Valant Posted August 10, 2010 Report Posted August 10, 2010 So I happened to find a cheap DMF5005N recently. I spent a few days getting it to work and it looks great. Despite what the datasheet said (or so I think it says), this is a negative contrast voltage display. Datasheet Here Notice that in section 3.1 of the datasheet that the "Recommended LCD Driving Voltage" has no negative signs anywhere! So, I assumed it needed a positive voltage for contrast. My power supply disagreed. Lo and behold, with a negative voltage, the LCD works! What gives? Is there a surefire way to determine what an LCD needs? Furthermore, I am playing around with another LCD, the DMC-50448N. Another Datasheet This datasheet, again under section 3.1, shows positive values for the voltages. But now I'm skeptical. Perhaps this one requires a negative voltage as well? Quote
Imp Posted August 10, 2010 Report Posted August 10, 2010 In both datasheets, there is a schematic under 4.2 which shows how to connect the driving voltage. It's quite hidden...I'd missed that one too.. The 2nd display needs positive voltage, as Vo is connected to a voltage dividing pot between Vdd(+) and ground. Quote
julienvoirin Posted August 10, 2010 Report Posted August 10, 2010 (edited) 1st LCD, section 2.1 Vcc-Vee = 23 if Vcc = +5V then that means that Vee is negative (remember some maths) Vee sets the contrast Section 3.2 : typical LCD Driving voltage : Vcc-Vee = +13V I let you solve this very complicated equation at 3rd degree. Your contrast is driven by a negative voltage. provide a source of -18V, DC 2nd LCD use the circuitry of the core to control it. it is a classical CLCD (toshiba driver). just use a 20K trimmer for contrast instead of 10K Edited August 10, 2010 by julienvoirin Quote
Valant Posted August 10, 2010 Author Report Posted August 10, 2010 In both datasheets, there is a schematic under 4.2 which shows how to connect the driving voltage. It's quite hidden...I'd missed that one too.. The 2nd display needs positive voltage, as Vo is connected to a voltage dividing pot between Vdd(+) and ground. Aha, I missed that too. Thank you! I will read datasheets more carefully. 1st LCD, section 2.1 Vcc-Vee = 23 if Vcc = +5V then that means that Vee is negative (remember some maths) Vee sets the contrast Section 3.2 : typical LCD Driving voltage : Vcc-Vee = +13V I let you solve this very complicated equation at 3rd degree. Your contrast is driven by a negative voltage. provide a source of -18V, DC 2nd LCD use the circuitry of the core to control it. it is a classical CLCD (toshiba driver). just use a 20K trimmer for contrast instead of 10K The maths - I recall! Vee = Vcc - 13 I read section 3.2 prior, but I misunderstood it as "Vcc-Vee", like it were some special symbol, but it really is a minus sign. Many thanks. This makes a lot more sense. Quote
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