Current consumption by a solenoid valve

Question

I want to calculate the voltage drop in a circuit loop consisting of a solenoid valve.

The valve power consumption is 16W. The valve datasheet mentions minimum voltage to operate as 18V.

I have a 24V PS to control the valve. The internal resistance of valve is given as 40 Ohms.

My confusion is regarding current. What value of current should I consider:

1. Given that it needs minimum 18 V to operate and internal resistance as 40 Ohms. The current is 0.45 A. However this gives power as (\$V\times I=18 \times 0.45= 8.1 W\$) which is less than the rated power.

2. If I consider the power rating at 16W, the required current is ( \$=16/18\$) is 0.8A.

Which current value shall I use for overall voltage drops in my loop. Please let me know if this is confusing.

Answer 1

Solenoid valves operate slowly at the low-end of their DC voltage range and so it's highly likely that at its rated voltage, it will operate much more quickly than when the applied voltage is only 18 volts. You need to take this into account.

If the typical DC resistance is indeed 40 Ω and, the typical nominal power is 16 watts, then the upshot of this is that the nominal voltage rating for the solenoid is: -

$$\sqrt{\text{Power }\times \text{Resistance}} = 25.3\text{ volts}$$

Without a link to the data sheet I'm speculating but, hopefully, this answer gives you the tools to work this out for yourself.

If I consider the power rating at 16W, the required current is (=16/18) is 0.8A.

This is unlikely to be true because you are taking nominal power rating and minimum voltage rating to calculate current. At the nominal power rating of 16 watts, the supply voltage is likely to be about 25.3 volts and, of course, that implies a current of 632 mA.

So, decide on your operating voltage (maybe 24 volts) and work out current based on ohm's law i.e. current = 24/40 = 600 mA.

But, without reading the data sheet, I'm partially guessing.