Why does this automotive blinker's N-MOSFET get so hot?

Question

It works fine in the protoboard, with 5, 10 or 21 W bulb (didn’t test it with more than one of such bulbs tho), however when I connected to a car’s (a clunker, may I add) turn indicator harness, it does blinks but the MOSFET gets very very hot. I didn’t run it for too long, perhaps 20 blinks and it was untouchable…why it got so hot?

I used an IFR630, 200V 9A N-channel MOSFET. Four 12 V/21 W car bulb may be consuming around 8 A. I didn’t test four at once but only 2. Could it be the car's harness loose ground, or positive to ground leak? Or it just can't handle the load? From the site I got this schematics, people say it worked for them just fine.

UPDATE: I have tried quite a few changes, many of these guided by the brilliant answers here, and once ok in the simulator I assembled in the breadboard for a test. They all worked fine as long as I don't add 21w x 4 + 2 x 10w + 2 x 5w bulbs as a full load test. All of the Mosfets I tried, both n and p, both on high or low sides turned into bread toasters! So seems a no-go project and instead decided to build a simple relay blinker using just a pnp transistor as driver. Of course, it works ok with all that load, since the relay power contacts can handle it. Not what I wanted but...well...works.

Answer 1

…why it got so hot?

The reason is that the NMOS FET cannot turn on fully. In order for tha to occur, the gate voltage must be at least 17 volts.

Could it be the car's harness loose ground, or positive to ground leak?

No.

Or it just can't handle the load? From the site I got this schematics, people say it worked for them just fine.

You should check back with the website. The circuit would work fine if it is used as a low side switch as shown in Figure 2. As a high side switch (Figure 1) not so much. See my diagram.

The "old clunkers" connect the negative of each lamp to chassis to return the current to the battery. All switching is done on the high side, no choice. As more electronics appeared in cars the lamp's current returned in a separate wire and so a low side switch would work. I suspect that this is the case for the folks on the website that you got the circuit from.

You need to look for a circuit that will fit between the battery and the lamp (high side). The circuit may work if you attach a heat sink as in GodJihyo answer.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

Your MOSFET is acting as a common drain amplifier, AKA a voltage follower. It behaves like a linear regulator -- the MOSFET and bulbs are in series, and any "extra" voltage that would bring the total bulb current is dropped across the MOSFET instead.

For example, let's say you connect one bulb with a resistance of 0.1 ohms, and your circuit provides 2 amps. The bulb's voltage is 2 volts. The remaining 10 out of 12 volts are dropped across the MOSFET. So the MOSFET dissipates 10 volts * 2 amps = 20 watts of power, which causes it to heat up quickly.

Even if it takes a much higher voltage to power the bulbs, you'll probably need 2-3 volts between the gate and drain just to turn on the MOSFET, which still means a lot of power.

If your bulbs can handle the full 12V of the car, switch on the low side instead -- connect one side of the bulb to the 12V line and the other side to the MOSFET's drain, then connect the MOSFET's source to ground. Now you can turn the MOSFET all the way on and it will drop a fraction of a volt.

Note that you might have to redesign your oscillator to get e.g. a 0-12V output; I don't have time to look at the oscillator in detail.

EDIT: Since no one else was doing it I went ahead and simulated your oscillator. Please ignore the artifacts at the start; I added a hack to make it start oscillating:

^{simulate this circuit – Schematic created using CircuitLab}

As you can see from the power simulation, the MOSFET dissipates around 30 watts on average when it's turned on. That's enough to heat up anything. You can also see that the source voltage (VO) is approximately equal to the gate voltage (VG) minus a Vth drop, as you would expect from a common-drain amplifier.

Answer 3

A number of things will contribute to this.

First, you don't mention a heatsink. Are you using a heatsink? You need to use a heatsink.

Second, it sounds like the bulbs you are driving are incandescent. Incandescent lights will draw considerably more current than their wattage rating would indicate while the filament warms up to full brightness. Once lit they may draw 1.75 A at 12 V, but for the time it takes them to get hot they might draw twice that. And since you're flashing them they're going to be constantly going cold/hot/cold/hot. Also, the voltage in a vehicle will be higher than 12 V when the engine is running, typically close to 14 V, so they'll pull even more current under those conditions.

Third, look at the datasheet for the specifications. 9 A is the absolute maximum rating for continuous drain current at a case temperature \$T_C\$ of 25 \$^\circ C\$ and \$V_{GS}\$ at 10 V. At 8 A you're close to the limit and if you don't have a heatsink keeping \$T_C\$ at 25 \$^\circ C\$ you'll be over the limit. Generally you design for a 20% safety factor so if you expect 8 A you design for 9.6 A, going to the next even value would be a 10 A device.

Drain-source on-state resistance is rated at 0.4\$\Omega\$ max, typically it will be around 0.2-0.3\$\Omega\$. taking worst case the power dissipated at 8 A would be \$8A^2*0.4\Omega = 25.6W\$, so yeah, it's going to get a bit hot.

This is of course beyond what the other answer says about the circuit topology, you'll have to straighten that out as well.

Answer 4

Here is the same circuit as shown by the OP, except with NPN and PMOS devices, and connected from the battery to the lamp, which is grounded.