Recommended
Related

Measuring Lithium battery voltage with nRF51

Stefan Birnir Sverrisson

Here is an example of a hardware setup to measure the voltage on a Lithium battery with a voltage divider and a connected capacitor. The Lithium battery typically has a voltage range of 2.7 - 4.2 V and we (Nordic) recommend that you divide the battery voltage with two resistors, R1 = 10 MΩ and R2 = 2.2 MΩ. After you do this you need to connect a 22 nF capacitor (C2 in the figure below) from your ADC analog input pin (AIN) to ground. With this configuration you can sample up to 20 Hz for 8-bit sampling and up to 8 Hz for 10-bit sampling without noticeable ADC output error. You should use the internal VBG voltage as reference (it’s fixed to 1.2 V), and no input prescaling (1/1).

image description

The ADC example code found here samples from analog input pin 6 and is set up for the hardware configuration mentioned above. This example outputs the ADC sampled result to port 1, i.e. pins 8-15.

With this setup the voltage divider consumes current of 4.2 V / (12,2 MΩ) = 0.35 uA.

Usable ADC resolution

The reason behind the specific voltage divider setup described above is to have the voltage on the ADC AIN input pin as close to 0.6 V as possible. This ensures that the current flowing in and out of the ADC is minimal and you can use a small capacitor. If the voltage on the ADC AIN pin is 0.6 V, then there is no current flowing into the ADC because the internal voltage source of the ADC is also VBG/2 = 1.2 V/2 = 0.6 V. If we assume that the voltage range of the Lithium battery is 2.7 V - 4.2 V, i.e. 2.7 V when empty and 4.2 V when fully charged, then the voltage range on the ADC AIN input pin is:

  • Maximum voltage: 4.2 V * (2.2 M/(2.2 M+10 M)) = 0.757 V
  • Minimum voltage: 2.7 V * (2.2 M/(2.2 M+10 M)) = 0.487 V
  • ADC value at 4.2 V - 10 bit setup: 0.757 V/1.2*1023 = 646
  • ADC value at 2.7 V - 10 bit setup: 0.487 V/1.2*1023 = 415
  • Usable ADC resolution - 10 bit setup: 646-415 = 231

It’s useful to increase the ADC resolution for measuring the battery voltage more accurately. This can be done by choosing a smaller resistance value for the R1 resistor in the voltage divider. By choosing a smaller resistor for R1, a larger capacitor is needed because there will be more current flowing into the ADC while sampling, which will further limit the maximum ADC sampling frequency. As an example, if the value of R1 is 6 MΩ, then the usable ADC resolution would be:

  • Maximum voltage: 4.2 V * 2.2 M/(2.2 M+6 M) = 1.126 V
  • Minimum voltage: 2.7 V * 2.2 M/(2.2 M+6 M) = 0.724 V
  • ADC value at 4.2 V - 10 bit setup: 1.126/1.2*1023 = 961
  • ADC value at 2.7 V - 10 bit setup: 0.724/1.2*1023 = 617
  • Usable ADC resolution - 10 bit setup: 961 - 617 = 344

Determine capacitor size

With the voltage divider setup above and R1=6MΩ instead of 10MΩ, a good capacitor value would be around 100 nF for 10 bit sampling and the sampling frequency should not be higher than 2 Hz.

To determine the size of a suitable capacitor for a custom voltage divider the calculation method given on this thread can be used. It should calculate a suitable size for the capacitor so that less than 1 bit error is observed on the ADC output for a specific sampling frequency range. However, often the easiest way to determine a suitable capacitor size for a custom voltage divider or any connected external circuit with impedance higher than 1kΩ, is to use trial and error method. Start with mounting e.g. 100 nF capacitor between the ADC input pin and ground. If you do not get adequate ADC accuracy with low sample rate (<1Hz), increase the size until error disappears. The trade-off is that if you have large capacitor, the maximum sampling frequency is lower then when you have a small capacitor. Typically you are limited to 1Hz - 10Hz maximum sampling frequency when using a capacitor.

  • in reply to Stefan Birnir Sverrisson

    Dear Stefan,

    I still want to know how you selected 0.6V as reference voltage i.e VBG/2.

    Values for resistor is in Mega ohm than how can we achieve or match impedance less than 1kohm as mentioned in reference manual of NRF51.

    Any equation to calculate impedance of ADC pin.Source impedance is mentioned less than 5kohm.

    Mean impedance values are in range of 120K to 400K ohm that changes depending on prescaling. How can i justify these values?

    Hope you can help to clear my doubts.

  • Is this information still valid for nRF52? What changes or values should we do for nRF52?

  • @prabhudurai When the ADC is not sampling, the ADC input is high impedance. When the ADC samples, the impedance of the ADC input is 130kΩ - 390kΩ, depending on your chosen ADC prescale setting. The result is that there is a voltage drop when the ADC starts sampling if the U_AIN input voltage is higher than 0.6V. However, there will be a voltage increase when the ADC starts sampling if the U_AIN input voltage is lower than 0.6V. When you have a capacitor, the voltage will remain stable during the sampling period, which is 20us for 8-bit sampling and 68us for 10-bit sampling.

    Of course the voltage will drop a little bit with a mounted capacitor when the ADC starts sampling. The goal is to have the capacitor large enough so that the voltage drop is less than what corresponds to 1 LSB on the ADC output during the sampling period. If you however have the capacitor too large, the capacitor will take longer to fully charge upp after sampling completes. A large capacitor therefore puts a limit on the sampling frequency. The larger the capacitor, the slower you must sample.

  • why do we need to use the 22nf capacitor across the voltage divider circuit? whether it is used to allow the current flow in both positive and negative cycle (charging and discharging)? please verify my answer?