Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The idea of this project is to have my car's high beams work as daytime running lights (DLR), with reduced light output. The DRL would turn on with the ignition on, and turn off, either when the parking lights are turned on or the ignition turned off. I found a PWM circuit online that I used for reference: The easy or hard way to build a PWM dimmer, but instead of putting the load on the high side, I put it on the low side.

The MOSFET I am using is BUZ11A.

Now, I temporarily assembled the circuit on my car to see whether it would work, and it did, everything worked as I planned, but the MOSFET I used got so hot that it started to burn off the tape which was wrapped around its legs, so I pulled the plug. I did use a heatsink.

Can anyone help me out with this, what am I doing wrong here, what am I missing? I want it to work as it did, but I can't have this this burning up things under the hood :). I'm no expert when it comes to MOSFETs, I do have an understanding of basic electronics, as I took electronics classes in college.

Here's the schematic: Schematic

(Full resolution image)

share|improve this question
Could you do a simple test? Disconnect the 555 from the MOSFET, and instead connect the GATE of the MOSFET to 12V. Does it get hot as quickly as when it was driven by the 555? – gbulmer Sep 1 '14 at 15:37
What SP said. Easiest solutions are (1) P Channel FET or (2) As GBulmer said - Add a battery as a high side supply for the gate (or a charge pump). Of these the PFET uses less complexity long term. The BUZ11A is a bad choice in your present configuration as it needs 6 to 8 V gate drive (see fig 5 in datasheet) to reduce Vds to under 1/2 a V. Using a very low Vgsth (gate turn on voltage) FET would allow your present cct to work at reduced heat. BUT a P Channel FET will be MUCH better. – Russell McMahon Sep 1 '14 at 15:59

The problem is that you need to give the MOSFET gate a voltage that is positive by 10V or so wrt the source to have it turn fully on. Your MOSFET is connected as a source follower so it will drop probably 5 or 6 volts when it is turned on, which translates into a LOT of heat loss.

You could try adding this circuit, which AC-couples the gate drive so that it's always positive wrt the source.


simulate this circuit – Schematic created using CircuitLab

It's still pretty hard on the MOSFET when it switches off (especially) and on at the beginning and end of the PWM (the MOSFET can dissipate up to 30W or so for a brief time) but it might be okay with a relatively rugged MOSFET.

Alternatively, simply swap the N-channel MOSFET out for a similar P-channel MOSFET (which reverses the gate drive 'logic' - low = ON).

You might want to hang a fat TVS across this whole thing.. 1500W type or more.. it might save the MOSFET gate and 555 when transients happen (eg. when jump starting the vehicle).

share|improve this answer

Looking at the circuit you've posted, it seems like the MOSFET is connected 'upside down' and on the high-side. An N-MOSFET is not going to switch that way round.

Either use a P-MOSFET as a high-side switch, or use switch the N-MOSFET by driving its gate with a voltage even higher than 12V (that is typically done using a circuit called a charge pump).

To help size the component you need, the two 55W bulbs will normally use 110W.

You don't say whether the bulbs are LED or incandescent. An incandescent bulbs resistance is significantly lower when they are 'cold', so the initial current is bigger. However, for now, let's assume 110W is an appropriate basis for calculation:

Power = I*V

I = Power / V = 110W / 12V ~= 10A

Ohm's Law V = I * R

So power in a MOSFET with on resistance Rds is

Power = I^2 * Rds = 100 * Rds

Pick a reasonable amount of heat dissipation, and you get the Rds(on) you're looking for.

Under a car bonnet/hood, where it is already hot, I'd prefer to limit the power dissipation, and hence temperature rise to something quite low, say limit the temperature rise to under 25C. The datasheet for the MOSFET usually says how much temperature rise per watt. Using a heatsink will improve on that that, but using the datasheets spec without including that factor will be safe.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.