Hi all, I am planning to change up the lighting on some aspects of my 4runner, and am inquiring about the safety of it due to amount of amps I will get through factory wiring. A lot of this is to just better understand circuits!
I am planning on changing out 3 sets of lights for LEDs
1: I am relocating my turn signals (currently hidden front and bottom of bumper AKA hard to see) to my running/parking light housings. I am using this kit: https://www.yotashop.com/toyota-4runner-pickup-tacoma-off-road-bumper-led-turn-signal-kit-kit-1026/. It lists no wattage or amp requirements.
I know that an equivalent brightness LED draws less amps, so I should be good to not burn up anything, but I can't really back that up from the supplied info! Any general consensus on this?
2: my reverse lights are useless, and I was planning to replace the current halogen with this: https://www.amazon.com/JDM-ASTAR-Extremely-Bright-Reverse/dp/B00MC6RQAM These are listed as "Max Capacity: 50W; Operate at 10W". Assuming they operate at 10 watts on a 12 volt system, they'll pull .83Amps. I believe my fuse is 5amp, so I'm safe with my wiring.
My question is "what dictates how many amps they pull?". Is it something in the light itself? I know a fuse would stop too many amps, I'm just having trouble or understanding what controls how many amps something pulls, if that makes sense.
3: I want to replace my fog lights with this: https://www.amazon.com/AutoFeel-6000LM-Single-Driving-Lights/dp/B08BY96QMD. Well that but in amber, can't find the actual link.
that one states 35watt for each light. Current halogen is 55 watt, so I'm good with wiring, because less amps at the same voltage. But the same question stands: what "caps" the light at 35 watts, or 2.9amps at 12V (35watt) vs 4.58Amps at 12V (55watt) exceeding what the LED is designed to handle?
what led me down this learning path was someone on a forum commenting that "you'll burn your wires". Well that makes zero sense to me since, as I understand it, excessive amps cause wires to burn up, and a fuse is what prevents that.
Well, you know- never take an Internet forum at face value, yet here I am lol!
To your point 1; the kit says that 50 watts will be used to light those LEDs. That's about 4 amps (50W/13.2V=3.78A and we just round up to 4A to save sanity). I don't know the vehicle wire gauge but I'm going to assume (beware!) that they used 20ga stranded wire. 4A should be fine there.
To your point 2; there's a lot going on with your question. Are the LED modules in question going into the stock reverse light sockets or are they replacing the sockets + bulbs? I'm seeing 50W max power draw here so the math is the same as in point 1. A circuit wired for a 5 amp draw is sufficient here assuming the information given is correct. The resistance of the load determines the amperage drawn for a given voltage. Ohm's Law dictates this and "Ohm's Triangle" gives a nice visual representation of the relationship going on here.
To your point 3; LEDs use a resister to limit current flow. An LED module is actually rather complex in comparison to a filament bulb. I do not have much experience with LED module construction so I can't help there. Lots written on the subject.
Fuses are used to ensure that the amperage limits of the wiring is not exceeded.
The electrical device itself dictates how much current it draws. When you are dealing with light bulbs it can get confusing since there is a lot of "equivalency" talk going on. An LED is more efficient than an incandescent bulb, so you will often see that the LED is, say, 7 watts and replaces a 50 watt incandescent. What they are really saying is that the LED is as bright (in lumens) when dissipating 7 watts as the incandescent is dissipating 50 watts. So you should never have an issue replacing an incandescent with an LED with about the same lumens and have to worry about overloading the wiring circuit and/or fuse. Now it you are trying to get much brighter LEDs than the incandescent in the same socket, then you need to be aware of how much power the incandescent was dissipating and you shouldn't exceed that. But if you replaced a 50 watt incandescent with an LED that actually dissipated 50 watts, it would be immensely brighter.
Also, on that kit for the turn signals, the LED isn't the biggest load there. They are actually putting a 6 ohm power resistor in parallel with the LED in order to fool the turn signal flasher circuitry into thinking that there is an incandescent bulb there. Most older turn signal flashers rely on the current drawn by the bulb to operate, and the LEDs don't draw enough current to get the flasher to work, so they have artificially loaded the circuit. Those resistors are going to dissipate far more power than the LED when the turn signal is on. P=E^2/R, so when the signal is on that 6 ohm resistor has to dissipate 12^2/6 = 24 watts. Don't touch it, it's gonna be HOT. Fortunately it is an intermittent signal.
In reply to pres589 (djronnebaum) :
Thank you for the reply- I've been avoiding electronics and even basic circuits my whole career, so I'm trying to learn now!
I understand that all for the most part, but there seems to be some contradiction. If "the resistance of the load determines the amps drawn", then why wouldn't an LED self regulate amps instead of needing a resistor?
Also for point 2- I am just replacing the lights. I actually decided against what I had originally linked, and found a 4watt LED.
My biggest problem is that I know not to go over OEM wattage, I just can't find what wattage they run at, or their amp draw to do the math myself. I consult the fuse box diagram and there's multiple draws on each circuit so I can't untangle it all in my head.
Thanks again for the help!
In reply to stukndapast :
Thanks! BTW your profile pic is me right now lol. I have been trying to go about it that way- don't overload the OEM wattage, and I'll be good. My problem is I can't parse out the OEM wattage or amp draw.
