Logic Level Power Switching

Discussion in 'The Projects Forum' started by Robert Seal, Feb 1, 2016.

  1. Robert Seal

    Thread Starter New Member

    Jan 22, 2015
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    View attachment PWR-CIR-FET.pdf

    The enclosed schematic is that of a 3.7v (LiPo) battery charging circuit that will derive it's power from a 5VDC wall wart (for recharging purposes only). To the far right of the circuit you'll see P-Channel Mosfet (Q1) which I would like to use to switch from battery power to 5VDC when it's available, then back to battery when the wall wart is removed.

    I have never used any sort of Mosfet before and while I vaguely understand how it works in the circuit, a lot of the terminology eludes me when choosing one for the job. I believe that since my operating voltage is 3.7-5VDC (and future load will operate at 5V), I will need a logic level Mosfet. I'm also looking for SMT that can be hand soldered so the two I've found are:

    http://www.digikey.com/product-detail/en/NX2301P,215/568-10314-1-ND/4162477 and
    http://www.digikey.com/product-detail/en/DMP2123L-7/DMP2123LDICT-ND/1964766

    Now my very basic understanding of how it works is that the Vgs (1.1v and 1.25v as linked) is the voltage differential that the FET will look for at the gate. Since the battery will power down to around 3.7V but operate up to 4.2v, there are times were the differential will be less than 0.8V, do I need a FET with a Vgs of <0.8V?
     
  2. Robert Seal

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    Jan 22, 2015
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    Pretty please?
     
  3. ronv

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    Nov 12, 2008
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    It is not clear to me that it works. The FET looks like a source follower, not a switch.
    What is the FET part number?
     
  4. ronv

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    Nov 12, 2008
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    Ahh, I see now.
    Flip the PFET over so the source goes to the battery and the +5 turns it off.
    Then it just needs to be a logic level FET so the 3.7 volts will turn it on.
     
  5. tracecom

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    Apr 16, 2010
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    Isn't the /PG output supposed to be used to control that function?
     
  6. Robert Seal

    Thread Starter New Member

    Jan 22, 2015
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    View attachment PWR-CIR-FET.pdf

    I've updated the schematic after reversing the Mosfet. Now again, I am a novice at this but this is how I see the circuit working as currently designed, please correct me if I'm wrong.

    1. The P-Channel Mosfet is normally closed between the Drain and the Source, so as long as the battery has power there should be 3.7-4.2V flowing through it to the load.
    2. If +VIN (+5V) becomes available through the wall wart, it will:
      1. Activate the MCP73833 to charge the LiPo battery.
      2. Provide the necessary power to the Mosfet to (open?) the gate, cutting battery power from the load.
      3. Flow through D4 to power the load.
    If the above is true based on the schematic uploaded here, then that is exactly what I'm looking for. What I'm not (seemingly) able to do is decipher the properties of a P-Channel Mosfet (Logic Level?) to choose the right one.

    I'm currently looking at this one: http://www.digikey.com/product-detail/en/FDN340P/FDN340PCT-ND/965604

    I believe the /PG is just a status indicator, the following is from the datasheet.

    From the Datasheet:
    3.5 Power Good Indication (PG) MCP73833 Only The power good (PG) option is a pseudo open-drain output. The PG output can sink current, but not source current. However, there is a diode path back to the input, and, as such, the PG output should only be pulled up to the input. The PG output is low whenever the input to the MCP73833 is above the UVLO threshold and greater than the battery voltage.
     
  7. tindel

    Active Member

    Sep 16, 2012
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    What is wrong with just running the load directly off of the battery? If the wallwart is connected it's charging the battery and providing power to the load - albeit through the drop transistor of the linear regulator.
     
  8. Robert Seal

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    Jan 22, 2015
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    Honestly I don't know that anything is wrong with that scenario. This is my first time working with Lithium batteries and I guess I'm just erring on the side of caution. Are there no risks or drawbacks to running the circuit off of a battery that is simultaneously being charged?
     
  9. tracecom

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    Apr 16, 2010
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    An output that changes state when the power is good could be used as only an indicator, but could also be used as a control signal to switch battery charging on and off, thus obviating the need for much of your peripheral circuitry.
     
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  10. tindel

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    Sep 16, 2012
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    Actually now that I think of it - driving the load through the charging circuitry is NOT what you want to do since the charge controller is controlling the current during the first part of the charge. I also know that you do not want to use the schematics that you have laid out thus far - they won't work. I do think you want to use the #PG pin but you may need a bjt to invert the logic. I've got to think about this one a little bit - and I haven't had my coffee so the brain cells aren't quite firing this morning yet.
     
  11. tindel

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    Sep 16, 2012
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    So I think this is more what you want. You probably need a second diode in series with D2 just to drop the voltage a bit more because there is a parasitic diode in the pMOSFET that you do NOT want to conduct. There will also be some timing issues... seems that there there should be a cap somewhere to control turn-on/off times and the such - but assuming you're not trying to make a sellable product - this will probably be okay.
     
  12. Robert Seal

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    Jan 22, 2015
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    So while I wouldn't question your intelligence level in electronics because it's obviously far superior to mine at this point, I am curious why the original circuit wouldn't work before redoing it all. The schematic that I've provided is straight out of Microchips AN1149 document - Page 2 with little to no modifications whatsoever.

    http://ww1.microchip.com/downloads/en/AppNotes/01149c.pdf
     
  13. tindel

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    Sep 16, 2012
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    Oh - that app note is very helpful... It makes a lot more sense... I also missed the pull-down resistor on the gate that sure helps the situation too. That circuit should work... but the choice of the FET is critical. I'd try the FDN340 and then the NX2301 that you were looking at.

    I also don't think that the polarity of the Source and Drain matter much in this circuit.

    Edit - oops - the source drain does matter and should be connect as shown in the app note - connecting it the other way will charge the battery due to the parasitic diode that I was concerned about in my circuit.
     
    Last edited: Feb 2, 2016
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  14. tracecom

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    Apr 16, 2010
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    Here is a link to the Adafruit product that uses the MCP73833. You can download the Eagle schematic for it. At first glance, it doesn't appear to use an external MOSFET.

    https://www.adafruit.com/products/259
     
  15. Robert Seal

    Thread Starter New Member

    Jan 22, 2015
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    First and foremost, thank you both very much for helping with this.

    I'm glad it made sense after the app note, you had me worried. Also, thank you for the recommendation of Mosfets to use as this was what I was originally trying to gather from this post. As mentioned in the OP, I do not fully understand how the gate works - Does it look for a certain voltage period before opening or is the voltage at the gate based on the differential between G and D? A certain voltage (ie: 5V) would be easy but if based on differential then I'm only looking at 0.8V in a perfect world (4.2V on Batt, 5V on Wall Wart).

    I could be over complicating it but I'd like to understand if you have the patience to enlighten me in lamens.

    Thanks for this, I didn't realize Adafruit posted their schematics. Out of curiosity though, what happens if I didn't use a switch and have both 5V and 3.7V running to a 5V booster/regulator? From the point where they meet, up to the booster/regulator is it just a 5V (as the higher of the two) circuit or do they combine for 8.7V?
     
  16. tindel

    Active Member

    Sep 16, 2012
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    pFETs turn on when a negative voltage is applied between the gate and the source. The more voltage - the more the turn on. You can also put a medium voltage on Vgs and the fet will not turn all the way on... it's somewhere in the middle.
     
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