Switching 36 V DC motor with IRF3205 MOSFET

Question

I'm attempting to switch a 36 V DC motor using an IRF3205 MOSFET, but I've now blown up two MOSFETs and I can't figure out why.

I've built the circuit as below, and it works correctly for about 30 seconds or so (the motor starts and stops as expected as the microswitch is closed and opened), but then the MOSFET is destroyed and the motor runs even without any gate input (i.e. with the switch open).

^{simulate this circuit – Schematic created using CircuitLab}

If I'm reading the data sheet for this MOSFET correctly, it has a gate threshold voltage of 4 V and and a maximum gate-to-source voltage of 20 V. I have confirmed that the gate-to-source voltage from the LM317 is ~13 V, which if my understanding is correct, should mean the MOSFET is fully 'on' while being well below the maximum gate-source voltage. The 100K pull-down resistor from the gate to the source appears to correctly switch the MOSFET off when the switch is opened (prior to the MOSFET failing).

To be fair, I don't have the specs of the DC motor on hand, but running under zero load I absolutely expect that it is drawing nowhere near the 200 W maximum power of the MOSFET, which has a TO-220 heatsink installed and is barely warm to the touch.

What am I missing here? Additionally - happy for suggestions on a simpler circuit to achieve the same result :)

Answer 1

As others have pointed out, without decoupling capacitors at the input and output of the LM317 regulator, it's likely to behave very badly, and the output will ring or oscillate, exceeding your calculated output of 13V.

Even with appropriate capacitors, switching the input of a regulator in this manner is not a good way of obtaining a signal to operate the MOSFET's gate. These regulators are designed to be very low impedance voltage sources, which are good for providing high currents at a fixed potential, but your application here does not seem to require this. All you need, to operate the gate, is a potential that changes between 0V and some value somewhat under \$V_{GS(MAX)}\$, which can be obtained with a simple resistor divider:

^{simulate this circuit – Schematic created using CircuitLab}

With switch SW1 open, R2 holds gate potential at \$V_G = 0V\$, and the MOSFET is off. When the switch is closed R1 and R2 form a potential divider producing this potential at G:

$$V_G = 36V \times \frac{R_2}{R_1 + R_2} $$

Using the shown values for R1 and R2, this produces a gate potential of \$V_G=+13V\$ with SW1 closed, enough to turn on the MOSFET without exceeding its maximum \$V_{GS}\$.

Zener diode D2 is optional, the resistor divider is sufficient to ensure that gate potential is capped at 13V. D2 provides another layer of security for the MOSFET, in case R2 somehow "disappears", causing gate potential to rise to the full +36V of the supply. D2 places two constraints on gate potential; it prevents \$V_G\$ from becoming more negative than -0.7V, or more positive than +15V.

This still may not be sufficient though. The gate driving circuitry here will switch the MOSFET on quite slowly, and for a brief time (especially when the motor starts from a complete stop), it may dissipate power beyond its safe operating area. You may need to drive the gate from a 0V/13V source of much lower impedance than R1 and R2, to switch the MOSFET on and off more quickly. Consider using a proper gate driver IC.

In the meantime, you can make use of the LM317 in a slightly different way; instead of switching its input, switch the output, like this:

^{simulate this circuit}

I've included C1 and C2 to address their absence in your own design, and keep the LM317 from oscillating. It's output is a fixed 14V, and this low impedance source of 14V is applied or removed from the gate by SW1. In this way we avoid any overshoot, ringing or slow output slew that the regulator may exhibit when its input is switched on.

Since we now have a much lower impedance, and constant source of 14V (much better than R1 and R2 provided in the previous design), we are able to obtain much greater gate charging current (up to 1A or so), and R3 here can be very small, permitting the MOSFET to switch on much more rapidly.

Answer 2

I have quite a bit of experience with motor circuits. If I was doing this circuit, I would use a PWM ramp for start-up of the motor. Using a dual 555 timer with one side for the PWM and the other for the ramp would make an elegant solution. It would also give the ability to vary the acceleration rate of the motor for smooth operation. The dual 555 takes little space.

Answer 3

You want to use the mosfet just as a slow switch on and off, as it will follow a user turning on and off, for this reason, you don't need a driver probably, as that will only speed up the turn on and off process.

You can just simplify the driving by using a voltage divider from your power supply. You can also add a zener to ensure the gate votlage is not beyond rating, something like:

About the device breaking, it would be interesting to know whether it is due thermal reasons or by switching. If you can leave it going and it just breaks by itselfs after X seconds then that's probably thermal breaking and you might need a larger mosfet or attaching a heatsink. If it is due to driving, you might be connecting the freewheeling diode backwards, your driving circuit might be messing with you (used my proposal instead).

Answer 4

You have several possible problems, the LM317 is probably oscillating like crazy as there is no decoupling on it. This could cause heating in the MOSFET and LM317 as the oscilationing is generally not at full amplitude. The 100K pull down resistor is not needed as you have that on the regulator output. The pull down resistor when used should go on the driver side of the series resistor so you do not divide the gate drive voltage. If you have extra voltage as you do then it is no problem. Insufficient Gate voltage will cause the MOSFET to not be properly enhanced. I would like to see the gate resistor in the 100 Ohm range to give a fast turn on/off.