NE555 Turn signal... would that work? :)

[betwixt]

Make sure you choose a P-channel MOSFET. LEDs draw a steady current from power up onward, incandecent (filament) lamps draw a surge until the filament has heated up and increased in resistance so you need to plan for more than calculated current. MOSFETS can usually handle short term overloads so it may not be a problem but it's better to plan defensively than regret it later.

Another point, the NE555 may 'misfire' and produce unexpected flashes because it's supply voltage is likely to dip when the lights come on. The dip is natural voltage drop in the wiring as more current is drawn but it can upset the internal operation of the timer. I suggest adding another 10uF capacitor directly across the GND and V+ pins of the IC, wired as close to the pins as possible.

Brian.

 

[MrPopples]

looking at fleabay, I'm considering the IRF 9540N with to220 package.
I calculated Tj at 8amps ... and I end up with 500 degrees ... so either I'm calculating it wrong or I badly need a heatsink
even dropping the "warning" functionnality and thus staying at 4 amps wouldn't work.

I'm using this formula:
Tj = P x Rja + Ta / Rja

which gives me: Tj = 8 x 62 + 25 / 62 = 496.4 degrees ... which is a bit more than the 170 degrees the fet can handle.

@betwixt: I don't know much about electronics and all, but I hope that the 14Ah battery wouldn't drop too much in voltage when lighting some lamps
I guess the capacitor can't hurt. I'm not mass producing anything, I can live with this capacitor. I will add it anyway.

 

[betwixt]

The IRF9540N has an 'on' resistance of 0.117 Ohms so the power it will dissipate is (I squared / R) = 64/0.117 = 7.5W and it's junction (where the heat is produced) to ambient (outside air) resistance is 62C/W so without heatsink it will theoretically reach 62 * 7.5 = 465 degrees. That's a tad too warm for comfort!!! However the 62C/W will drop dramatically with even a small heatsink, just make sure it is insulated from the metal tab or anythnig else nearby as the tab and drain pins are joined internally. Also bear in mind that the current is only flowing while the lights are on so assuming equal on and off times, you only dissipate half the calculated power.

The voltage drop I referred to isn't at the battery but at the flasher unit, in other words after some length of cable. The battery shold hold fairly constant but the drop due to the resistance of the wiring when 8A flows might be enough to upset the 555. The capacitor should help to fix it.

Brian.

 

[Audioguru]

The datasheet shows that when a Mosfet heats then its on-resistance increases which causes it to heat more which causes its resistance to increase more which causes it to heat more which causes its resistance to increase more which causes it to heat more which causes its resistance to increase more ....
it is called "Thermal Runaway". You need a heatsink larger than you think.

 

[MrPopples]

I contacted multiple fleabay sellers about the Rth on the heatsinks they sell ... and they are completely clueless :-/

So, I'm looking the other way around... what about IRF4905?
with Rds(on) = 0.02 that makes 1.25 Watts in the transistor and thus (with same to220 package) 70 degrees at the junction without heatsink.
perfectly acceptable? I think I have enough margins knowing that It'll be "on" only 0.5 seconds at a time

I don't see any reason not to go the IRF4905 way actually. they are not that much more expensive.


also, any idea how to calculate that voltage drop? just for my own understanding.
I'm lacking in both practical and theoretical knowledge

 

[Audioguru]

Ohm's Law:
Voltage Drop= current times the resistance.
Current= voltage divided by the resistance.
Resistance= voltage divided by the current.

Power Law:
Power= voltage times the current.
Power= voltage squared divided by the resistance.
Power= current squared times the resistance.