EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Mp3 on February 11, 2018, 05:03:32 am
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I have a 2x7 style lcd to replace, and have two here which have a single row of 16 pins. However, the lcd i am replacing only has 14 pins. I know 15 and 16 are for the backlight. Is it safe to simply wire them to pins 1 and 2 or do i need a resistor in line? Thanks in advance.
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I'd use a resistor.
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If you have a bench supply, set it for 5V and a low current limit (say 50 mA) and drive pins 15 and 16. If there's no ballast resistor in the module, your supply will current limit.
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No, use a series resistor in addition to current limit. Many (most) power supplies have an output capacitor that can cause a current spike above the current limit setting if you touch the wires to the pins while it's turned on.
Unless you can guarantee that you have the pins solidly connected before you turn on the PSU and wind up the voltage, then a resistor is a sensible and easy precaution. It's not worth the risk.
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A lot of these 16x2 led boards have a current limit resistor on the backlight already. The PSU test will verify that. Mine did and pulls around 11mA directly connected to 5V.
The other issue you could run into is how the contrast adjust resistor is wired. If your 14 pins include the return voltage from the divider (or single resistor) it may or may not be compatible. You may need to adjust it with a resistor. If you need more resistance you can simply put one in series with the pin, if you need less, you will need to review the schematic of the LED board and see if you can get a resistor in parallel with it's internal one or supply a determined voltage on the pin.
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If you have a bench supply, set it for 5V and a low current limit (say 50 mA) and drive pins 15 and 16. If there's no ballast resistor in the module, your supply will current limit.
What can I do if i do not have a bench supply? And i am not sure how to determine what value resistor i should use for the display.
Sorry for my ignorance, i am trying to teach myself this stuff. I know now i need a bench supply, but i should be able to figure this out with just a multimeter, can't i?
Thanks have been given ;D I appreciate everyone's help so much
These here are the LCD boards I have. The generic "QC1602A" that can be bought anywhere: https://www.ebay.com/itm/391721594261 (https://www.ebay.com/itm/391721594261)
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Do you have a potentiomter and a multi-meter?
Use the POT with pins 1 and 2 as your resistor. Twiddle the pot until it looks bright enough. Then measure the resistance between pins 1 and 2, pick a resistor that is roughly that value.
To check if it's going to pull too much current, with the potentiometer in place, wire the DMM in series with the pin on mA range.
V+ --> Meter --> POT --> Backlight PIN
Then turn the POT all the way down, but do not exceed around 20mA. If it goes higher than 20mA you will need a resistor. If it doesn't it probably already has one in it.
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The other way is to guess there is no resistor and assume a few things.
Assuming:
2.3V forward voltage
5V supply
20mA max current.
Therefore 2.7V drop across resistor.
V=IR ... R=V/I ... R=2.7 / 0.020 R=135 Ohm
So put a 220 Ohm in there and see what happens. If it's too dim try a 100 Ohm. If it's still dim, just remove it.
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The other way is to guess there is no resistor and assume a few things.
Assuming:
2.3V forward voltage
5V supply
20mA max current.
Therefore 2.7V drop across resistor.
V=IR ... R=V/I ... R=2.7 / 0.020 R=135 Ohm
So put a 220 Ohm in there and see what happens. If it's too dim try a 100 Ohm. If it's still dim, just remove it.
I happen to have 220 and 2.2k ohm resistors so i will try with 220's an see what happens! Thanks sir! :-+