Today I have made a tear down and reverse engineering of a cheap 12V LED light that is designed to substitute a 12V halogen lamp. Let's see what it is inside.
|12V LED Light|
The LED light is composed of 21 small white LEDs. The package says it's 1W. Applying 12V we I get 87mA so it is trully 1W. But what efficiency we get on the circuit?
The insidesAfter breaking the enclosure we get the lamp internals.
The lamp is internally composed on two boards. One round board holds the 21 LEDs and a smaller octagonal board holds the control circuit.
The LED board is painted white at the LED side to give a better reflectance of stray light.
Before continuing the tear down, I liked to check the curren on one of the LEDs. To do that, I disordered a LED and re soldered only one terminal to put a multimeter in series.
|Current meas setup|
Measuring the LED current when operating it gives 10.5mA. Those are quite simple LEDs. More modern LED lamps use high current devices. That's why it needs 21 LEDs to get 1W of power.
Continuing the tear down, I have separated the two boards and removed all LEDs from the led board. The front side has markings so we can see where should be the anode and cathode of each led.
|LED board: Front|
The back of the led board holds all the connections. From them we can see that the 21 LEDs are arranged in seven parallel groups of 3 LEDs in series. As one white LED can have a forward current of up to 3.5V, we cannot use more than three LEDs in series if we use a linear control circuit. We also don't want to put less than three LEDs because that will give less efficiency.
|LED board: Back|
As a result of the tear down we get also a bunch of leds.
|Bunch of LEDs|
Reverse engineering of the circuit
The circuit used to power the LEDs is shown in the next figure.
There are two input cables for the AC 12V input source and two output nodes (+) and (-) to connect the seven stripes of three LEDs.
To get the circuit schematic I have taken a photo of the back of the board. I have mirrored it to have the same positions as the component side and I have written down all the components present on the board.
|Component distribution on the board|
The equivalent schematic of the board follows:
Nothing fancy. Just a linear current control regulator.
The four diodes take the 12V AC input source and rectify it to get, with the help of the capacitor E1, a constant voltage between Vdd and GND.
The diodes have a marking RS1M. Those seems to be standard 1A silicon diodes. The voltage at 87mA should be around 0.7V. As we have two voltage drops, that gives a Vdd-Vss voltage of about 10.6V.
The LEDs are connected to the collector of the big transistor Q1 (D882). The base current of Q1 is fed by the resistor R2.
The base of Q1 is at two Vbe drops over Vss, so, about 1.2V. That gives a maximum base current of 3.1mA. That means that we can drive the 87mA current of the 21 LEDs with a minimum forward beta (current gain) of 28.
We need, of course a way to limit the current at the LEDs. That's the mission of Q2.
The collector current in Q1 is similar to the emitter current that goes to R1. As the base-emitter junction of Q2 is connected to this resistor, when it reaches about 0.6V, Q2 will turn on and drain the base current of Q1. The equilibrium point of the circuit is where Q2 is just starting to turn on and the current at Q1 will be at about 0.6V/R1 that gives 80mA. This theoretical result is quite similar to the measured current of 87mA.
I have measured one of the LEDs and it gives about 3V at 10mA. For a series of three LEDs we get 9V. The voltage at R1 is 0.6V. That gives a voltage drop of about 10.6V - 9V - 0.6V = 1V between the collector and emitter of Q1. Enough room for voltage drop changes on the diodes as function of the temperature.
If we consider that we get a total voltage of 9V on the LED strips from a source voltage of 12V, that means that the circuit efficiency cannot be better than 75%. It could be better using a DC/DC voltage to current converter instead of using linear circuit but what can you expect from a 3€ LED lamp?