0
\$\begingroup\$

I made a circuit that uses the DMP2305U P-Channel enhancement mode MOSFET. I connected the pin of the PIC microcontroller directly to the gate pin of the MOSFET. The purpose of this MOSFET to enable the power rail that an Xbee is connected to. I'm running the whole board off of a CR2477 lithium battery or external power with a TPS63001 buck/boost. The problem I'm running into is the CR2477 cannot run the Xbee, the PIC starts browning out and things just stop working. If I connect two AAs to the battery terminals, the board runs fine. From external power it runs just as fine. I can enable and disable the MOSFET by manipulating the pin on the PIC.

Xbee Power Rail

I believe the problem I'm running into is the buck/boost is providing 3.3V and the DMP2305U has a -0.7V drop which means there's not high enough voltage to run the Xbee which requires a minimum of 2.8V.

Assuming it's okay to run the MOSFET directly from the microcontroller, why might I have an issue running from CR2477 but not 2xAAs?

A friend pointed me to this write-up: Using MOSFETs in Load Switch Applications (AND9093 by On Semiconductor)

This tells me I need a transistor in front of the MOSFET. My success on external power and 2xAAs tells me I don't, that the 25mA I can source from the PIC pin is enough. Even if I put a transistor in front of it, I'd still have a voltage drop across the MOSFET, right?

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Which Xbee module are you using? Even one with just 1mW transmitting power draws 40mA of current at 3,3V, which is a lot for a coin cell. \$\endgroup\$
    – jms
    May 31, 2015 at 21:41
  • \$\begingroup\$ Hi jms, it's the lowest power one. \$\endgroup\$
    – Stateful
    May 31, 2015 at 21:52
  • 1
    \$\begingroup\$ Point of interest, MOSFETs have an internal channel resistance, not a fixed voltage. Your MOSFET, even driven at just 1.8V below source has such microscopic resistance, you can assume it to be zero for any load under 500mA. \$\endgroup\$
    – Asmyldof
    May 31, 2015 at 22:44

1 Answer 1

5
\$\begingroup\$

I think it's nothing to do with the MOSFET- the CR2477 simply has too much internal resistance to drive enough current to run your circuit. As the battery terminal voltage drops, your boost regulator will try to draw more and more current to keep the output voltage stable and at some point cannot work. According to the datasheet, the voltage is starting to drop substantially even at 1mA.

http://na.industrial.panasonic.com/sites/default/pidsa/files/crseries_datasheets_merged.pdf

That is typical for lithium coin cells.

\$\endgroup\$
5
  • \$\begingroup\$ +1: since the MOSFET has <100mohm of Rdson at 3.3V, it is totally unlikely that there is a drop of 0.7V across it - that would mean the load is consuming 7A! \$\endgroup\$ May 31, 2015 at 23:59
  • \$\begingroup\$ @LaszloValco Agreed. 0.7V means that either it's wired backwards or (my guess) Vgs << 3.3V. Since it works on the AA cells, Occam's razor would suggest the supply is fading. \$\endgroup\$ Jun 1, 2015 at 0:00
  • \$\begingroup\$ Actually, an average Lithium coin cell is hardly able to provide a few milliamps... \$\endgroup\$ Jun 1, 2015 at 0:07
  • \$\begingroup\$ @LasloValco if D/S swapped it would drop 0.7V only when 'off' \$\endgroup\$ Jun 1, 2015 at 0:08
  • \$\begingroup\$ Did any of you actually look up the CR2477? It's a huge beast of a coin cell. I very much doubt it would have such a high internal resistance as to be the problem. Even so, I suspect that this problem would be solved by putting a large value capacitor (470uF or similar) across the power supply to the XBEE - those things pull rather high current pulses. \$\endgroup\$ May 11, 2016 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.