240x64 LCD replacement - Page 1
Forum section: electronics / beginners

author: Mp3
author_rank: Frequent Contributor
timestamp: April 15, 2018, 06:40:27 am
content: Hi everyone, I have a new question about replacing an old LCD.I have a replacement for a Toshiba T6963C 240x64 screen. It still works but is very dim and difficult to read. I have two questionsi was following this guide:http://wp.visuanetics.nl/kawai-k5-display-upgrade-cleaning/and bought one ofthese displays(product page)First questionSo that guide the guy replaced his with another display that has 20 pins like the stock LCD. My display has 22 pins though. I tied font select to VCC as the guide says. Checking the pinouts everything matches up.Pin 21 is backlight +5v, pin 22 is backlight negative 0v. Currently neither pin 21 or 22 are hooked up.After checking all my connections and continuities, i plugged it in and turned it on, nothing happened, so i unplugged it right away. I am not sure if i should have expected to see the display work but without any backlight, or if the nothing happens when the backlight is not connected. Right now i am just hoping nothing is broken.Second questionPins 1 and 2 are ground.Pin 3 is VCC +5v.Pin 20 is negative voltage supply 0v.Would i be correct to assume pin 21 (backlight +5v) connects to pin 3, and pin 22 "Backlight cathode 0v" connects to pin 20? or do i connect it to ground?Lastly, Is there any chance anything is permanently damaged?

author: Twoflower
author_rank: Frequent Contributor
timestamp: April 15, 2018, 10:07:27 am
content: The 20Pin displays seem to have a slightly different pinout:http://www.newhavendisplay.com/specs/NHD-24064CZ-NSW-BBW.pdf. The only differences I see are:Pin# 20Pin 22Pin9 VEE NC20 LED+ VEE21 - LED_A22 - LED_KAs you connected the pin 20 to the supply voltage for the LED (probably +5V) and the new display provides there a negative supply voltage of about -8.5V there might a chance that the charge pump got damaged. But I don't think your instrument got damaged. Usually the charge pump provides only very little current compared to the +5V.

author: Mp3
author_rank: Frequent Contributor
timestamp: April 15, 2018, 05:06:00 pm
content: Quote from: Twoflower on April 15, 2018, 10:07:27 amThank you and sorry for my follow up question but is the charge pump built into the display or the power board?I also ended up finding the datasheet for the display i replaced. (i think)https://cdn.sos.sk/productdata/ef/bd/dc8ae086/bg-24064a-yplhnt-1.pdfA problem i noticed:New display Old display9. Not Connected 9. Vee Negative Voltage output -13.6 V (like the Newhaven you found)20 Negative voltage supply 20. N.C No connectionSo perhaps, i didn't send negative voltage to the wrong place, i sent it to..... no place?Would it be safe to tie pin 9 to pin 20 or is that not advisable? The way the wires are soldered in, it might not be very easy to swap where wires 9 and 20 are, but i will do it if it's the right way to do things.Another graphic of a compatible pin-out:

author: Twoflower
author_rank: Frequent Contributor
timestamp: April 15, 2018, 06:29:41 pm
content: The charge pump is located on the LDCisplay driver.In case your information is right, then you might be lucky and everything is fine.With a bridge 9<->20 I would be careful. Either check that the pin 20 is in fact going nowhere on the PCB or measure it: Remove the LDC, power on the instrument and measure GND-Pin20 if there is no voltage present.And be careful with the not loaded inverter for the EL-Foil of the old display. If no EL-Foil is connected it might generate an extreme high voltage probably not deadly but very unpleasant. Also running it unloaded the voltage is that high that it could kill the inverter. See if you can disable it. It is part of your instrument to prevent some side effects (noise in the audio signal). At least add a good insulation to the leads and fix them at a position that doesn't come close to other electrical conductive parts.Edit: Added 'could' as the inverter must not get broken but could get broken.

