I had an accident last Monday on my way to meet balearicdynamics.
No one got hurt, but my side-view mirror is damaged.
There's a blinker in the mirror housing that's now exposed. A good opportunity to see how it's designed.
Here's a view of an intact mirror. The part in the photo above is the horizontal construct you see in the middle.
image source: snippet from Mercedes commercial portal, fair use.
The majority is plastic light piping. There's a PCB inside with two power LEDs.
The LEDs are at the far outer side of the construction.
Their light is diffused across the whole mirror width by that pipe. That pipe is molded in a way that the light appears as narrow vertical lines.
The two power LEDs are at opposite sides of their PCB. The pipe is wrapped around them and combines, then directs, the light of both LEDs.
The device is built up with two PCBs. One, mounted flush with the mirror, has the driving circuit.
The other one, mounted at a 90° angle facing in the direction of the traffic, has an LED on each side.
The PCBs fit into each other like 2 puzzle pieces and are soldered together. Both as mechanical and electronic joints.
Both boards are double-sided. The driving circuit has all components on one side.
It's not hard at all to derive the schematics. All traces are beefy and easy to follow (see photo below).
I could identify most of the components. The ones that I don't have a product code for, I know what type they are.
The result is a very simple circuit. As you can see, this is an imperfect but good enough current driver.
There's no blinking logic on-board. That's part of circuitry inside the car.
- two capacitors at input (the first white one is probably input protection instead of a cap edit: I think it's a Würth or Murata TVS diode) are 95 nF together.
- The diode in the 12 V path has markings PJ811K. Search brings me to the LittleFuse website, but I can't find a match.
- The transistor is a bipolar , power NPN
- The diode in the bias circuit is a SOT-23, with marking Z1W. I think it's a 4.6 V Zener (edit: it's a 4.7 V Zener: )
- The bias resistor is 1K2.
- The power LEDs have no identification
- The emitter resistor is made out of 4 82 Ohm ones in parallel.
source: photo of the same design, from my own copy of The Art of Electronics
The LEDs take about 190 mA,
(the measurements in the drawings are what I really measured. The text here uses the theoretical values. It's only 0.1 V difference).
The base of the transistor is kept at 4.7 V always, by the Zener diode.
The 1K2 resitor is there to bias the Zener. To give it the right current so that it is in its operation range.
That Zener bias current is: (11.4 V - 4.7 V) / 1200 Ohm = 6.7 V / 1200 Ohm = 5.6 mA (ignoring the base-emitter current that eats away a tiny bit of that).
5 mA is a typical Ir for these Zeners. The circuit biases the diode perfectly.
The emitter of a transistor, in this "common-emitter" design, is always one diode drop (the base-emitter diode) lower than the base.
4.7 V - 0.7 V = 4.0 V
So we know that the voltage over the emitter resistor (in reality 4 resistors in parallel), is 4.0 V.
The resistor is 20.5 Ohm.
Ohms law says that I = U/R = 4.0 / 20.5 = 195 mA
Because the current through the collector is the same as the one through the emitter, we know that 195 mA will flow through the LEDs (ignore the base-emitter current. It's neglectable in this calculation).
Equation solved. This current driver circuit drives the LEDs at 195 mA.
Why four 82 Ohm resistors instead of a single 20.5 Ohm one?
The power dissipated by the resistor is 4.0 V * 195 mA = 780 mW.
It's cheaper to mount 4 resistors that can dissipate 250 mW than a single 1 W resistor.
And the 82 Ohms is just a plain E12 series resistor.
How much power is dissipated in the transistor?
(For this calculation I use the measured values)
The collector is at 5.55 V. The emitter is at 3.9 V
That gives a voltage drop of (I'm rounding here) 5.6 V - 3.9 V = 1.7 V.
The current is 190 mA.
The transistor dissipates 1.7 V * 190 mA = 323 mW.