Autoranging circuit simplifies hardware and software
Dana Romero, Phonex, Midvale, UT -- EDN, July 22, 1999
A natural approach to autoranging is to use an inverting op amp and switch in one of N feedback resistors at a time for a gain of A=-RF/RO (Figure 1). However, the inverting configuration suffers from low input impedance of ZIN=RO because the negative input is at virtual ground. Also, another inverter is necessary to provide positive voltage ranges because of positive signals at the input.
An alternative approach is to use a noninverting op amp that has a high input impedance and a gain equation of A=1+RF/RO (Figure 2). This circuit augments the typical 0 to 5V input range of a µC with ranges of 0 to 1V and 0 to 10V. The circuit in Figure 2 uses a 68HC16 µC, but the results generally apply to any µC with an ADC. The unusual arrangement of circuit components provides three input ranges that use only a single op amp, two spdt relays, and resistors with standard values. Also, the circuit simplifies the autoranging software because after K2 switches, the status of K1 is irrelevant.
With the relays in the positions that Figure 2 indicates, RF equals R1, which in turn equals RO. Thus, the amplifier gain is 2. Combined with the voltage divider of R4/R5, the overall circuit gain is 1/2, which is the gain necessary for a 0 to 10V input at VIN to provide 0 to 5V at VAD.
When K1 switches, RF becomes R2+R1=2RO+RO=3RO for a gain of 1+3RO/RO=4. Dividing by 4 gives an overall gain of 1, which is suitable for a 0 to 5V range at VIN. At this point, it becomes obvious why the voltage divider sits before the amplifier: With VIN near 5V, multiplying by 4 would saturate the op amp.
When K2 switches, the position of K1 is irrelevant, and RF equals R3+R1=18RO+RO=19RO. Thus, A=1+19RO/RO=20. Now, the overall gain is 20/4=5, or exactly the value necessary for a VIN range of 0 to 1V to produce an output of 0 to 5V.
Although the gain calculation works out exactly with standard 5% resistor values, for greater accuracy, you can switch the resistors to the nearest 1% values. Or, even better, sort through a batch of 5% resistors and use a good multimeter to find resistors that are within 1% of nominal. (DI #2381)
