recently I have found an interesting MPPT Lipo charging IC with an integrated 3.3V LDO

So I have bought the Buck-Boost eval. Board(Datasheet) for few tests.

For my tests I used a 6.2V power supply, deactivated the LDO and connected a 3.7V Lipo to see how high the charging current is. With my Ampere meter I measured 30mA battery charging current which seems to be a little bit to low. So i put a 2.2Ohms resistor in series to the DC/DC inductor to measure the current of the inductor(schematic page 4 and 5). The pic below shows the voltage across the resistor. My calculations for the inductor current are

$$\Delta I=\frac{376mV}{2,2Ohm}=170mV$$

$$I_{DC}=\frac{1}{2}\Delta I=\frac{170mA}{2}=85mA$$

Inductor current sensed with a 2.2Ohms Resistor

Now as you can see, my battery is charged with 30mA but my DCDC converter provides about 85mA. So how comes the difference and what happens with the remaining 50mA?

any ideas?


1 Answer 1


The 2.2 ohms you have added reduces the inductor's ability to extract energy from the incoming supply quite significantly and therefore, your buck-boost circuit has to take more current (i.e. operate at a fatter duty cycle) to provide a stable voltage at the output under its load conditions. Try using something much lower like 0.22 ohms or 0.1 ohm if you can find one.

Another problem is the way you have used the o-scope to get the picture of the voltage waveform across the 2.2 ohm resistor. Oscilloscope probes are not floating differential inputs and, in effect, you have grounded one side of the resistor with the wire attached to the o-scope probe. It's hard to say how much this affects things but it doesn't help.

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    – Dave Tweed
    Jul 24, 2017 at 21:18

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