I need to replace a 1602L LCD (winstar i think) with an ACM1602S LCD.

The original lcd had this resistance RA enter image description here

enter image description here

also the end of the white wire from the switch was connected to the part that is marked J16, but not I can remember which of the three points. I think the other connections are equivalent. Full gallery here.

If anyone can help me, I would be grateful

Datasheet for the original lcd is here


1 Answer 1


Assuming you have a led backlight, and the lcd is taking 5V power... Seems the lcd was already replaced once? And the backlight was not originally used by the main board, resulting in a hack.

RA was added on as a series resistor for the backlight, likely led. This was added because the voltage for the backlight is higher than the backlight wants (4.2V based on the LCD). It's 56Ω, so the current is likely around <30 milliamps? The data sheet says it normally takes 260mA, so I'm guessing the power supply is limited.

J2, the backlight Anode jumper makes me think the backlight is powered via the main power for the LCD, instead of Pin 15. That it has a solder blob and an smd resistor that didn't reflow properly says a lot about the hack that built this (no offense).

J16 is the backlight Cathode jumper. The left side goes to pin 16, the right is likely connected to the A pad on the left of the board. See feature 5 on the data sheet where it says LED can be driven by pin 1/2 (Main Ground and VDD), pin 15/16, or A/K. The wire likely connected to the center pin, a direct connection for ground to the backlight. The wire goes to a switch that disconnects ground to turn off the backlight, obviously.

To replace it with a new one, the data pins 1-14 will be the same. Since R8 is missing, you have to provide power via Pin 15 (or A) the Positive for the LED. Since the power supply is limited, you may want to add that 56Ω resistor in series or it will try to pull a few hundred milliamps. Since R7 is there (6.8Ω), the backlight may be on as soon as led + power and normal ground are connected. To use the switch, you will need to remove R7, and connect that switch to Pin 16 (or K).

enter image description here

A more detailed data sheet for a similar version of the ACM1602s confirms it. R7 ties it to normal ground (Pin 1), while R8 would have connected it to VDD (Pin 2), like your existing lcd uses. Instead of RL1 being the current limiting resistor, it has a 0 ohm jumper and uses R7 to limit the current, likely to 120 to 240 mA.

enter image description here

Side note, RL3 is useless. 1 megohm across the led for no reason? Weird.

  • \$\begingroup\$ believe me, no offense at all. i have nothing to do with that. i need to ask you something: these two buttons used to command the device (things like mode and intensity , if I remember correctly ) . does this make any difference? So, just to make sure : 1) pin 1 to 14 as in the original scheme 2) remove R7 (SMD resistor) 3) find 5v connector, connect it to 56ohm resistor , and then connect the resistor to pin 15 . 4) connect the switch (the wire hanging in photo 2) to pin 16. \$\endgroup\$ Jul 21, 2016 at 14:18
  • \$\begingroup\$ if you need more info or pictures about the device, please let me know. \$\endgroup\$ Jul 21, 2016 at 14:21
  • \$\begingroup\$ @simplyelectronics yes, that should be all that's needed. The white thing connected by the blue wire, is that an LED or a button? \$\endgroup\$
    – Passerby
    Jul 21, 2016 at 14:50
  • \$\begingroup\$ both were switches (this kind canadarobotix.com/image/cache/data/products/800/…). One served to control mode and the other to control intensity(i think) but neither had anything to do with lighting or power on / off. i replaced one of them with one of these (g01.a.alicdn.com/kf/HTB14idFIXXXXXcjXXXXq6xXFXXXi/…) just to test purposes (didn't workout though). you can zoom at this image (postimg.org/image/y8dea60v5) to see it. does that makes any difference ? \$\endgroup\$ Jul 22, 2016 at 12:02
  • \$\begingroup\$ i would really appreciate your last comment on this. can i go with your original answer @passerby ? \$\endgroup\$ Jul 26, 2016 at 2:28

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