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I have a circuit based on 555s that flashes a 12 V, 10 W lamp. I am using an IRF540 MOSFET to drive said lamp. I have the output of the 555 (12 V) pulsing (on/off) once per second. The drain is connected to a 12 V battery, the source to the lamp. The problem is that the source only outputs about 8.5 V, which means the lamp is dimmer than it should be.

Now, I know that I can connect the lamp directly to the drain, but I can't connect it in series because the lamp already has another ground that can't be changed.

Currently I have it connected like in the first circuit. If I connect it like in the second circuit, the lamp will always be on because it will be connected directly to the 12 V positive.

I have the following transistors available to use: IRF520, IRF540, and BD681. I don't want to use relays, only the transistors.

enter image description here

WORKING but low voltage (above)

enter image description here

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    \$\begingroup\$ You need to drive the gate with a voltage above 12 V to turn on an N-channel FET on the high side. It's not the gate-to-ground voltage that matters, it's the gate-to-source voltage that matters. \$\endgroup\$
    – Hearth
    Commented Oct 1, 2022 at 17:29
  • \$\begingroup\$ What’s your measured gate voltage in the low side example? \$\endgroup\$
    – winny
    Commented Oct 1, 2022 at 17:32
  • \$\begingroup\$ I haven't actually tried the low side example yet, but the gate voltage in example one is 12v. \$\endgroup\$
    – Grant Lee
    Commented Oct 1, 2022 at 18:44
  • \$\begingroup\$ the second image, if correct, that circuit would blow out the MOSFET (or a fuse) when switched on, as its getting full power supply voltage across its channel! \$\endgroup\$
    – Rich S
    Commented Oct 2, 2022 at 0:34
  • \$\begingroup\$ You are probably right, I guess it would need some kind of load. Anyway, gonna try a PFET instead. Regards \$\endgroup\$
    – Grant Lee
    Commented Oct 3, 2022 at 2:32

2 Answers 2

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You need a "high side driver", which is most easily done with a different kind of transistor from what you have (a P-channel MOSFET)

Here is a way of doing it with the transistor you have, which has a couple issues. First is that it is not very simple. Second is that it does not start up with full brightness due to the need to run the charge pump. That could represent a hazard as a turn signal since it delays the full brightness by one entire cycle. The second issue could be solved by running the charge pump continuously with another 555, but that would draw some current continuously which might not be acceptable and would be the opposite of simpler.

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

With a P-channel MOSFET, ideally the circuit is pretty simple:

schematic

simulate this circuit

enter image description here

Note: Automotive '12V' electrical systems can contain nasty transients which require a bunch of additional components and design considerations to be reliable. The above circuits do not take those into full account.

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    \$\begingroup\$ As far as I understood, the 555 outputs a 12V waveform, so it can drive the PFET directly. \$\endgroup\$
    – tobalt
    Commented Oct 1, 2022 at 19:19
  • \$\begingroup\$ Ok, so definitely easier the PFET way. I'll be buying a couple to test out on Monday then. Great explanation by the way. Learning a lot on here. Thanks \$\endgroup\$
    – Grant Lee
    Commented Oct 1, 2022 at 19:30
  • \$\begingroup\$ Yes, you can drive it directly if you don't need HIGH= ON and the 555 is powered from the same 12V supply. \$\endgroup\$ Commented Oct 1, 2022 at 21:43
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The default way is a P-MOSFET in a high side drive layout. I.e. source at +12V, drain to the lamp.

But as you mention battery:

Turn the battery around to create -12 V and ground. Then use your 555 circuit between -12V ( negative supply) and ground (positive supply).

For the lamp you can then use a standard low side drive layout: Lamp between ground and drain of the mosfet. Source at -12 V.

The lamp will be driven with -12 V instead of +12 V, which is irrelevant for incandescent bulbs, but might not work for e.g. LED types.

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  • \$\begingroup\$ They said in the question that they can't do that. One end of the lamp is grounded and they have no access to that. \$\endgroup\$
    – Hearth
    Commented Oct 1, 2022 at 18:00
  • \$\begingroup\$ @Hearth thanks..Adjusted hoping that this wil be possible \$\endgroup\$
    – tobalt
    Commented Oct 1, 2022 at 18:13
  • \$\begingroup\$ Hi, it has to be 12v negative ground. It's on a motorcycle. So I need a 12v positive feed to the lamp which already is grounded. \$\endgroup\$
    – Grant Lee
    Commented Oct 1, 2022 at 18:46

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