# Using a single power source for both the gate and source of my transistor?

I am using an IRFZ44N MOSFET (n-type) transistor to regulate the current of a Lithium-ion battery, and I'm trying to use the battery to supply power to both the drain and gate of the transistor. My goal is to allow current to flow from drain to source when I apply voltage to the gate (when I attach the (+) wire to the gate), and stop the current flow when I connect the transistor gate to ground (when I attach the (-) wire to the gate). In my schematic I left the circuit open, but depending on what I want my gate voltage to be I connect either the (+) or the (-) wire momentarily.

My problem is that I don't understand why I'm reading certain values on my ammeter. When I connect the (+) wire to supply voltage, I should see a current of 3.7V/10ohms=370mA, but I'm only reading a value of around 30mA. However, if I use a separate power supply for the gate of the MOSFET with everything else unchanged, I get an ammeter reading of 370mA as expected.

I know the gate-source voltage is the same in both scenarios because I measured it with a voltmeter (measured about 3.5V), and when I use a separate power supply I'm also using a 3.7V battery.

Why does using the same power source to supply the gate voltage only let 30mA through?

simulate this circuit – Schematic created using CircuitLab

• It's a MOSFET. There is no base. Or emitter. Or collector.
– user16324
May 4, 2018 at 23:27
• And as drawn your body diode is forward biased so you won't be able to decrease the current below about .3A. Not to mention that you need Vgs to be >Vt which you can't get by using a voltage equal to or less than the source potential. May 4, 2018 at 23:36
• Nchannel fets connect to the Negative battery side. The little arrow thing in the symbol actually goes straight across S->D and it conducts that way (S+>D-) always. Only the other way is switched (D+>S-) May 5, 2018 at 0:03
• Why don´t you use a transistor to switch between positive and negative in mosfet gate? Also you need a greater votlage source as JOHN D said. May 5, 2018 at 0:25
• when you "short" the battery the voltage must drop below VgsTh, thus increasing the FET's resistance and limiting current in that regard. In short, you'll float source-vgsth, which might not be above vgsth unless the battery is double the threshold. May 5, 2018 at 13:37