As electronic devices become more self-aware, the need for voltage scaling is also increasing. I am not talking about artificial intelligence, such as Hal in "2001: Space Odyssey." I mean electronic devices with more self-tests, which require reading a wide range of voltages.
Scaling the input voltage is not always as easy (or complicated) as the first time. In this article, I will show you how to solve this challenge in recent signal chain designs that need to shrink the +/- 10 V signal to the 0 to 2.5 V range to match all other signals to the analog-to-digital converter (ADC). The transfer function to achieve this goal is linear: VOUT = VIN / 8 + 1.25V.
Solution 1:
My first thought was to use a non-inverting op amp (op amp) circuit. After some fast arithmetic, I determined the circuit, as shown in Figure 1, requiring a 1.43V bias supply with a feedback/ground resistance ratio of -7/8.
Figure 1: Solution 1 simulation is good, but impossible
The gain of the non-inverting amplifier is (1 + RF / RG). If the gain is +1/8, the resistance ratio is negative. I can't buy a -7k resistor, so this is a big problem. The input common-mode range of my op amp needs to be as low as -10V; this is also a problem because I don't have a negative supply available. Obviously, in this case, the non-inverting op amp circuit is not compatible, but it does work when the required voltage gain is greater than one.
Solution 2:
The five-resistor op amp circuit shown in Figure 2 is a differential amplifier with its inverting input to ground and a noninverting input of 1.25V. The gain is set to 1/8. The input common-mode range is 0V to 2.22V, so a single-supply op amp can be used.
Figure 2: Solution 2 is useful, but is there a better solution?
Solution 3:
I don't need an operational amplifier to attenuate the signal. I can use the three resistors - A, B and C - and voltage source V to perform the required scaling tasks. See Figure 3.
Figure 3: This simple solution uses only three resistors and an existing power supply
In my example, the gain is 1/8 and the offset is 1.25V. I will use the letters G and Z to represent the gain and offset (the output of the zero input); therefore, G = 1/8 and Z = 1.25V. My power supply voltage V is 3.3V.
So what is the best way to solve the values ​​(or ratios) of resistors A, B and C? I can calculate G and Z using Equations 1 and 2 using the resistor divider rule VOUT = VIN * RI /(RG + RI):
The || symbol means "parallel"; for example, x || y is x*y/(x+y) or 1/(1/x+1/y).
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