# Solid State Relay Input Resistor

I have a set of solid state contactors that I want to turn on and off with 24v PLC outputs.

That contactor takes 4.5 - 15V inputs ( Changing the relay to the 24v input model is not an option right now). I need some help calculating the size of the series resistor that I should use, or some other method to reduce the 24V output to 4.5-15v.

DP4RSC60D20B - http://www.crydom.com/en/products/catalog/d_p4_r.pdf

Logic Supply Voltage Range [VDC]: 4.5-15
Minimum Logic Supply Current mA: 16
Maximum Logic Supply Current mA: 20
Minimum Control Input Current @ Min voltage [mA]: 0.20
Maximum Control Input Current @ Max voltage [mA]: 1.0

• Have you tried using a transistor? – Ignacio Vazquez-Abrams Jun 23 '15 at 17:07

Your data sheet lists a maximum control current of 1 mA at 15V, which means the input acts like roughly a 15 kOhm equivalent resistance. What you want is a voltage divider that takes your 24V down to 15V, with a lower resistor of 15 kOhm.

$$\frac{15 k\Omega}{15 k\Omega+x}= \frac{15V}{24V} \\ x=9 k\Omega$$

Now, that's your lower limit. Any less series resistance, and the voltage across the input will be over 15V and out of spec.

The data sheet also lists that the controls will pull a minimum current of .2 mA at 4.5V, which is a 22.5 kOhm equivalent resistance. In that case, you get a maximum dropping resistance of 97.5 kOhm. Any more than that and your voltage will be too low and out of spec.

The problem here is that we don't know what happens in between. The control current at 15V will not exceed the 1 mA spec. The data sheet doesn't say it can't be less than 1 mA at 15V, though. So what happens then? Suppose the control current is .5 mA. Now you only get a 4.5V drop across your 9 kOhm resistor, and your voltage is 19.5V, out of spec. The same problem applies to our maximum resistance if the controls draw more current than the specified minimum: your voltage will drop below spec.

Unfortunately we can't make any guarantees based on the information in the data sheet. All we can do is make some reasonable assumptions.

Let's assume the input response is linear between the two points provided. That means it's basically a diode with a resistor in series, which is not unreasonable. Taking the two voltages and currents they provide, you can model the input as a 1.875V constant drop and a 13.125 kOhm resistance. This lets us compute the terminal behavior at any operating point, not just the two provided. Of course, its behavior could be totally different, but it's the best we can do

Now we can ignore the terminal voltage and just worry about keeping the current between .2 mA and 1 mA. So you have a voltage drop of 24V - 1.875V = 22.125V across (13.125 + x) kOhm. We'll target .6 mA, middle of the range. A 23.75 kOhm resistor will hit that target.

I'd use a 22 kOhm. Even 1/16W should be fine.