Wiring information needed for CFAG128128B

Status
Not open for further replies.

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)

Thanks in advance.
Looking for additional LCD resources? Check out our LCD blog for the latest developments in LCD technology.
 

CF Support2

Administrator
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
 

CF Support2

Administrator
Yeah, I got carried away with my zeros :eek: - I was testing you... yeah... testing you... that's it.

And I think you made a mistake with the calculation. It should be 1.5 / 0.08 ~ 18.75Ω
All you need is a pot, or some other way to control the bias voltage. The module has the circuitry on-board to get you the voltage you need. Your job is to get it there!

So you dont really require an external circuit even though the datasheet mentions it ?
 
Status
Not open for further replies.
Top