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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.

enter image description here

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.

enter image description here

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    \$\begingroup\$ a MOSFET in this situation would usually get hot if you didn't turn it completely on \$\endgroup\$
    – user253751
    Mar 31, 2023 at 20:17
  • \$\begingroup\$ Can you simulate this circuit? I suggest LTspice because it's easy to determine wattage of each component. However, most simulators do not model an incandescent lamp very well, so you may need to create a sub-circuit to provide the cold resistance at start-up and then the hot resistance after 100 ms or so. Take a look at circuitlab.com/forums/general-electronics-discussion/topic/… and electronics.stackexchange.com/questions/534086/… A capacitor in series with a resistor might also work. \$\endgroup\$
    – PStechPaul
    Mar 31, 2023 at 22:07
  • \$\begingroup\$ You might look at the circuit in this post, but it may also dissipate a lot of wasted power due to the slow linear gate drive. A proper gate driver or a Schmitt trigger device might improve it. electronics.stackexchange.com/questions/659919/… \$\endgroup\$
    – PStechPaul
    Mar 31, 2023 at 22:22
  • \$\begingroup\$ IFR630? Not IRF630? \$\endgroup\$
    – Hearth
    Apr 3, 2023 at 3:08
  • \$\begingroup\$ Did you actually try the circuit I showed in my answer? According the simulation, the MOSFET should dissipate much less than 1 watt. What is the total load? 114 watts? That would be about 10 amps. I'll check my results with a 1 ohm load. \$\endgroup\$
    – PStechPaul
    Apr 6, 2023 at 18:39

4 Answers 4

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…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.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Yes, but Fig 2 shows a not viable option since the bulbs are already grounded to chassis, like in Fig 1. \$\endgroup\$ Apr 6, 2023 at 1:52
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    \$\begingroup\$ As staed in my answer, You need to look for a circuit that will fit between the battery and the lamp (high side) Figure1. The circuit may work if you attach a heat sink as in GodJihyo answer. Figure two just demonstrates the difference between high side and low side. It won't work for you. \$\endgroup\$
    – RussellH
    Apr 6, 2023 at 2:37
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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:

schematic

simulate this circuit – Schematic created using CircuitLab

Transient voltage simulation Transient current simulation Transient power simulation

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.

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    \$\begingroup\$ The OP's circuit looks like an attempt to create a 2-wire blinker circuit to replace an electro-mechanical (bi-metallic strip) type. For a pin-compatible replacement circuit only one supply connection and the load connection are available. It may be possible to redesign to steal and hold power (diode and capacitor) from the bulb circuit while the lamp is off. \$\endgroup\$
    – Transistor
    Mar 31, 2023 at 21:52
  • \$\begingroup\$ The problem described in the first and fourth paragraphs applies to circuits when the gate is given a certain voltage relative to the bulb's ground, which is not the case in this circuit, where the source acts as the ground for this circuit. The second paragraph also does not make sense because it is not clear why the circuit should only provide 2 amps. The third paragraph does not make sense because it is not clear how you know the asker's circuit doesn't supply 2-3 volts between gate and drain. Overall this seems like a rote answer to a different question than what was actually asked. \$\endgroup\$
    – user253751
    Apr 1, 2023 at 9:23
  • \$\begingroup\$ @user253751 I don’t think the driving circuit (which, as I said, I didn’t talk about) changes my answer. What makes it a common drain amp is that the load is connected to the source. Since the gate voltage ultimately comes from the same 12V supply, the source must be at least Vth volts below the 12V at the drain, which means the MOSFET will dissipate significant power. The point of my answer is that if you want the MOSFET to stay cool you should switch on the low side. \$\endgroup\$
    – Adam Haun
    Apr 2, 2023 at 0:02
  • \$\begingroup\$ @Transistor Good point. A PMOSFET would be a better switching element, then. \$\endgroup\$
    – Adam Haun
    Apr 2, 2023 at 0:04
  • \$\begingroup\$ @AdamHaun no, "the load is connected to the source" is not what makes a common drain amp. It is actually that the drain is in common between the input and the output - and in this circuit, it is not. If the blinker is on the negative side of the bulb you could consider it common-source; if the blinker is on the positive side, then nothing is in common between the input and the output. \$\endgroup\$
    – user253751
    Apr 2, 2023 at 10:20
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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.

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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.

Blinker simulation

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