I get an LCD from a control panel (miele wash machine) and maybe someone can identify it for me.
I tried to google it but its hard to find some usefull info.
All important lines will go from the connector to the T6963C. The picture is not so clear in this area, but it seems possible to trace the pins back from the T6963C. For example the C/D line (left side, lowest pin of the T6963C) goes to the lowest two pins of the connector. Maybe with an Ohm-Meter you can finally figure out the pin assignment of the connector. Specification of the T6963C can be found via google (look for "T6963 pdf", page 3 of the data sheet).
Another note: T6963 often do not have the -15V source which is required for this display. But your display seems to have a circuit for it (which is very good). Can you identify the 5 or 6 pin IC left of the label D1 (upper middle)? This area looks like a negative voltage generator.
Edit: The output (-15V) of the negative voltage generator usually is also available on the connector.
Thanks for the info but im a kind newbie with arduino.
U11 is an LTES which can be an LT1611 - Inverting 1.4MHz Switching Regulator if google is right.
I tested out the pins and i will try to connect to the arduino next time.
This looks already similar to the T6963 displays, that i have seen before. All connections, which go to the Arduino, are already identified.
Pin 1 often is "Frame Ground".
The remaining problem will be to identify the contrast input pin. Often, this is pin 4 (e.g. see here: noodlehed.com).
My suggestion is to do this:
Apply 0 and 5V to the two power pins, which are already identified. The NVI should now start to work and the -15V..-10V output should be visible somewhere (maybe pin 9 or 20). If this is the case, then it is very likely that pin 4 is the contrast input pin. The contrast input pin should go to some resistors and caps on the PCB. Once you have identified the contrast input pin, you can connect a variable resistor:
one end of the var res goes to 0V
other end of the var res goes to the NVI output
wiper must be connected to the contrast input.
Powering the display with 5V, you should be able to turn on and off all pixels by moving the wiper from one end to the other.
Warning: Choosing the wrong contrast input might destroy your display.
Oliver
Edit: Variable resistor value should be between 10K and 22K
I connected the pins but no negativ voltage and no backlight. Maybe the lcd is functioning but its too dark to see it.
Pin 23 and 24 is the backlight pins to the LEDs but they are dark... I tried with 5V from the arduino. Is it enough amps in the arduino +5V output to drive the backlight? 16 SMD LEDs in the backlight. Why is it two pins for the LEDs?
This datasheet is the nearest one but its only 20 pins... i have 24
I shorted out! ]
The pin 23 is the backlight 12V+ and 24 is backlight ground. I dont understand why need an extra ground for the backlight???
Then i connected a 1.5V AA batteri for some negativ voltage and i can see my Hello World on the screen.
So the panel doesn't have negativ voltage generator!
Thx for the nice library...
It seems to be normal, that the LED power has independent power pins, but i wonder, that only one 1.5 AA is enough. For my own tests, i had to use a 9V block battery as negative voltage source to see something.
zsiti4:
The pin 23 is the backlight 12V+ and 24 is backlight ground. I dont understand why need an extra ground for the backlight???
Perfectly straightforward really - so that you can control it either by switching the positive supply or the negative - as latter you would using a NPN transistor or N-FET.
I tried this NE555 circuit to get the negativ voltage for the display but the output is just -6volt and dropping.
What can it be the fail? Is this a good circuit for the job?
That circuit should work as far as I can see, but you could do just as well by arranging one Arduino pin to be continuously generating a square wave output and driving the "pump" with the two capacitors and two diodes.
Since the current requirement is so low and you can "pump" at a high speed, you could use 1N914 or similar diodes and 1µF capacitors.
Paul__B:
That circuit should work as far as I can see, but you could do just as well by arranging one Arduino pin to be continuously generating a square wave output and driving the "pump" with the two capacitors and two diodes.
Since the current requirement is so low and you can "pump" at a high speed, you could use 1N914 or similar diodes and 1µF capacitors.
Thx for the information but my question is still upp. Why the negativ voltage dropping in the circuit? I drive it with a transformator powered 12V 1A standard power supply. I dont want to use en arduino pin for this purpose.
According to my mental arithmetic, with the components you specify, that circuit will oscillate at upwards of 50 kHz. If you are using silicon power diodes of the 1N4004 sort, they cannot switch at (anywhere near) that speed, so they capacitively load the circuit instead of rectifying. I think you might do a lot better using 1N914/ 1N4148 diodes and 1µF capacitors and use a 0.047µF capacitor for the time constant.
In any given design, you may have an Arduino pin that is continuously being clocked at a relatively high frequency which you could use to drive the "pump".
Paul__B:
According to my mental arithmetic, with the components you specify, that circuit will oscillate at upwards of 50 kHz. If you are using silicon power diodes of the 1N4004 sort, they cannot switch at (anywhere near) that speed, so they capacitively load the circuit instead of rectifying. I think you might do a lot better using 1N914/ 1N4148 diodes and 1µF capacitors and use a 0.047µF capacitor for the time constant.
In any given design, you may have an Arduino pin that is continuously being clocked at a relatively high frequency which you could use to drive the "pump".
Thanks again but i have not the knowledge for the replacing. Im good at the programming but not with the circuits itself. Can i ask you to draw a circuit with your components? Or replace them in my circuitdrawing? Sure i used 1N4001 diodes. Is it better with this diodes? http://media.digikey.com/pdf/Data%20Sheets/Rohm%20PDFs/RB050LA-40.pdf
And I'm not so good at circuit drawing (which is to say with whatever software you might have used, I am not familiar).
Given the circuit you posted, the capacitors that are too large and could be replaced with 1µF are the two on the right hand side, while the 0.047µF capacitor would be that on the left hand side.
I cannot say whether the Schottky diodes you cite are suitable or not as the data sheet make no mention of recovery time. I presume however that means they are not suitable, and you should use 1N914/ 1N4148 s.