LCD – Issues with LCD Backlight Boost Converter Circuit

backlightboostdisplaylcdvoltage-regulator

I have been working with a dev board with the LCD backlight circuit shown below. The boost converted used is the TPS61041. On page 18 of the datasheet, there is an example of the backlight circuit configuration.

I am trying to drive a NHD-4.3-480272EF-ASXN# 4.3" TFT. The TFT backlight requires 40-50 mA at ~25 volts. To make the circuit compatible, I have replaced R2 with a 27 ohm resistor for ~45 mA regulation. I have also removed DZ1 to prevent it from clipping the output at ~16 V. The TFT backlight still does not power on. I have checked all of the TFT connections and they are correct. I also have the EN pin set high. Where am I doing something wrong?

Best Answer

Look at the front page of the data sheet: -

The internal switch that causes the inductor to "charge up" is limited to 250 mA for the TPS61041. This means that going "balls-out", the maximum power that can be taken from the 3.3 volt supply is 0.825 watts. Inefficiencies might mean only 0.75 watts is delivered to the output load.

You want to drive 40 mA at 25 volts and that's a power requirement of 1 watt. Do you see the problem?

What is wrong with this LCD backlight boost converter circuit

Probably nothing - misapplication is the likely cause of the problem.

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LCD Debugging – Initialization of NT7605 Driver

Oh, duh, I didn't read your code carefully -- your send SendLcdByte is wrong:

    void SendLcdByte( unsigned char command_0_or_data_1, unsigned char command)
{
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Replacing a sound module’s LCD/EL backlit display with a LCD/LED backlit display

Your LCD's LED typically accepts 4.2~4.6V. It should be self regulating, so no further resistor needed. You have 11V or 5V available. Below are the three options available. First is use a Adjustable regulator from 11V. The Second is use a fixed 5V regulator and then a regular silicon Diode (1n4001 or similar) to bring it down another ~0.7V. In either mode, the regulator wastes 1 Watt of heat, a heatsink might be recommended (attach the regulator to the module's metal casing is one method).

Or as mention, tap off the already existing 5V regulator, and just throw on the diode. 0.160A is not exactly a large load, it should be fine.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Best Answer

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