MDBT42Q and 5v water pump

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  • Hi,

    I'm trying to do the same thing as here:­sPs&t=312s

    The only difference is that I'm using MDBT42Q and it's +/- bat pins with 5v and not an external power source. (without a diode for now)

    The switch pin in my case is D18. (See attachment photo)

    What could be the problem?

    Thank you,

    1 Attachment

    • IMG_20201020_135035__01.jpg
  • Hi - that all looks great. Are yo using the exact same FET? I guess some of them may have different pinouts.

    And what happens? Nothing at all?

  • Hi Gordon,

    Thank you for your quick reply.

    I'm using this FET.

    Nothing happens.

    If I connect it directly to the board +/- 5v it works fine so no problem with the pump it self.

    1 Attachment

    • Screenshot_20201020-145014.jpg
  • Honestly what you've got there looks perfect.

    And doing D18.set() does nothing? Maybe you could check the voltages on each of the FET pins with a volt meter?

    Is it possible you could have wired the FET up wrong previously and blown it up?

    I just wired up your circuit on breakboard, with that exact P36NF06L FET and the exact same water pump, running off a ~4v LiPo battey, and it works great.

  • Hi,

    I did digitalWrite(D18, 1), now I tried D18.set() and D18.reset() same result.

    I have 5 pcs of those new FETs I'll be surprised if they are all broken, what is the proper way to check/test them with multimeter?

    Thank you.

  • This is the test I did on FET

    Video with length of 44 seconds­VVsWkoX506Xw6I3b0pwM3S2ny/view?usp=drive­sdk

  • Hi - sorry, when I meant check the voltages I meant to check the voltages with it all plugged in to the MDBT42Q.... So to see if the voltage from D18 was getting to it for instance.

    It could just be a dodgy connection on the breadboard. As I say I have wired up the circuit exactly as you have done to test, and it works great for me.

    1 Attachment

    • IMG_20201020_135319.jpg
  • Hi,

    This is what happens every on/off of D18

    D18 = 0 - 0v
    D18 = 1 - 5.08v
    D18 = 1 With LED - 2.61v
    D18 = 1 LED removed - 3.46v

    Thank you.

    4 Attachments

    • D18_0.jpg
    • D18_1_WithLoad.jpg
    • D18_1.jpg
    • D18_1_AfterLoad.jpg
  • @michael_101

    I'm not sure what you measure is what @Gordon is asking for.

    Have D18 connected to the Gate of the MOS FET and pump con connected. Then with for both states - D18.set() and D18.reset() - measure - with Multimeter set for DC Voltage measurement - the Voltages from following points to GND:

    • D18 pin / Gate of FET
    • Source of FET
    • Drain of FET

    Then power off everything (disconnect from power source) and replace wire from FET Drain to Pump with Multimeter with setting for Current measurement. Then power all on and measure the current with D18.set() and D18.reset().

    With breadboard contacts can be flaky. Therefore you measure at the FET Gain pin and compare what you measure at the MDBT42Q Pin 18. To catch eventual contact issues with the wiring, use also other points on the bread board. Of course you can also measure MDBT42Q Breakout Pin 18 output voltage with Pin 18 not connected.

    Your Multimeter has an auto setting which will not always measure what you want to measure.

    Comments to your measurements so far... they look good to me... may be with one exception: what is the difference between picture 2 and 3?

    Also, can you show 'the other side' of the FET? see the orange, white and eventual other connections that happen?... (you can also just bend down the FET so the pins and wirings and breadboard points become visible and update the pics.

    Btw, 'never ever' run an LED directly... use a resistor 'in Series'. A forward bias voltage lower than the source (5.08V) is like a shortcut of the source (2.61V) and if the source (and eventual controlling circuit - FET) is strong enough it will heat and eventually burn up the LED, or, if the controlling circuit - FET - is 'weaker' than the LED, it will heat and eventually burn up.

  • So by the look of it you're saying that if you connect a LED across the wires, you can control and LED using D18 as expected? Just not the pump?

    Also as @allObjects says, what happened in the final picture D18=1, LED removed? In that case, what's the actual voltage across the voltage input? Is it still 5V?

  • Hi,

    Thank you @Gordon and @allObjects for your detailed answers I learned a lot from them and sorry for the delay.
    I am a software developer and I am new in all electronics thing.. so thank you for your patience with me ;)

    I got it to work, apparently all 5 FETs I had were not good.

    I did the tests as you asked me with this results:


    • (D18 pin / Gate of FET) with GRD => ~3.3V
    • (Drain of FET) with GRD => 5V


    • (D18 pin / Gate of FET) with GRD => 0V
    • (Drain of FET) with GRD => 5V

    About the current measurement from FET Drain:

    • D18=1 => 0A
    • D18=0 => 0A

    So I bought 25 new same FETs and I did the same test (20% of them were not good too).
    The good ones worked fine with same results, except the current measurement that was ~100mA, and the FET Drain was ~8mV when D18=1

    Thanks again.

  • Wow. That's a really bad failure rate.

    I stopped buying my stuff from aliexpress and started using digikey for bulk purchases (since they do free shipping for over $50) or element14 for smaller batches (which is slightly more expensive per item but free shipping all the way).

  • Wow, yes, that is bad. Where did you get the FETs from?

    People do actually counterfeit electronic components sometimes, and it's possible that you actually got some counterfeit parts.

    It may be that the FETs do work, but they just need higher voltages. I'd chosen the P36NF06L for some examples because it's one of only a few 'big' FETs that conducts a useful amount of power at 3.3v gate voltage - but it's possible someone repackaged some less good FETs in a P36NF06L package!

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MDBT42Q and 5v water pump

Posted by Avatar for michael_101 @michael_101