battery voltage measurement

Thread Starter

harinikrishna

Joined Apr 3, 2020
10
Hi,

We are using a voltage divider circuit to measure the battery voltage of li-ion batteries of 3s2p configuration ( NCR18650B)

The parameters are as following
Vin- 12.24V
r1= 604 k ohm
r2 = 191 k ohm
V out( expected) = 2.9V
Vout (obtained) = 2.4 V.

We measured voltage using multimeter at the resistor divider and found out that the measurement is the same as that of ADC readings. ( all the measurements were made on the PCB)
What are the factors that need to be considered to measure accurate battery voltage?
Do you have any suggestions on different methods to measure the battery voltage?
Can you suggest an open-source software where I can simulate the battery-related measurements?
 

crutschow

Joined Mar 14, 2008
24,711
It would appear that one of you resistor values is off.
Have you measured them?

Several here use the free LTspice analog simulator from Analog Devices.
 

Thread Starter

harinikrishna

Joined Apr 3, 2020
10
It would appear that one of you resistor values is off.
Have you measured them?

Several here use the free LTspice analog simulator from Analog Devices.
Thank you for the suggestion
We measured the values of the resistor using a multimeter. We did not find any resistor value to be off.
We chose resistor values with 1% tolerance.
 

AlbertHall

Joined Jun 4, 2014
9,628
So what you are measuring the voltage with is loading down the divider and reducing the voltage. I calculate that about 600k in parallel with the 191k would it.
 

Thread Starter

harinikrishna

Joined Apr 3, 2020
10
So what you are measuring the voltage with is loading down the divider and reducing the voltage. I calculate that about 600k in parallel with the 191k would it.
thank-you for the reply.
The following is being observed:
when the battery pack is not connected, the resistor values are r1= 590ohm and r2= 195kohm.
and when the battery pack is connected, the resistor values read as r1= 686kohm and r2 does not show any value.
the measured battery voltage is 11.6v
voltage measured across r1 is 7.91V and voltage measured across r2 is 2.5V
Is it because of the Board layout problem?
We have the battery + terminal connected to MOSFETs and regulators. The board has a common ground (negative terminal of the battery pack is connected to common ground).
 
Last edited:

AlbertHall

Joined Jun 4, 2014
9,628
1. You cannot measure the value of a resistor in a circuit which is powered.
2. What are you using to measure the voltages (make, model)?
 

Thread Starter

harinikrishna

Joined Apr 3, 2020
10
That meter seems to have a 1M input resistance that will load down high source resistance potential measurements.

The easy solution is to reduce the resistor values of the divider.
Thank you for the suggestion.
With reduced resistor values, the circuit works fine.
but we want to have less current to flow through it.
Instead of measuring with a multimeter when the divider was hooked to ADC channel, there was no difference in the measurement. it was showing the same values as the multimeter.
So, is it that ADC channel also has got input resistance of 1M ohm?

Thank-you
 

nsaspook

Joined Aug 27, 2009
7,181
Thank you for the suggestion.
With reduced resistor values, the circuit works fine.
but we want to have less current to flow through it.
Instead of measuring with a multimeter when the divider was hooked to ADC channel, there was no difference in the measurement. it was showing the same values as the multimeter.
So, is it that ADC channel also has got input resistance of 1M ohm?

Thank-you
With a 3400mAh battery you should have some reasonable measurement margin in the power budget for accurate measurements. You can use a single software correction factor if you have stable input impedances across the needed voltage ranges.
 

nsaspook

Joined Aug 27, 2009
7,181

crutschow

Joined Mar 14, 2008
24,711
Is there any way to calculate impedance from the datasheet?\
The >1MΩ is the resistive portion of the input impedance.
The other parts would the in input capacitance (usually in he neighborhood of a few picofarads) and input inductance (which is normally negligible.
 

Thread Starter

harinikrishna

Joined Apr 3, 2020
10
The >1MΩ is the resistive portion of the input impedance.
The other parts would the in input capacitance (usually in he neighborhood of a few picofarads) and input inductance (which is normally negligible.
So, to make this work, is it a trade-off between resistor values and current?

Thank-you
 
Last edited:

crutschow

Joined Mar 14, 2008
24,711
So, to make this work, is it a trade-off between resistor values and current?
Basically a trade-off between resistor values with the resulting current versu measurement accuracy.

If you are using a micro to measure the voltage, you could always add a correction fudge-factor to the value to correct for the error due to the divider.
Thus if the measurement is 2.4V but the actual value is 2.9V, then multiply the measured value by a factor of 2.9 / 2.4 = 1.2.
 
Top