Wiring information needed for CFAG128128B

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kasunt

New member
Hi guys,

I followed the wiring diagram on the datasheet closely. I'm wanting to connect the LCD to a MCU like AVR or PIC.
What is the purpose of having an external contrast adjustment if there is already circuitry internally to adjust the contrast (Ref. page 8 of datasheet) ?

And what is the required value for resistor B/L drive circuit ? (Ref. page 14 of datasheet)

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CF Support2

The CFAG128128B family has an onboard negative voltage generator - but you still need to connect up Vo so that it can be driven and the contrast adjusted. The simplest way is to use a 10KΩ to 20KΩ pot across Vcc (+5v) and Vee (NVG out), with the wiper to Vo (contrast).

Since the backlight LED characteristics are the same for the white and yellow-green LEDs, the same current limiting resistor is calculated. You use the voltage dropped by resistor in the circuit:

Typical current = 80mA
Forward voltage of LEDs = 3.5v
Supply voltage = 5v
V / I = R
(5 -3.5) / .080 = R
1.5 / 0.80 = R
1.875 = R

For power rating:
I*V = P
0.080 * 1.5 = P
0.120 = P

The minimum you would want to use would be a 1/4W 1.875Ω resistor.

kasunt

New member
Thanks a bunch for this.

So you dont really require an external circuit even though the datasheet mentions it ?

kasunt

New member
The CFAG128128B family has an onboard negative voltage generator - but you still need to connect up Vo so that it can be driven and the contrast adjusted. The simplest way is to use a 10KΩ to 20KΩ pot across Vcc (+5v) and Vee (NVG out), with the wiper to Vo (contrast).

Since the backlight LED characteristics are the same for the white and yellow-green LEDs, the same current limiting resistor is calculated. You use the voltage dropped by resistor in the circuit:

Typical current = 80mA
Forward voltage of LEDs = 3.5v
Supply voltage = 5v
V / I = R
(5 -3.5) / .080 = R
1.5 / 0.80 = R
1.875 = R

For power rating:
I*V = P
0.080 * 1.5 = P
0.120 = P

The minimum you would want to use would be a 1/4W 1.875Ω resistor.
And I think you made a mistake with the calculation. It should be 1.5 / 0.08 ~ 18.75Ω