circuit challenge - Replacing obsolete LCD display with current technology

Thread Starter

jimkk

Joined Mar 11, 2019
32
The challenge here to to provide a way for non-technical people to replace a failing display with a identical looking current technology equivalent without needing a soldering gun or any need to dig into the electronics. Much of this was easy. Physical layout and logic connections are the same, so in theory, the display could be replaced with a screw driver and unplugging 2 connectors, so I decided to see if I could design a "plug and play replacement." The display is basically a standard 4002 (40x2) LCD display. The original 1985 product uses the same logic configuration and supply voltage as the current products. The original backlight was much different- driven by +12vdc to the Anode and -12VDC to the cathode (24vdc). Current displays looks for 5V difference between the A and the K. I resolved that with a simple potted regulator circuit so the user could still use his +/- 12vdc source connector, but deliver 5vc to the backlight. Now I have everything resolved but the contrast circuit which unfortunately, is important. The original display uses -12VDC (minus) source for contrast control to pin 3 (yes, the logic circuit is still 5 vdc). Current technology uses +5 at pin 3. See attached schematics. The original display is a Sharp LM402B02 which I cannot find ANY data on except one "simplified" one page data sheet which lists a "secondary support voltage" requirement of -12.

So.... I think my question is.....is there a simple circuit I could put in line to pin 3 that would convert the existing variable -12vdc signal to a variable +5vdc signal at the display? Any thought? Thanks.
 

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MaxHeadRoom

Joined Jul 18, 2013
25,988
If it conforms to the Hitachi HD44780 based display, pin 3 requires a 2.5k pot across 5v and the slider to pin 3 (contrast) .
Max.
 

BobaMosfet

Joined Jul 1, 2009
2,028
That's what BestBuy is for. Letting non-technical people change from LCD displays to OLED displays. :)

Beyond that, the only way you get a negative voltage in a DC environment is by creating negative potential below ground. If you have 24V you could use an OpAmp and provide +/- 12V allowing you to feed either way. Sorry to say, you might need to create a DC DC convert to step 5V up to 24, so you can use an OpAmp to create your negative rail at 12V, just for the contrast pin...

Not saying it's the best idea, but it will get you there.
 
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Thread Starter

jimkk

Joined Mar 11, 2019
32
Allow me to clarify as your response seems like I haven't been clear. The original product includes -12v in its supply and uses it in a circuit that controls and dynamically adjusts contrast by sending the subsequent voltage to the Vo pin (3) on the display.....the final voltage (roughly -8 to--12) depending on heat and a couple other things. Modern displays and the one I'm trying to retrofit controls contract exactly the same and where I understand it would be easy to bypass the -12v part of the original circuit, I' trying to avoid this so I can use the original citcuit and control.. I've looked at a couple OpAmps, but the ones I'm familiar with, I can see problems. To sum up the perfect solution would be a module (build or buy) that would take a negative input (variable -8 to -12) and convert it to a variable positive output that would change along with the input. Examples : variable input: -8vdc to -12vdc converts to variable output: +4vcd to +5vdc or variable output: +8vdc to +12vdc. Current needs are negligible as the display is merely looking for the difference to Vcc.
 

Thread Starter

jimkk

Joined Mar 11, 2019
32
Yes! It looks perfect which I think might just be good enough! I wish all this virtual software came out before I retired. This is all soooo cool! So, just because I'm not smarter than a fifth grader and have been away from this for so long......allow me to restate the obvious. In your schematic, V- is my -12Vmax in. I assume U2 is merely referring to the LM324. V+ on the LM324 is simply my 5VDD supply and the "-" on the LM324 ties to ground. "Out" is my +V result. Looking at your graph, even the I/O ratios are correct! Thank you so much! Geeeez, Now it looks like I better go spend some time on Ebay and buy a breadboard kit. Thanks again! Jim
 

dendad

Joined Feb 20, 2016
4,219
I've found in practice, just adding a 1N914 or 1N4148 diode from pin 3 to 0V works quite well.
Anode to pin 3, cathode to 0V.
 

crutschow

Joined Mar 14, 2008
30,801
In your schematic, V- is my -12Vmax in.
It's the -12V power from the original system.
I assume U2 is merely referring to the LM324.
No.
U2 is the designation for the potentiometer.
U1 is the op amp.
V+ on the LM324 is simply my 5VDD supply
No, it's the +12V supply from the original system.
If you must use 5V for the op amp power, then you will need to change U1 to a rail-rail output type op amp, such as an OP777.
the "-" on the LM324 ties to ground.
Yes.
That's the "-" power input, not the (-) signal input.
"Out" is my +V result.
You got it.
It's the blue trace in the simulation.

If interested, the simulation was done with LTspice, which is a free download from Analog Devices.
 
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Thread Starter

jimkk

Joined Mar 11, 2019
32
Crutschow, thanks for taking the time to help and the correction s above. I'll be experimenting with "side A" of a LM358N. Just to be thorough, I've attached a pdf which shows my understanding of pins assignments.

Dendad, I'm not sure I understand your suggestion. I'm guessing from your references that you're talking about a standard 44780 based display where the backlight power connections are typically named anode (A) and cathode (K) and pin 3 is where the display receives an adjustable voltage for contrast control (typically 3-5v). Putting a diode between A and 3 would produce a voltage of Vdd (5v) minus about .7, so 4.3 volts on the contrast control pin which would actually be just about perfect and a really good idea (simple solution) for applications where contrast was not to be adjustable. Is that what you're referring to?

Thanks again everyone. You guys have been great!

PS I'll be downloading LTSpice next.
 

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Thread Starter

jimkk

Joined Mar 11, 2019
32
So....it works perfectly! Of course, now that I've learned this, I'd like to make a change.

Input now is -4 to -7. Output is +4 to +7 Supply can either be +5 or plus +12. Stay the same at +12 for simplicity. I've tried the circuit as is but the output way too low, down around 2-3v.

Meanwhile, I've started playing with LTSpice (not very intuitive....yikes!), but am not far enough along to run simulations. I'm still trying to figure out basic housekeeping. Youtube videos have been helpful. Any other learning suggestions would be appreciated.

Thanks again.

Jim
 
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