Resistor in series with Solenoid Valve (for voltage reduction)

Question

I have a question regards the use of a resistor in series with a latching solenoid valve. Specifically, I have a valve that wants to see 14 Volts, but my power supply is 28 Volts. So if I put a resistance equal to the steady-state valve resistance (which is 9 ohms), then the valve will never see more than 14 Volts. However the resistor will also limit inrush current, which is not necessarily a good thing, if it is needed to get the valve moving. So my question is whether valve performance will be degraded by having the resistor in series, versus what these valves normally see, which is essentially an infinite source powering them (i.e., low series resistance)?

Any help would be greatly appreciated.

Thanks, Jim

Answer 1

Inductors don't have inrush current.

The current thru a ideal inductor changes proportionally to the applied voltage. When you first apply a voltage to the solenoid, its current is zero. It will then ramp up. Due to the DC resistance of the coil, the current doesn't keep increasing infinitely, but exponentially decays to the steady state value. This steady state value is the applied voltage divided by the DC resistance of the coil.

Put another way, the current thru the coil exhibits the opposite of inrush since it ramps up over time.

To answer your overall question, yes, putting a resistor in series with the solenoid equal to the solenoid's DC resistance allows driving the whole thing with twice the voltage the solenoid is rated for. Of course you're going to use twice the power too. The solenoid dissipates the same power as before, but the resistor also dissipates this power too.

Answer 2

When given 14V into an 9 ohm load would give you 1.55A.

If you double both, 28V into an 18 ohm load give you the same 1.55A. The DC resistance of the coil is what is going to be the limiting factor to the inrush. All of that works.

The problem is the power. The resistor is going to be dropping 14V. That means that the power dissipated is going to be 14^2/9 = 21.7W That is a lot of power to dissipate.

Answer 3

Instead of using a resistor you can use a DC to DC Adjustable Regulator LM2596, its very cheap and easy to use, because i think using a series resistor will cause the latching solenoid to misfunction.

Answer 4

Providing you are only pulsing the solenoid (it is specified as a latching device) then perhaps you have the opportunity to speed up operation too.

Most solenoids have a broad voltage specification, and you can overvolt them for a short time to speed up actuation. I assume here that the 14 V specified is the continuous voltage rating for the solenoid. If you look at the datasheet you may find a duty cycle limited voltage specification too.

There is no inrush current by the way. This is a simple LR time constant (no matter how short the time), so the current rises from zero to a maximum defined by the circuit resistance.

^{simulate this circuit – Schematic created using CircuitLab}

In the RHS schematic above the solenoid is given a 1 mS 'kick' of higher voltage. This provides a higher aiming voltage and shortening the time to reach a given current through the solenoid. You could increase the value of the capacitor to increase the time more than 14 V is applied to it. Steady state is just the same as the LHS when the capacitor is charged to 14 V.

Remember that either of these cases absolutely depends on this being a pulse applied to the solenoid (in your comment you said it was 150 mS). Steady state power dissipation would be very high (22 W in each). Providing you know you will always pules the solenoid, the R value could probably be a 9 Ohm 2 - 5 W wire wound.