Although maybe I should be doing it from Lumens.. or I'm just overthinking it.
iIn the end, I'll just blow a fuse before I burn it all down... right? Lol
The OEM amp draw is limited by whatever the OEM fuse is. You can trust that the OEM engineers designed the circuit to operate safely at as long as the load is less than the fuse, in amps. To determine Watts, you can use a nominal 12V battery voltage and find the maximum wattage in a circuit using Ohms Law P=I*E, power equals current times voltage. If the circuit has a 15 amp fuse, at 12V battery voltage that circuit can handle a load of 12*15=180 watts MAX. You probably shouldn't run it at more than 80% capacity, so call it 144 watts to have a safety margin. The whole purpose of the fuse is to protect the wiring and keep it from overheating due to excess load. So yes, it will just blow a fuse before it all burns down, provided everything works as designed.
russ_mill said:
If "the resistance of the load determines the amps drawn", then why wouldn't an LED self regulate amps instead of needing a resistor?
An LED cannot self-regulate in it's basic form. It is a semiconductor that drops a certain voltage at a certain current. Lets say it drops 3.5V at 50 milliamps (determined from the datasheet for the LED). If you put that LED across a 12V battery, the current wouldn't be 50 milliamps, it would actually be a whole E30 M3 load for an instant until the LED burned up. There needs to be some other element in series with the LED that will limit the current and drop whatever voltage the LED doesn't drop, which in this case is 12V - 3.5V = 8.5V. That element is a resistor. The value of the resistor determines the current through the circuit, not the LED. So to get 50 milliamps and drop 8.5V, use ohms law to determine the resistor value. R=V/I so R = 8.5V/50 mA = 170 ohms. So putting a 170 ohm resistor in series with that LED will make 50 mA go through the LED and the resistor, the LED will drop 3.5V and the resistor will drop 8.5V and all of the applied 12V will be accounted for. A more efficient solution would be to put three LEDs in series, along with a resistor. Now you get three times as much light output (3x the lumens) and don't waste much power in the resistor. Now, since the three LEDs together drop 10.5V (3.5+3.5+3.5), the resistor only has to drop 12V-10.5V = 1.5V. To keep the same 50 mA current you need a resistor that is 1.5V/50 mA = 30 ohms. The resistor is still regulating the current, and dissipating some power, but not nearly as much as in the first circuit even though the TOTAL power dissipated in the first and second circuit are identical (which using ohms law is 12V * 50mA = 0.6W). The only thing that a resistor makes is heat, while the LEDs make both heat and light. If you are going to dissipate (use) power, you might as well get something for it, unless you are intentionally making a heater, that is.
In reply to stukndapast :
Ok, I'm starting to get it! I understand the use of the resistor, in theory.
my wiring diagram for the LED turn signals has the resistor in parallel with the LED
That's a pic reference since the wiring diagram isn't handy.
What you were saying about having it in series makes sense- basically a resistor removes energy from a circuit (in the form of heat), and therefore the LEDs perform at optimal watts.
If they run parallel... how does that work? Wouldn't power/energy/electricity run through where there is least resistance, which in this case is the LED, and completely ovoid the parallel resistor, making it useless?
I didn't read through this whole thing but this might be a useful article to read through if you have the time;
https://www.instructables.com/LEDs-for-Beginners/
I don't know why a simple LED would be wired in with a resister in parallel. Here's more discussion on current limiting resistors and LED's;
https://eepower.com/resistor-guide/resistor-applications/resistor-for-led/#
When they are in parallel, both legs have the 12V across them. So with the turn signal circuit, that LED that they are supplying is designed for 12V operation, in other words, there is a resistor already built into the LED device itself. It isn't a bare semiconductor, it is a circuit itself, designed to operate on 12V. So when the turn signal is on, there is 12V across that big gold resistor, which is 6 ohms, and also 12V across the LED element. The current in the two legs is not the same. The current through the gold resistor is large, I=V/R=12/6 = 2 amps, which is a lot. Power equals I*V so that resistor is dissipating 2 amps * 12V = 24 watts!! That is a lot for a little device. If you have a continuous current through that resistor it is going to very hot. But, the current through that resistor is independent of the current through the LED. You can figure out how much current that is if you know the power rating of the LED element. Lets say they say the LED is 6 watts. In order to dissipate 6 watts with 12V across it it has to draw a current which is calculated as I=P/V where P is the power. So in this case, I=6/12 = 0.5 amps. So there is 2 amps going through that gold resistor, and 0.5 amps through the LED device for a total of 2.5 amps, and the whole mess is dissipating 30 watts. Note that if this is on a circuit that was designed to use a 15 amp fuse, you are no where near the limit of its capacity.
pres589 (djronnebaum) said:
I don't know why a simple LED would be wired in with a resister in parallel. Here's more discussion on current limiting resistors and LED's;
I had mentioned before why they have that resistor in parallel. This is a retrofit for a turn signal on a vehicle that was not designed to use LEDs. The turn signal flasher operates on current, and when you replace a high current incandescent lamp with a low current LED, there is not enough current flow for the flasher to work. So they just stick a resistor across the LED to draw a bunch of current when the turn signal is on to fool the flasher into working. It is very inefficient, and defeats the whole purpose of an LED if the intent was to save power, but it is a simple way to not have to change the old-fashioned flashing circuit.
In reply to stukndapast :
Ok that makes a little more sense. Thank you! I've got a lot of studying to do lol
Agreed, forgot about faking out the flasher unit.
How did you wire your lights up I bought the same kit and it didn't come with any instructions and have been trying to figure it out
