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Help needed with High Power LED driver for Car

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I have no idea on how to design a LED driver, so I decided to look for one online and found this. **broken link removed**




I will be using another circuit to switch the LED's on and off with the music.

So will this circuit be able to switch on and off extremely fast? And will it be able to handle the voltage spikes in the car?

Also, In my filter/amp circuit, on the output FET, I have LED anode's connected to V+ and am switching the cathode on the drain.

How should I use the filter/amp circuit to switch the high-powered LED's in this circuit?

Thanks
 
If you add another transistor like Q1 connected from Q2's gate to ground you can use it to turn Q2 on and off. Put a resistor in series with the base of the added resistor (no greater than 10 times the value of R1). Connect this resistor to the filter/amp circuit. A high voltage at this resistor will turn off the LEDs. Less than 0.7V will turn the LEDs on.

The spikes it can tolerate are determined by the voltage ratings of Q1 and Q2. To be on the safe side add a small resistor in series with the + power and add a large capacitor (100µF or larger) from the circuit side of the resistor to ground. Select the resistor size to generate a 1-2V drop at the LED current level (R = V/I).
 
Thanks for the reply.

I have little idea what is going on in this circuit, so if anyone could give me a little walkthrough I would appreciate it.

As to the voltage spikes, I'm not worried about the transistor blowing up, but the $40 LED's.

I need to ensure that the LED's can only drop 4.3V at 1000mA
 
You don't concern yourself with the voltage drop across the LEDs. They will drop whatever they need. You are concerned about the current and that is what this circuit regulates.

The two transistors are configured as a constant-current source (current independent of voltage). Initially Q2 turns on due to the positive bias voltage from R1. When the voltage across R3 due to the current through the transistor and R3 reaches about 0.7V, Q2 turns on and reduces the voltage at the gate of Q1 to maintain 0.7V across R3. Thus the current through the LEDs stays a value of about 0.7V/R3.

You select R3 to give the desired current. In this case you want 0.7V/1A = 0.7 ohms. If you can't find that value you can parallel two or more resistors to get the desired value.

Be aware that Q2 will be dissipating over 8W when operating and will need to be on a heat sink.

You need to be concerned about the transistors blowing because if they blow, that can cause the LED current to increase and blow them also.

But as an alternative circuit I recommend using an LM317 regulator in a current limit configuration (see this). It requires only one resistor and the regulator automatically protects itself against over-current and over-temperature. It also will require a heat sink. A resistor capacitor filter at the power input as I previously described is also recommended.
 
Putting 2 in series will reduce the wastage in the regulator.
Here is a suitable circuit:
**broken link removed**

This constant current circuit is designed to drive two 3-watt Luxeon LEDs. The LEDs require 1,000mA (1Amp) and have a characteristic voltage-drop across them of about 3.8v. Approximately 4v is dropped across the LM317T regulator and 1.25v across the current-limiting resistors, so the input voltage (supply) has to be 12.85v. A 12v battery generally delivers 12.6v.
The LM 317T 3-terminal regulator will need to be heatsinked.
This circuit is designed for the LM series of regulator as they have a voltage differential of 1.25v between "adj" and "out" terminals.
 
Thanks. I have already made the original circuit and it is working quite well, well except for the fact that I can't find a 1/2W .5ohm resistor laying around. :)
 
If you add another transistor like Q1 connected from Q2's gate to ground you can use it to turn Q2 on and off. Put a resistor in series with the base of the added resistor (no greater than 10 times the value of R1). Connect this resistor to the filter/amp circuit. A high voltage at this resistor will turn off the LEDs. Less than 0.7V will turn the LEDs on.

That would work, but it would be a pain since the circuit is already designed to turn a load on with a high voltage. Instead, I think I am going to add a transistor in series with the 100k R1 and the collector of Q1 and gate of Q2. This should work, right?
 
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