# Pump does not get enough current via MOSFET

I am trying to build the v4.3 circuit described here: Will this schematic work?

I am building just one branch out of the decoder IC to test on a breadboard. Component info is as follows:

1. The optocoupler is PC817 i.s.o TLP621;
2. The MOSFET is IRF630B;
3. The flyback diode is 6A4;
4. The resistor between MOSFET gate and ground is 5.2K;
5. The peak power rating of the pump is 2.7A @12V.

Now when power is on, RasPi is up and the input to the decoder is set correctly, the pump does not work.

I tried the following to troubleshoot,

1. Connected the pump directly to 12V input, it works. Connected a multimeter in series and found the current draw to be approx 3A to start and then dropping down to 1.8A;
2. Connected an LED via a suitable resistor across the output side of the optocoupler (no mosfet/pump) and it turns on and off depending upon logic input to the decoder;
3. Removed LED and measured the voltage across the output side of the opto-coupler to be ~12V;
4. Added the mosfet and measured the voltage across the drain and source with correct logic input to decoder. This measured ~11-12V;
5. Added pump (w/o flyback diode). Now with correct logic input to decoder, it does not work;
6. Connected a multimeter (in current reading mode) in series between the cathode of the pump and drain. It measures only 0.63A;
7. Replace the pump with LED and proper resistor and enable the decoder, the LED lights up.

It appears to me that connecting the pump via the MOSFET somehow impedes the current flow to a great extent even when it is biased according to specs.

I thought that the MOSFET in question is a voltage controlled device. So when a proper $V_{gs}$ is applied, it should simply conduct the current required by the load.

Why does it conduct only 0.63A when the starting current requirement is around 3A? Am I missing anything?

• What Vgs of the MOSFET can/did you measure?
– Fizz
Oct 14 '15 at 19:29
• Breadboards are not designed for carrying currents of that magnitude. Try soldering the cables for the pump to the MOSFET and power supply directly. Oct 14 '15 at 19:30
• I doubt that's the main reason though. For the breadboard to drop the current to about 1/5, it would have to have a lot of resistance (around 15 ohms), and would have started to smoke with 10W dissipation.
– Fizz
Oct 14 '15 at 19:51
• I assume you have a heatsink on your FET ? 36watts is enough to damage it. Also try checking the FET with the diode checker on your Multimeter and make sure it still functions and you have your pin orientation right.
– BenG
Oct 14 '15 at 19:55
• Ok, I connected my multimeter directly across the mosfet before & after the 5.2K resistor. Assuming Iam measuring Vgs correctly and with the decoder enabled, I see that Vgs starts at 10.68v and slowly rises to 11.25. I stopped there because I could sense that the mosfet has become very hot. My apologies to @BenG, I wasnt looking close enough the last time. Oct 15 '15 at 7:14

Step (3)after edits. If there's 12V across the opto output, there's about 0V across the 5K gate resistor and the FET is OFF...

Try again with the optocoupler ON (probably 10mA through its LED. max 2V across the opto, so that Vgs=10V and the FET turns on.

• Phototransistor current will be limited to about 2.5mA because of the 4.7K resistor. That will cause the MOSFET to turn on rather slowly so there's risk of burning it, but it should turn on... eventually.
– Fizz
Oct 14 '15 at 20:05
• Yeah but he's saying there's practically no photocurrent (12V across the opto output), not 2.5mA. If that's so, the problem lies earlier. Oct 14 '15 at 22:56
• Who knows? He says "Connected an LED via a suitable resistor across the output side of the optocoupler (no mosfet/pump) and it turns on and off depending upon logic input to the decoder; Removed LED and measured the voltage across the output side of the opto-coupler to be ~12V;" I didn't exactly grok what he measured there (was the driving signal on or off?) He seems to have enough current to light up a LED on the opto's phototransitor side.
– Fizz
Oct 14 '15 at 23:06

Keep your freewheel diode .It protects the mosfet from voltage spikes at turnoff.Use a logic level fet or anything that is properly specified at lower Vgs .Substitute a lower RDS on fet which will be easy to find for VDs of say 50Volt.