I'm using MOSFET transistor IRFP260N for switching on and off 12V light bulb. The light bulb is between + of 12V battery and drain of a MOSFET, and a source is on the minus pole of the battery. A voltage between gate and source is pulsating around once a second and I can measure about 5,5V there.

When I'm using 12V 6W bulb circuit works fine - a light bulb is switching on and off, but when I changed the bulb with 12V 21W the bulb doesn't come on (filament gets red just slightly and doesn't switch on completely. Also, MOSFET starts heating up a lot and that was not the case with 6W bulb. Can somebody help me to understand why I can not use MOSFET to control switching of 12V 21W light bulb? enter image description here

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    \$\begingroup\$ What's limiting the gate voltage to 5.5 V? Post a schematic. \$\endgroup\$ – winny Nov 19 '16 at 11:13
  • \$\begingroup\$ I posted schematic to original question. \$\endgroup\$ – Jakob_ Nov 19 '16 at 11:58
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    \$\begingroup\$ Just increase R2 to 10k or so. \$\endgroup\$ – Brian Drummond Nov 19 '16 at 12:32
  • \$\begingroup\$ I think you should learn how a transistor amplifies, or switches a larger current. The current being amplified is the one that comes out of the battery, through the base and out of the emitter and back to the power source. Here that current has to go through all 3 resistors, so the current would be little in Q2 and the MOSFET would still be pulled to low. You most likely need a PNP. \$\endgroup\$ – Bradman175 Nov 19 '16 at 13:05

filament gets red just slightl + MOSFET starts heating up a lot

This means that the MOSFET is not fully on. The Vgs_max of the IRFP260N is 20 V so you can just feed the +12 V to the gate.

The transistor Q2 is also used in a pretty useless way. Let me propose a different schematic:


simulate this circuit – Schematic created using CircuitLab

If M1 still does not fully turn on, then try replacing R1 by 100 kohm.

You don't need Q2 from your original schematic since there is no need for current amplification. The MOSFET's gate is basically a capacitor, it just takes longer to charge with a small current. But since this is for a lightbulb, it does not matter if charging the gate capacitor takes one second. The lightbulb is just as slow.

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The problem is with the circuit you used. The NPN transistor acts as an emitter follower as does the opto coupler. You lose 2x 0.7V before you get to R1 and R2. Then these two resistors half that voltage giving a gate voltage of about 5.3 volts. (assuming 12V supply)

enter image description here

The circuit could be improved by using a PNP BJT instead. This would give a much higher gate voltage, about 11.6V. If a lower voltage is required (e.g. a logic level MOSFET used) then an extra resistor, R4, could be added with it value calculated using the potential divider formula.

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  • \$\begingroup\$ Simply increasing R2 (say to 10k) would be a much simpler way to generate the required voltage swing. \$\endgroup\$ – Brian Drummond Nov 19 '16 at 12:33
  • \$\begingroup\$ @BrianDrummond or simply remove R1 but you still lose the 2 x Vbe drops and I consider it a much better design to switch the signal properly rather than use an emitter follower. \$\endgroup\$ – JIm Dearden Nov 19 '16 at 12:38
  • \$\begingroup\$ @JIm Dearden Thank you for an answer. But I would still like to understand why switching with 6W bulb works perfectly and with 21W not. Gate voltage is in both cases the same as the circuit is the same so also opening should be the same; or I'm wrong? With 5V on the gate and 6W bulb MOSFET opens fully and with 5V on the gate and 21W bulb MOSFET is not fully opened? Is opening of the MOSFET dependent also from the resistance of the load? This is just a test circuit for me as a primary intention was to have a load with very low resistence (that circuit is in a little different configuration). \$\endgroup\$ – Jakob_ Nov 19 '16 at 14:48
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    \$\begingroup\$ I think that you're just on the edge of saturation for the 6 W bulb and the low Vgs you have in your original circuit. Look at Figure 1 in the datasheet of the IRFP260N, to get the lowest Vds drop you need at least a Vgs of 6V, which your circuit cannot provide. So for the 21 W bulb the MOSFET will not have a low Rdson as the 21 W bulb tried to draw more current than the low Vgs will allow. So the MOSFET operates in saturation mode and not linear (or triode) mode (which is where you want it to be). \$\endgroup\$ – Bimpelrekkie Nov 19 '16 at 15:13
  • \$\begingroup\$ @Jakob_ Take a look at the mosfet datasheet. You will see that for a certain gate voltage, there is a limited amount of current the transistor can pass. This is called saturation. \$\endgroup\$ – winny Nov 19 '16 at 19:11

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