Electrical – Low-value digitally set resistor

If you wire all three switches to give a single analog output, you require 0.1% resolution. Since common precision resistors are 1%, and ADCs typically have a couple of counts of uncertaintly, I don't think this technique is practical.

However, if you wire each switch independently to an ADC input, with resistors selected to give evenly spaced voltages, it would be much easier to get reliable readings, and would even work with 5% resistors.

You could use three ADC inputs, or a single ADC with analog switches to read the input switches one at a time.

4 - 20 mA is common in industrial systems and has a few advantages over 0 - 10 V or 0 - 5 V control.

• It is less sensitive to induced noise.
• Voltage drop along the wires is accommodated. The transmitter adjusts the voltage to maintain the signal current. Long cable runs are possible.
• Since signal "zero" is 4 mA we can tell the difference between "zero" (4 mA) and a cable break (0 mA).
• A fault condition can be signalled by sending, for example, 2.5 mA.
• The remote sensor can be powered by the loop provided it can work at a minimum of 4 mA. Only two wires are needed.

Since you want to experiment the voltage control is a much simpler option.

The source will be 24 VDC and I want the input to be manually adjustable (assuming with a potentiometer). The input impedance for the 0-5 VDC input is 150 kohms.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Deriving 0 - 5 V from 24 V supply.

We don't want the input impedance to load the pot significantly or the response will not be linear. If the PLC input is 150 kΩ we should go for 1/10 of that for the pot. 15k pots aren't common so we'll go for 10k - even better. The maths is easy too: 2 kΩ per volt of output.

Now we just need to calculate the value of R1 to drop 19 V across it. Using 2k/volt again we get 38 kΩ. 39 kΩ is the nearest standard value.

One more check: the power dissipated in the resistor will be given by $$ P = \frac {V^2}{R} = \frac {29^2}{39k} = 21~mW $$

A 1/4 W resistor will last forever.

... or the input impedance for the 4-20 mA input is 200 ohms.

Almost certainly 250 Ω (not 200) because you would be jumpering in the resistor across the 0 - 5 V input so that 20 mA gives 5 V. \$ R = \frac {V}{I} = \frac {5}{20m} = 250~\Omega \$. The 4 - 20 mA will be read as 1 - 5 V by the analog input.