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LED Backlight Driver Circuit?

njwtech

New member
Does anyone have a good circuit for driving an LED backlight? I have a CFAG240128L-YYH-TZ that has an LED backlight rated at 900ma. This needs to be backlit for an automotive application, but running a resistor to supply 13W of power for the backlight will be big, hot and nasty, not to mention pretty inelegant. Anyone have a chopper drive circuit or something that can drive the LED without so much heat? I was thinking maybe if I can't find anything else I might try a stepper motor chopper drive and put a choke in line. Any help would be appreciated.

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Norm
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To run the backlight in an automotive environment (I assume 13~14 V), you will need a 10 ohm, 16 W resistor, to run at 900 mA. Possibly you could use a lesser maximum current for adequate viewability. The chopper circuit (PWM) will only dim the B/L, you still need the resistor to limit the current during the on-cycle of the PWM. Warning: automotive power is famous for having nasty transients and voltage variations, so you need to use protective circuitry (e.g. zener diode or tranzorb, and reverse polarity protection).

One alternative is to get a cheap DC-DC converter to bump the 14 V down to 5 V. Then you could use a 1 ohm, 2W resistor (plus the PWM for dimming). You'd also want to use the protection circuit on the DC converter input.

The PWM driver circuit can be as simple as a n-channel FET in series to ground with the B/L, driven from one of the I/O pins of your cpu.
 

njwtech

New member
Thanks for the help. What I had in mind was actually a circuit like that in a stepper motor driver. It doesn't use PWM as such, but rather turns on until a set current is reached and then drops off and then back on until the current is reached and then off. It does this usually at about 10Khz or 20Khz. It has a current sensing circuit that keeps it pulsing to maintain a particular amperage. In stepper motors that is how they get away from having to have very large wattage resistors when running low current motors at high voltage. They use a combination of an STS L97 and L98 chips to do it. For instance, the steppers in my cnc equipment are running through one of these chopper drives with no resistors in the circuit, and I am running 48V on a motor rated for 2.8A @ 6V. If you ran it without the driver circuit at that voltage it would eat the motor in about 3 seconds.

It was just a thought anyway. I thought maybe there was a driver like that around that could do the same thing with an LED. You would probably just have to insert an inductor in series with the LED to maintain the current and fool it into thinking it had motor windings.

Thanks again. The DC converter is probably the way I will have to go.

..
Norm
 
You would probably just have to insert an inductor in series with the LED to maintain the current and fool it into thinking it had motor windings.
Yes, that is exactly what you'd have to do, the current rise time thru the inductor (LR time) would control pulse width. It is a good idea, you should try it and let us know how it works. To be sure of not over-currenting the LEDs, you'd probably want to have some way to measure the peak current to see that it is within the spec. But buy doing this method, you have no way to control dimming, unless you can change the pulse repetition rate, and then it would probably be limited in range, as stepper motors don't operate the way PWM dimmers do. So you'd have to do dimming by varying the current set resistor or voltage.

After thought: you'd have to be sure the magnetic field collapse from the inductor doesn't reverse-polarity zap your LEDs, by using a diode across the LEDs or inductor.
 
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njwtech

New member
Thanks again CV. When I have time to play a little more I will try it. For right now though I have to get this thing installed and lit. I wonder if you could help me just a bit more. The specs for the LED on that display say that the supply current is between 720 and 1350 ma, and the voltage between 4.0 and 4.4 with typical at 4.2. The math that you are using for your values is unclear to me. I would have thought it fairly straightforward. Ignoring the voltage and using the current to rule, I would have thought I would use ohms law simply to say R = E/I or R = 14V/.9A so R = 15.555 ohms giving a wattage of 12.6W. So too, if I use 5V would I not use R = 5V/.9A or 5.6 ohms at 4.5W? Your use of a 1 ohm resistor confuses me. Or am I missing something like subtracting the voltage drop of the LED or something. Sorry if it is a stupid question. I am not used to using LED's at this much current. Thanks again.

..
Norm
 
Or am I missing something like subtracting the voltage drop of the LED or something.
Yes, or something. The voltage drop across the resistor would be (14 - 4.2) / 0.9 = 10.9 ohms; (5 - 4.2) / 0.9 = .888 ohms.

You might consider using one of the white LED B/L types, they only use 128 mA.
 

njwtech

New member
Thanks for the information. I had a look. The LED driver costs more than half as much as the display. I think I am going to buy a white LED model. Then I can run it off the existing logic regulator with a surface mount resistor. Thanks everyone for your help.

..
Norm
 
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