0
\$\begingroup\$

My boost-converter circuit

I'm designing a circuit to convert the 3.7-4V Li-ion battery output to a steady 5V DC power supply. The boost-converter I chose is a synchronous boost converter SGM6611C with a fixed switching frequency of 1 MHz (I'm too lazy to change the component's name, since 6611B and 6611C are basically the same in pin-out). The datasheet of SGM6611C could be found here: SGM6611C.

I followed the example design shown in the datasheet, and the outcome looks like something shown above. However, after the board is done, I found the input voltage PWR, which was supposed to be ~4V, was only 2.5V. Since the minimum input requirement of SGM6611C is 2.7V, the whole block seems to not work properly. The output voltage is only 2V.

If disconnected from the converter, the PWR voltage is correct, which means the upper-stream design should be OK. It seems that my design accidentally pulls down the input voltage so that the whole block is not working at all. I suspect that it has something to do with the inductor, since the voltage becomes normal again if I detach the inductor. Or have I made some stupid mistakes somewhere? Thanks for your guys' suggestions!

For reference, the inductor is shown in the figure below.The inductor I chose

And for the debug, I've found out that it's not caused by over-current on the input side. The input DC impedence bewteen PWR and GND is around 300k. The interesting point is, the SW pin is supposed to deliver a 1 MHz on-off switching signal, but the signal there is just the same with PSW, except for some unknown ripple.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ How does the output voltage look? Is it at 5V or below? How much current is flowing? Does anything get hot? I would expect that the boost draws too much current, resulting the Li-ion battery to drop its output voltage. \$\endgroup\$
    – jwsc
    May 11 at 6:29
  • \$\begingroup\$ Indeed, what is the output impedance of the battery ? \$\endgroup\$
    – tobalt
    May 11 at 6:51
  • \$\begingroup\$ I've checked, the output is only 2V, and the input impedance on the converter is bewteen 200k-300k based on on-board measurement. Nothing is getting hot, it seems that it's just the converter not working properly. \$\endgroup\$
    – Keyan
    May 11 at 7:46
  • \$\begingroup\$ can you switch out the battery with a lab power supply or similar? Just to rule out an insufficiently loaded battery or too high output impedance of the battery? \$\endgroup\$
    – jwsc
    May 11 at 8:03
  • \$\begingroup\$ and the input impedance on the converter is bewteen 200k-300k No. The DC input impedance of your converter while it is running is low. It is a power converter! And your battery DC impedance is by far not 0. So if it's a high power converter your battery can well be loaded down to 2.5V. @jwsc has asked all the right questions in his first comment. And I second the suggestion of using another PSU instead of the battery. \$\endgroup\$
    – tobalt
    May 11 at 8:28

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.