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 (R_{5}) that affects the feedback for the LCD-bias-voltage
(V_{6}) regulator in IC1. Decreasing temperature, for example, causes
an increase in R_{5}'s resistance and a consequent increase in the V_{6}.
R_{4} linearizes the effect of R_{5}, and R_{3} adjusts
the temperature coefficient of R_{5} to that of the LCD (other temperature
coefficients require different values for R_{2} and R_{3}.

When calculating R_{2} and R_{3}, first note that V_{6}
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 R_{3} and the parallel
combination of R_{4} and R_{5}. In equation form, it's as follows:

V_{6} = V_{D/A} - (5V - V_{D/A}) R_{T} / R_{2}

Therefore,

R_{T} = R_{2}(V_{D/A} - V_{6})(5 V - V_{D/A})

Solve for R_{T} at the extremes of V_{6} (-10V and -15V) using
the midrange value for V_{D/A} (0.625 V): V_{6} = -10 V, R_{T}
= 2.43R_{2}; V_{6} = -15 V, R_{T} = 3.57 R_{2}.

Equivalent expressions for R_{T} are based on its definition: V_{6}
= -10 V, R_{T} = R_{3} + (R_{5} @ 50°C) || R_{4};
V_{6} = -15 V, R_{T} = R_{3} + (R_{5} @ -20°C)
|| R_{4}.

From the R_{5} data sheet, R_{4} = 277 kΩ (choose 280
k, 1%), R_{5} @ 50°C = 52.7 kΩ, and R_{5} @ -20°C
= 250.1 kΩ. With that information, substitute those values in the previous
equations, equate corresponding expressions for R_{T}, and solve for
R_{2} and R_{3}. As a result, R_{2} = 172 kΩ (use
169 kΩ, 1%); R_{3} = 365 kΩ (use 365 kΩ, 1%).