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 V_D/A and R_T. V_D/A is the output of the internal 5-bit DAC, which enables the user to digitally adjust the LCD bias voltage. Also, R_T is the sum of R₃ and the parallel combination of R₄ and R₅. In equation form, it's as follows:

V₆ = V_D/A - (5V - V_D/A) R_T / R₂

Therefore,

R_T = R₂(V_D/A - V₆)(5 V - V_D/A)

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

Equivalent expressions for R_T are based on its definition: V₆ = -10 V, R_T = R₃ + (R₅ @ 50°C) || R₄; V₆ = -15 V, R_T = 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 R_T, and solve for R₂ and R₃. As a result, R₂ = 172 kΩ (use 169 kΩ, 1%); R₃ = 365 kΩ (use 365 kΩ, 1%).