The car halogen headlight is rated at 12V 100W.
Will it be okay if I connect it to 60V dc supply, and using pwm make sure the average power doesn't exceed 100W?
Context:
I have just built a pwm hobby circuit using Arduino+irf540 MOSFET, which I want to test using high power.
I have 60V lion battery from my electric scooter and a 12V 100W halogen bulb.
I want to discharge the 60v battery over 12v bulb using pwm.
The lamp filament is likely to just blow up.
Let's assume for simplicity that the lamp is completely resistive 100W load at 12V. It will thus draw 8.33A, and thus lamp resistance can be assumed to be 1.44 ohms.
If connected to 60V at any given time, for a second, or microsecond, as the time does not matter, it will draw 41.67A and will dissipate 2.5 kilowatts.
As the filament is very thin and heats up quite rapidly, the PWM pulses would need to be extremely short to not blow up the filament, as they drive 2500 watts momentarily to a 100 watt lamp, so on average you only need 4% of maximum power to get 100W. The PWM pulse width would need to be quite fast to not momentarily drive too much power into the lamp, and that on time with repetition rate would determine the average.
What is even worse that cold resistance of the lamp is a lot smaller, and cold current can be approximated to be 10x the average nominal current, so it would need extremely soft ramping of power on to the lamp.
So you can try but don't expect it to work, as you might just blow up your switching component as well.
Well use 5 of those bulbs in series, that would work.
One bulb even allowing for charging variation on the car will be 15V, so 60V is too much.
Just to be clear, if you connect that 12V bulb directly to your 60V battery, then the filament will burn out.
If you have some resistances handy you could make a voltage dropper.
Yes it will work, but not the way you set out to do it.
The PWM-ing needs to be done by a switching power converter in a buck configuration. It will act as a PWM, and its inductor will smooth the current going into the bulb, so the bulb won't get destroyed.
One cannot do this PWM-ing without the inductor, and to keep the voltage stable, you need a control loop. I doubt you feel like designing switching converter control loops on an Arduino, so it's best to use a chip that will do the job. Most likely the chip will drive an external MOSFET, i.e. it will be a switching controller.
This calls for a 100W buck converter from 60V to 12V, and adjust the buck's feedback ratio to regulate the lamp brightness.
Don't even think of putting such a buck converter together on a breadboard. Either buy a premade module, or design a PCB for it, and add detailed layout images in a new question, where you'll be able to get an idea of how close to potentially working it may be.
Do some very simple math to get an indication of whether it would work.
At 12V the bulb will draw 100/12 --> 8.3 Amps. The pre-supposes that the bulb is warmed up. It will draw perhaps 5-10 times as much current if it is cold. As pointed out earlier the filament has a PTC characteristic.
You can obviously use a PWM to reduce the AVERAGE power into the bulb. eg In terms of voltage across the bulb a 50% duty cycle would look like 30V and a 25% duty cycle would look like 15V ....on average (the same as RMS in this case). However the filament is just a resistor and we can see from Ohms law that the bulb has approximately 12/8.3 --> 1.4 Ohms when heated up.
Put a 1.4 Ohm resistor across a 60 V supply and it will draw 42 Amps. That's an instantaneous power dissipation of 2520 Watts. A PWM driver results in a much lower average power dissipation, but it all comes down to how much current can pass through the filament before it simply acts like a fuse.
I'd suggest that if your PWM was high frequency (> 10 Khz) and you you can slow start the bulb (ie start at a very low pulse width and slowly increase) you might just get a way with it. However if you are using an Arduino PWM (490 or 980 Hz) then your chance of success is much lower. I'd also add that you have absolutely no protection using an open loop system like this. You are not considering the initial surge currents when the bulb is cold for example.
You could put an NTC (surge suppressor) in series with the bulb, that might help you, but it sounds like you are using only what is on hand. You also need to consider your PWM controller, you've not presented any details for that. For example, the IRF540 is only suitable up to a peak of about 80A (for <10us), so you'd need to have at least 2-3 in parallel to handle the peak currents expected.
In short, if you aren't able to design what you want to build, you shouldn't be doing it.
V * sqrt(0.5) the voltage, 70.7% of the voltage. As Sphero's answer says, 4% duty cycle gives you 12V RMS.
– Peter Cordes
Apr 25 '22 at 06:28
In theory a 4% duty-cycle PWM would give you 12V RMS but the peak currents would be very high and it would likely emit EMI etc to a large degree. Without some kind of mitigation the peak cold filament current would be enormous and likely to damage the lamp or the switch.
If you add an inductor and maybe some smoothing capacitors you will have a DC-DC converter, which is perfectly fine and is feasible, but might be more of a challenge at 60V/100W than you want to take on. There are controller ICs that will handle some of the task.
If you just want to test your PWM into a resistive load, a resistor would be the best kind of load. Even a bit of stray inductance (long wires or a wire wound resistor) can cause issues if you don't have a diode across the load since the IR540 is only good for 100V.
