can be measured when the battery is not connected to any load (or source).
We will decide on a known load like a resistor. The exact value of its resistance (R_load) is not important but lower values give more accurate results. But a low resistance value discharges faster the battery (though the discharge time can be made very short) and the rated power of the resistor (or combination of resistors) should be higher, approximately:
P_load = (V_bat*V_bat) / R_load
While we are measuring (V_bat) we connect (R_load) and read the new value of the battery voltage (V_dis) then remove the load.
R_bat = R_load * (V_bat - V_dis) / V_dis
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That´s a good dicussion, due internal resistance means some heating into battery while charging.
There are several approaches to get account of this resistence, and bellow, follows a particular study wich correlate dynamic impedance and life cycle of battery.


You can not treat battery as a resistive load while charging. you can make analogy between a capacitor and battery for understanding. As Initially uncharged, it draws high current and reduces as plates get saturated.
For discharging the battery internal resistance comes into play. first measure battery open terminal voltage say Vopen.simple, if you have two digital multimeter having "hold data capability". take a small high watt resistor and connect them. Use one meter to measure voltage across the resistor and second to measure current through it. Push both "hold data" buttons at the same time. and use Ohms law to calculate the voltage across the resistor, subtract it from the Vopen, use the same current to calculate battery resistance.

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