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How to drive the Backlight for the S64128MX

I am trying to drive the back light for the S64128MX display. My total system is connected to a Li Ion battery. This battery is regulated down to 3.0 Volts to support the display worst case operation voltages. To drive the backlight I created a simple circuit using a MOSFET connected to the 3.0V supply thru a resistor (in series) with a capacitor to reduce the ripple. The MOSFET is controlled by a microprocessor (PWM frequency TBD). I am concerned that this display will have issues with this circuit. For example a normal LED would have a forward drop of 1.2V and a current of 10mA. I could scale my resistor by using the following equation Res = (V power supply) - (Vf Diode drop) / I current Res = 3.0V - 1.2V / 10mA The display has a large Vf of 3.2V. Not only does this exceed my supply, but I can not put a series resistor in the circuit without reducing the voltage to the backlight below its operating value. Can you help me with this circuit.


Guest's picture
July 17, 2009

I have a few suggestion for you - I hope they help...

Suggestion #1
To keep things simple, you may not need a resistor.
DisplayTech LED backlights typically put 5R1 resistors in series with each LED in the pack.
If you run the backlight directly from 3.0V it will take around 30mA and still be reasonably bright - no external resistor required!
If your micro has high current pads, you can even drop the FET, and drive the backlight directly from the micro PWM port.
BEWARE - with no external limiting resistor don't operate the LED above 3.2V otherwise it will be driven in over-current and may be damaged.

Suggestion #2
Use a 12R5 resistor, and connect directly to the battery.
This is another simple and safe method, but the backlight intensity will dim rapidly as the battery discharges.
You'll still be running at 30mA when the pack reaches it's nominal voltage of 3.6V
R = (4.2V - 3.2V) / (80mA) = 12.5 Ohms

Suggestion #3
Change the backlight color.
Using a yellow/green LED in the backlight allows you to the 3.0V for the full range of the backlight.

Suggestion #4
This allows the backlight intensity to be regulated as the battery discharges.
But it adds a bit of complexity for accurately measuring the battery voltage, and compensating for it with PWM
Additional series resistor 4.2 Ohms
Design for 75% intensity of the LED (corresponds to 60mA which works quite well for me).

Here is a quick table for the battery condition and PWM fraction to maintain 75% intensity (this is a starting point - you'll need to make some actual measurements to refine this)
4.2V PWM 25% Fully charged
3.6V PWM 63% nominal voltage
3.4V PWM100% Minimum voltage for 75% intensity
3.1V PWM100% Current will drop to ~30mA (about 40% intensity)
3.0V PWM100% Current will drop to ~20mA (about 25% intensity)

When the battery is fully charged if you operate the PWM above 33% you run the risk of blowing the LED.

Li-Ion batteries (depending on chemistry) have a peak voltage during charge of 4.2V, a nominal voltage o 3.6V and a discharge limit of 3.0V.
The backlight is isolated from the LCD electronics, so there is no mandated to drive the LED from the same voltage.
The backlight is white, and this needs a higher voltage (~3.2V)to operate than a red LED (1.5V).
The LED current consumption is stated at an average of 80mA.
Assuming that the backlight is constructed from 4 LED's, then the peak current will be around 4x60mA = 240mA
If the input voltage is 4.2V and the LED can take 3.2V at 240mA peak, this sets the resistor at ~4R2.
For a voltage of 3.6V The LED would take (3.6 - 3.2) / 4.2 = 95mA peak.
If we want 75% intensity, then we need an average current of 80mA x 75% = 60mA
60mA/95mA = 63% PWM
To calculate the table I used:
PWM% = 100% * 80mA * intensity * 4.2R / (Vbatt - 3.2V)
A minimum of 100KHz PWM will maintain a 10uS max pulse width for the LED at maximum current