author: Mp3
author_rank: Frequent Contributor
timestamp: April 15, 2018, 07:14:39 pm
content: Quote from: Twoflower on April 15, 2018, 06:29:41 pmMany thanks againTwoflower!!!My idea was to remove the high voltage inverter, replace with a 100 ohm resistor, but i thought i should test if everything works before i go to remove it. But, i will remove the inverter and replace with a resistor next time i open the instrument up.I have some 5 watt 100 ohm resistors, i hope those are okay?And many thanks for the advice about bridging pins 9 <-> 20, i will check before attempting to tie them together

author: Twoflower
author_rank: Frequent Contributor
timestamp: April 15, 2018, 07:31:35 pm
content: I don't understand why you want to replace the inverter with a 100Ohm resistor. Is there a reason why you want to add the resistor?In case you keep the inverter in place and want to load the output of the inverter to 'simulate' an EL foil you need a resistor that 'consumes' about 80mW at 110V (you can find the value in the display datasheet for the power consumption for the EL-Foil): R = (U * U) / P = (110V * 110V) / 80mW = 151kOhm. So 100Ohm is far to low at this place ;-)

author: Mp3
author_rank: Frequent Contributor
timestamp: April 15, 2018, 11:44:41 pm
content: Quote from: Twoflower on April 15, 2018, 07:31:35 pmAh, i thought it was pretty low a value as well, it was on the guide/blog post i linked:"On the PSU board the inverter can be desolderded, and replaced by a 100-Ohm resistor (between pins IN and OUT)"I have some 100kOhm resistors around, maybe i should go higher for this purpose? If the power supply is fine with no dummy load in place of the inverter, that is fine with me!

author: Twoflower
author_rank: Frequent Contributor
timestamp: April 16, 2018, 04:51:35 am
content: Ah, so you're really want to de-solder the inverter. That's good. I haven't read the guide in detail. In that case you don't have to follow my calculation as it was based to keep the inverter in place. So a 100Ohm resistor for whatever reason (see below) is more the right value. The voltage across the resistor is 5V. With that the power is P = (U * U) / R = 25 / 100 = 0.25 Watt.But I'm still not sure if that resistor is required. As the power supply schematic snipped at the project page shows a normal regulator and not a SMPS and it has some other load attached as well. I would give it a try if you don't apply the resistor at all.

author: Mp3
author_rank: Frequent Contributor
timestamp: April 16, 2018, 08:19:57 am
content: Thanks again Twoflower!In the meantime i got some good emails from the display makerFrom the translation between Cantonese and English i think the 2nd line implies the 9th pin is truly not connected. I will inspect the PCB to check anyway.Quote

author: Twoflower
author_rank: Frequent Contributor
timestamp: April 16, 2018, 05:47:50 pm
content: You should double check that the tin 20 on your instrument is not connected even if the old LCD does not have anything there that does not mean that the instrument does not use it. If in doubt, un-solder the wire of pin 20 on the instrument side and insulate the end. This allows to make the bridge direct on the PCB of the LCD without the fear that the instrument uses that pin.

author: Mp3
author_rank: Frequent Contributor
timestamp: April 17, 2018, 12:54:33 am
content: Quote from: Twoflower on April 16, 2018, 05:47:50 pmExcellent idea

author: Mp3
author_rank: Frequent Contributor
timestamp: May 23, 2018, 11:10:43 pm
content: Hi everyone,sadly i am back to ask for more help.I've measured the power supply unloaded - seems fine? So i don't think i've destroyed it, yet. I am not sure how to measure it with everything plugged in, since i don't want to damage something else by my ignorance.Still no power with the new LCD screen. If the charge pump has been damaged already, then it makes sense why it isn't working still, but it's been wired up properly now.On the power supply, 2 fuses are good (2x 500mA 250V). 1 is blown (1A 250V). I will replace the fuse, but not sure why it blew.Unfortunately the screen is the only indication the instrument is getting any power - no power LED or similar....

author: Twoflower
author_rank: Frequent Contributor
timestamp: May 24, 2018, 04:18:07 pm
content: As starter you can measure if the LCD gets it supply voltage: Measure between pin 2 (GND) and pin 3 (VDD) of the display. Careful not to make any shorts with the tips of the multimeter. You should measure approx. 5V. If that's correct the keyboard should at least have some parts powered.Next would be to check if the charge-pump in the display is still working: Measure between pin 2 (GND) and pin 9 (VEE). You should measure roughly -10V. This voltage is probably not very accurate, that's OK. And if completely missing that's a problem as it is required to drive the crystals between the glass plates of the display. As mentioned earlier the display is generating the -10V and is reduced to approx. -8.5V (depending on the contrast setting) using a voltage divider. And feed back to the display at pin 4 (V0). It shouldn't be too low. Otherwise the contrast is low.