MDBT42Q and 5v water pump

@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? ...to 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.