Electronic Design

Temp Compentsation Levels LCD Contrast

The variation of LCD contrast with temperature can become a problem in some applications. For instance, when working in a meat-packing plant, you might go from scanning sides of beef in a refrigerated locker to downloading data from your LCD bar-code reader in a heated office. That type of drastic temperature change would definitely require an LCD-bias adjustment.

By combining automatic bias adjustment with manual-adjust capability, the user can compensate for LCD viewing angles and manufacturing differences (Fig. 1). IC1 is a power-supply chip for portable systems, and it includes two other switching-regulator controllers, plus circuitry for backup=battery switchover, low-voltage warning, and power-fail reset. When using this setup, the circuit provides a linear bias change with temperature, from -10 V at 50°C to -15 V at -20°C (Fig. 2).

The automatic compensation is supplied by a negative-temperature coefficient resistor (R5) that affects the feedback for the LCD-bias-voltage (V6) regulator in IC1. Decreasing temperature, for example, causes an increase in R5's resistance and a consequent increase in the V6. R4 linearizes the effect of R5, and R3 adjusts the temperature coefficient of R5 to that of the LCD (other temperature coefficients require different values for R2 and R3.

When calculating R2 and R3, first note that V6 is a function of VD/A and RT. VD/A is the output of the internal 5-bit DAC, which enables the user to digitally adjust the LCD bias voltage. Also, RT is the sum of R3 and the parallel combination of R4 and R5. In equation form, it's as follows:

V6 = VD/A - (5V - VD/A) RT / R2

Therefore,

RT = R2(VD/A - V6)(5 V - VD/A)

Solve for RT at the extremes of V6 (-10V and -15V) using the midrange value for VD/A (0.625 V): V6 = -10 V, RT = 2.43R2; V6 = -15 V, RT = 3.57 R2.

Equivalent expressions for RT are based on its definition: V6 = -10 V, RT = R3 + (R5 @ 50°C) || R4; V6 = -15 V, RT = R3 + (R5 @ -20°C) || R4.

From the R5 data sheet, R4 = 277 kΩ (choose 280 k, 1%), R5 @ 50°C = 52.7 kΩ, and R5 @ -20°C = 250.1 kΩ. With that information, substitute those values in the previous equations, equate corresponding expressions for RT, and solve for R2 and R3. As a result, R2 = 172 kΩ (use 169 kΩ, 1%); R3 = 365 kΩ (use 365 kΩ, 1%).

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