If the bulb has two filaments forget it, the filaments will arc paralleling the two filaments, causing them to be parallel from there common point until the arc. This overpowers the remaining filament causing it to blow in a very short time. We learned this when PWMing 12V lamps on 42V for future electrical vehicle systems. We designed a very fast circuit that detected the fault and shut the power off for the remainder of the PWM cycle. see: https://patents.justia.com/patent/20030218426 A much simpler solution that is safe for your expensive bulb is to use normal household tungsten lamps. In the US they are 120V so you will not need as many. As you are PWMing them the cold resistance will come into play and help you.
The power used by a resistance of R Ohms connected to voltage V is
P = V^2 / R
So if we raise the voltage from 12 to 60 V, we have 5 times the voltage and thus 5^2 = 25 times the power. So for a hot filament, the power would raise from 100 to 25 * 100 = 2500 W.
If your PWM is on for only 10 microseconds, you need an of time interval of 240 µs.
It might work using a 24 V battery instead of a 12 V. So the PWM may be on for 1 ms and off for 3 ms.
But the resistance of the cold filament is about 10 times smaller than the hot ones resistance. So the PWM should start with on time 100 µs and a period of 4 ms and then ramp up slowly over about a second to 1000 µs on time with the same period of 4 ms.
But it would not be easy with 24 V. The maximum power peak for the first pulse to the cold filament would be about 100 * 4 * 10 = 4000 W.
A 12 V 100 W auto bulb will draw about 8 amps when at operating temperature.
Its cold resistance will be about 1/10th of the hot resistance, so expect about 80 amps when cold.
At 60 V, expect about 400 amps when cold.
The FET you have chosen has an absolute max peak pulsed current of 120 amps (absolute max, not some reasonable operating spec).
Adding inductance and a flyback diode to the load could keep your current lower, but then you'd be building your own homebrew buck converter.
I want to discharge the 60v battery over 12v bulb using pwm.
If you simply want to discharge the battery using PWM, then use a load that doesn't overload your FET.
You could use the same characteristic of filament lamps to get a reasonable current draw from 120 V or 230 V lamps at 60 V, if you have any to hand. An electric kettle is an excellent heat-dumper, and has typically 10 to 30 ohms resistance depending on whether you're in 120 or 230 V land.
You can, but:
This is important both because something could go wrong and because of the inrush current of the bulb.
At least use a fuse on the battery side.
Keep in mind that a lot of things can go profoundly wrong with the power electronics.
This is the difference between regulating your PWM at ~4% (when you feed the raw 60V pulses over the load) and 20% (when you feed a smoothed-out voltage to the load).
In some use cases, the inductance is skipped from the circuit because the intended load (e.g. motor) has its own inductance.
PWM at 20% is much easier to control than PWM at 4%. How much is the granularity of your control? 1%? 2%? You see the problem.
Is your bulb internal wiring capable of 60V without arcing somewhere? Most probably it is, but you don't know and you will find out the hard way. See p. 0. See also p. 3.
Do you have any means of slow start?
The inrush current/power of the incandescent light bulbs is ~10 times the rated one. For headlight bulbs, the current and power settle for about half a second.
If you set the PWM right at 4% (or 20%, see above), your current pulses will go well above the switching transistor rating and/or the battery rating - for long enough to burn them out.
100W headlights are not road-legal, meaning that their quality is not governed by any standard. This, in turn, means that their quality is generally low and you can expect any kinds of problems - including, but not limited to, exploding the bulb into a cloud of flying red-hot glass particles.
Keep your eyes and your skin off such bulb, especially when trying non-standard things.
How can your experimental setup be improved with a minor investment?
Get two bulbs 55W/24V or 70W/24V (24V bulbs are for heavy trucks/busses, 70W are also road-legal for them) and connect them in series.
Two 24-volt bulbs in series are perfectly capable of running at 60V even without PWM. A 24V truck/bus battery routinely goes up to 30V with the engine running and the bulbs are designed with this expectation.
It would probably be easier to buy 5 20W 12V bulbs and holders, and connect them in series. G4 halogens would work well and the holders are cheap. Then you can test your PWM knowing that 100% on won't destroy your test bulb.
For a very first test, you could even use a 120V incandescent bulb. It will light dimly at 60V, and will dim further with PWM. You could connect a few in parallel to increase the current you draw.
I like the idea of using a different way to discharge the battery. Read through all of my answer to accurately see the stupidity in this redundant question. The problem is wanting to discharge a 60v battery. Everyone is forgetting to inquire what its amp-hour load rating is. If this guy discharges that 60v Lipo(or lith-ion) battery too quickly this will cause thermal breakdown within the cells of either lith-ion or lipo types essentially turning a fairly large battery into a chemical bomb. Now if you guys want to continue to tell him how he can turn this into a bomb then be my guest but in my opinion it is best for him to discharge said battery in the way it is designed to and JUST GO RIDE THE DAMN SCOOTER and forget about this other nonsense before you end up blowing you and your house up!!!
