# In two-phase boost application, the master stage draws more current than the slave stage

I designed a dual phase converter based on the LM5122(LM5122) boost controller. In two-phase application, the master draws more current than the slave. Current sense resistance is 2.2 mΩ for both section. The input voltage is 4.1 V and the current drawn from the output is 10 A and output is 9.96 V. When I measure the voltage on the current sense resistor on the slave section, it is 32.8 mV, while the voltage on the current sense resistor on the master section turns out to be 35.80 mV.

So the master draws 1.54 A more current. Is this normal?

Also, I measure the two terminals of the current sense resistor with a multimeter. (Is this a good way to measure the current that two different phases can give?)

• Schematic please. Commented Dec 9, 2022 at 9:23
• What's the inductance on the first and second phase? Commented Dec 9, 2022 at 13:02
• inductance is same for both of two phase ,and 2..2uH Commented Dec 10, 2022 at 18:02

It's impossible to give an answer without seeing the schematic with component values/models. But I'll give some rough info about possible causes of difference.

When I measure the voltage on the current sense resistor on the slave section , it is 32.8mV, while the voltage on the current sense resistor on the master section turns out to be 35.80mV.

The voltages you measured are basically the reflection of input currents. You are paralleling two converters which cannot show equal effective impedances to the source, and cannot have equal efficiencies. So the input currents cannot be equal. Also, just by looking at the input currents we cannot say whether the phases share the load equally or not. So it's quite normal to see some difference. In your case this difference is 8.5~9%. Is that level of difference normal? Who knows? Cannot say anything without a schematic.

1- Input Current

The input current includes inductors' charge currents. Remember the inductor current-voltage relation:

$$v_L(t) = L \ \frac{d}{dt}i_L(t)$$

Although $$\v_L\$$ is equal for both thanks to parallel operation, and $$\dt\$$ is equal as well thanks to the synchronisation, they cannot have the same $$\L\$$. Therefore their $$\di_L\$$ cannot be equal. This will bring some difference. If the inductors are 95% matched then the peak of $$\i_L\$$ for both will have 5% difference.

2 - Losses

It's impossible to get perfectly equal drop across the MOSFETs/diodes for the same current. So even if they share the load perfectly equal the losses can be different. This difference can be anything. Impossible to say without schematic with component values.

3 - Measurement

This includes sense resistors' tolerances and your meter's accuracy.

• Resistor tolerances can be anything between 0.1% to 20%. So, even if the input currents are equal for both phases the voltage drops can be different due to the different resistances. We don't know what type and tolerance you selected. If you selected sense resistors with 1% this will bring 1% measurement difference at the same current.
• Remember that the input currents are not pure DC i.e. they have ripple. Therefore the voltage drops across the sense resistors have ripple as well. If you measure the voltage at DC voltage measurement mode then the meter averages the measurements and shows the result. I don't know your meter's accuracy at mV ranges. Plus, as I stated above, the AC components due to the switching inductor currents can be different. This yields different averaging and therefore different result.

I measure the two terminals of the current sense resistor with a multimeter. (Is this a good way to measure the current that two different phases can give?)

Measuring input current does not give you a clue about equal sharing. What you should do is to measure the input and output currents properly.