SLUAAY8 October   2024 BQ25300 , BQ25302 , BQ25303J , BQ25306 , BQ25308 , BQ25620 , BQ25622 , BQ25622E , BQ25628 , BQ25628E , BQ25629 , BQ25630 , BQ25638 , BQ25640 , BQ25960H

 

  1.   1
  2.   Abstract
  3.   Trademarks
  4. 1Introduction
  5. 2How to Measure Efficiency Correctly
  6. 3The Impact of Incorrect Efficiency Measurement
  7. 4Measuring Efficiency in a Buck Charger
  8. 5Measuring Efficiency in a Switch-Cap Charger
  9. 6Summary
  10. 7References

The Impact of Incorrect Efficiency Measurement

Here are a few examples of how measuring input and output voltages incorrectly can affect the efficiency. If the applied input current to a battery charger is 2A, but the resistance from the adapter to the input pin of the battery charger is 100mΩ, then what is the impact? By Ohm’s Law, this is a 200mV drop from the adapter to the input pin. Assume a 92% efficiency for example. If the input pin is actually 5V, with the adapter being 5.2V, and the battery is charging at 4V/2.3A, then the recorded efficiency is approximately 88.5% with the 5.2V adapter voltage, as shown in Equation 2. That has a huge impact on the evaluation of the thermal performance of the part. Second, take the same example (92% efficiency, 5V/2A input, 4V/2.3A battery), but now make the measured battery voltage 3.8V at the terminal of the battery. This presents a 87mΩ drop, making the calculated efficiency being 87.4%, as shown in Equation 3.

Equation 2. Efficiency=PoutPin=4.0V x 2.30A5.2V x 2.0A=88.5%
Equation 3. Efficiency=PoutPin=3.8V x 2.30A5.0V x 2.00A=87.4%

In both cases, there is an incorrect 200mV measure offset as seen by the adapter or terminal of the battery that lowers the efficiency by more than 3%. Proper care must be taken to accurately measure the efficiency and, thus verify the performance of a battery charger.

Having a significant digit off in a sense resistor voltage measurement can have a big impact on the efficiency, as mentioned in Section 2. Use the same example as before (92% efficiency, 5V/2A input, 4V/2.3A battery) but now with a 10mOhm sense resistor in the battery current path. The correct sense voltage for the battery current is 23mV for 2.3A of charge current, as shown in Equation 4. However, just 1mV off in the sense voltage leads to an observed current of 2.2A, as shown in Equation 5. This leads to an observed drop in efficiency to 88.0% as shown in Equation 6. Just 1mV off in a voltage measurement can lead to over 10% error in a current measurement and 4% drop in efficiency.

Equation 4. Sense Current=VsenseRsense=0.023V10m=2.3 A
Equation 5. Sense Current=VsenseRsense=0.022V10m=2.2 A
Equation 6. Efficiency=PoutPin=4.00V x 2.20A5.00V x 2.00A=88.0%