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I made a board recently and just started checking it out yesterday. I used a typical lithium coin cell and a boost converter (Microchip MCP1640). With a battery plugged in I get no voltage out of the boost converter (i.e. 5V to GND measures 0V). Worse yet, I measure very nearly 0V from BATT_PWR to GND with the battery in circuit.

So I removed the battery and fed 3V from a benchtop power supply into the FTDI_PWR net and sure enough was able to measure 5V. The PSU indicated I was drawing about 50mA. I experimented with lowering the voltage, and the booster held 5V, but current draw increased as input voltage decreased (which I believe is expected).

So, is the peculiar battery behavior described above due to the battery current limiting? Is there likely to be some more insidious problem at work here - how might I determine that? If this is the wrong way to boost a coin cell to 5V, what's a better way?

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    \$\begingroup\$ Please post a schematic here. \$\endgroup\$ – Brian Carlton May 13 '13 at 16:28
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    \$\begingroup\$ Are you able to measure the bootstrap current needed by the converter at start-up? If it is too high for the coin cell to provide instantaneously, that may be a problem. Also, any reservoir cap at the output will increase this bootstrap current requirement, so I'd play with that to check. \$\endgroup\$ – Anindo Ghosh May 13 '13 at 16:30
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    \$\begingroup\$ Re. "what's a better way?", the Texas Instruments BQ25504 Boost Converter will bootstrap from as little as 450 mV and 50 microWatts, and operate down to 80 mV once cold start is complete. \$\endgroup\$ – Anindo Ghosh May 13 '13 at 16:44
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    \$\begingroup\$ 50ma is quite a lot for a coin cell. If the startup current is higher than this, the converter may never start up. \$\endgroup\$ – Brian Drummond May 13 '13 at 19:51
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    \$\begingroup\$ Which coin cell are you using? \$\endgroup\$ – Adam Lawrence May 13 '13 at 20:07
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  1. Schottky diodes don't make great blocking diodes for the purpose of power rails. They have often have high reverse leakage currents.

  2. Your initial in-rush current is probably high causing the battery voltage to dip below the part's UVLO. To verify this put a scope on your battery and set it to trigger on it dipping more than 100mV below it's nominal value. Make sure you are looking at things at micro-second resolution (like 10 us/div). Post your results here.

Also, if your current requirements are low (<150mA) you can get away w/ a charge-pump.

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protected by Kortuk May 14 '13 at 1:52

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