Continuing my last post on how I fixed a faulty AFC trailer converter I thought it might be useful sharing my thoughts about how I built a trailer light tester for one buck or less. So this is rather a DIY project.
The short story in pictures:
- What you would expect if you pay a visit to a car workshop (it would cost $200+) :
- What you could normally buy from eBay for only a fraction of the price:
- What you could do with less than $1:
So, my version of trailer light tester - DIY comprises the following parts:
- a bunch of red, green and yellow light emitting diodes (aka LEDs)
- few resistors with respect to the Ohm's law
- a 12V power supply
- preferably a solderless bread board and some assorted 24-AWG jumper cable wires which would tolerate ~3.5A for our test
Create a pair of 5mm LEDs connected in series for each lamp you want to test. For instance, I assigned the 2 x 4 greenÂ LEDs for the tail lamps, 2 x 4x yellow LEDs for the turn signal lamps and 2 x 4 red LEDs for the stop/brake lamps. The leftmost column (the one with green-red mixed LEDs) is for my reverse/back-up lamp (I would I had a white/blue LED but I hadn't).
Now don't jump to power your LEDs right away because you might just burn them. First make sure they can tolerate the voltage you supply.
First check the LED's spec data-sheet for the forward-voltage and forward-current. Usually the forward current (If) is about 20mA and the forward-voltage might vary from 1.8 to 2.2-2.5V. From my experience it would be safe to assume that a 5mm LED would tolerate ~20mA at a ~2V. We mark this info with (A). Read more here.
To make sure we don't burn our LED's we want (i) to make sure that the first LED would tolerate the power supply voltage and (ii) also that the last one (in series) would light.
Ok, but how to calculate the resistance? The Ohm's law comes into rescue: the current (I) that flows through a conductor is proportional with the voltage (V) across the two points. Now, if the conductor would be of Mercury (Hg) its conductivity would be far smaller than if the conductor would be let's say Copper (Cu) or even better Silver (Ag). So some conductor tends to resist more than others. Thus the Ohm's law can be restated as the current (I) is directly proportional with the voltage (V) and inversely proportional with the conductor's resistance (R), thus I=V/R. We mark this info with (B).
Note that after each LED the voltage drops a bit (due to the LED's resistance) so the circuit voltage after the last LED will be smaller than what it was for the LED in front of it, which was also smaller than the one positioned in front of if, and so on.
According to the (A) the voltage drop (aka forward-voltage drop) for each LED is about 2V. Thus the entry voltage for the 2nd LED would be with 2V less that what was for the 1st LED. The entry voltage for the 3rd LED would be 2V less than what was for the 2nd LED, ie 2V+2V less than what was for the 1st LED, etc.
It would be reasonable to say that the electrons that start flowing from the positive terminal should be equal with the one that arrive at the negative terminal if they complete the whole circuit, right? They cannot vanish like that, right? It's sensible. So with other words the current is going to be the same, no matter how many LED's we would connect in series. The same number of electrons would travel through each LED thus the same current would flow though each LED. So the current through each LED would be, as we've already said earlier (A)Â I=20mA.
The total voltage drop for 4 LED in series would be 4 LEDs * 2V ~ 8V. If the electrical potential betweenÂ the positive and the negative terminal of our car's battery is 12V, if our series of LEDs can resist to about 8V, then we have to mount a resistant barrier in front of the first LED such that when the first electron hits the first LED its electrical potential to be at max 8V (the 4x LEDs cannot stand together to more than 8V, remember?). With other words right after that barrier the voltage must drop from 12V to 8V, ie. with 12V-8V=4V. The so called resistant barrier is nothing else than a resistor that, by opposing to the current flow, it could drop the voltage with 4V when a 20mA current flows through it.
So finally, from Ohm's law (see B) we can state that the required resistor is R=(12V-4V)/20mA=8V/20mA=400Ohm. I could just give you this 400 value but you wouldn't understand why 400Ohm and not 1million Ohms. Now, hopefully, everything is clear. And sensible.
OK, so here are the electrical components we need (which would cost less than a buck, except the power supply which we assume it's the car itself):
- 5 x 400Ohm resistors (the metal-film type would be just fine).
- 2 x 4 red/yellow/green 5mm LEDs.
- optionally 2 x 2 green/red 5mm LEDs for reverse/back-up or any other color, it really doesn't matter as long as you know their meaning.
- a 12V power supply.
- optionally 4 switches which would help us to simulate the car's switches; in reality we could do that without switches by just connecting/disconnecting the wires from the power supply line.
Although we have 7 columns of LEDs we use only 5 resistors. Why is that? Do we really want to send an excess ofÂ 4V through some two LED columns and burn them? No, we do this on purpose and here is why.
Normally the trailer's converter module* supplies the car mounted trailer's socket with voltage on different terminals designed for different lamps:
- on a common terminal (54, pin-6)Â for both the left and right stop lamps
- on two distinct terminals, one for the left turn signal (L, pin-1) and one for the right turn signal (R, pin-4)
- on two distinct terminals, one for the left tail lamp (58L, pin-7) and one for the right tail lamp (58R, pin-5)
- a common pin for the Earth/ground (31, pin-3)
- optionally on a single terminal (pin-2) for the rear fog lamp - which our device doesn't cover yet
- optionally on a single terminal (pin-8 on a 13-pin trailer connector) for the reverse/back-up lamp
(*) a trailer converter module is an electronic device used for wiring the car and the trailer together to power and synchronize their lighting systems.
Since two trailer stop lamps are supplied by a common terminal we are using also 2 columns of red LEDs supplied by a 12V power supply. Due the fact that the forward voltage of each red LED is 1.8V the 8 LEDs would require 8*1.8V=14.4V. So in this case there is no resistor required since we already exceeded the available voltage (12V). The solution would be (1) no resistor and (2) fewer LEDs. With 7*1.8V we would need a 12.6V battery which is exactly the voltage a car battery normally has.
So in case of these two red LED columns we won't need a resistor, thus we only need 5 resistors in total for the other columns.
If we really want to simulate the car's switches and everything then a testing breadboard for our little DIY project would look like this:
Ignore the 9V in the scheme above, it's just a limitation of the CAD software I've used for drawing the schematics.
This is exactly the design of my $1 trailer light tester device as shown on the picture on the top. And honestly, it took my more time to write this article than to make that "device". Hopefully it could help someone else some day.
Now, if you think that this article was interesting don't forget to rate it. It shows me that you care and thus I will continue write about these things.