# Using AC Coils on DC Power

Here's an alternative to rewinding AC coils for operation on DC circuits. It's relatively simple - all it takes are a few calculations and a little bench testing. Want to skip the time and expense of rewinding AC coils so they can operate on DC? Or are you wondering if it's even possible to use AC coils on DC circuits? If you answered "yes" to both of these questions, there is a method you can use.

Here's an alternative to rewinding AC coils for operation on DC circuits. It's relatively simple - all it takes are a few calculations and a little bench testing.

Want to skip the time and expense of rewinding AC coils so they can operate on DC? Or are you wondering if it's even possible to use AC coils on DC circuits? If you answered "yes" to both of these questions, there is a method you can use.

Remember, a holding coil, or solenoid, is simply a current-operated device. The device doesn't care what voltage (AC or DC) you apply to it as long as the applied voltage level doesn't exceed the voltage rating of the coil's magnet wire insulation. You can apply a DC voltage to an AC coil. Here's how to go about it:

• First, apply a DC test voltage to the AC coil you want to use, until you draw the same magnitude of operating (holding) current as when the coil operates on AC.

• Then, while applying this DC voltage, insert a fixed but adjustable resistance to get the desired coil dropout characteristics.

There is one important thing to consider here: Make sure you use a bridge that's vastly oversized, so it can withstand the heat generated and any transient overvoltages that may occur. Generally, you can use a 1000PRV bridge on a 120V circuit, but you should use one having a rating about four times the coil's rated current.

The DC voltage noted above usually will be within 30% of the AC voltage. The DC resistance noted in the second bullet usually will be in the same range (30%) of the AC resistance. The DC pickup voltage also will be in this same range of the AC pick-up voltage. However, the DC drop-out voltage will be within 10% of the AC drop-out voltage.

Are there any benefits (aside from saved rewinding costs) by using AC coils on DC circuits? You bet! You can expect up to three times the extended coil life. That's due to the approximate 70% power reduction. By using AC coils on DC circuits, you can expect clean, solid make-and-break operation with no contact chatter, hum, or buzz. Let's go through the procedure in detail. The conversion procedure is simple. All you have to do is follow these four steps.

Step 1: Take some measurements. First, measure the AC current drawn by the specific coil at rated AC voltage. Then, measure the coil's AC pick-up and drop-out voltages. Finally, measure its DC resistance.

Step 2: Do some calculations. First, calculate the coil's AC impedance (Z subscript AC). You do this by dividing the coil's AC voltage (E subscript AC) by its AC current draw (I subscript AC), or

Z subscript AC = E subscript AC/I subscript AC (equation 1).

Then, calculate the coil's AC power (P subscript AC). You do this by multiplying the coil's AC voltage (E subscript AC) by its AC current draw (I subscript AC), or

P subscript AC = E subscript AC x I subscript AC (equation 2).

Step 3: Do some bench testing. Gradually apply a DC voltage to the coil until you see a DC current (I subscript DC ) having the same magnitude as the AC current (I subscript AC) measured in Step 1. Then, measure the DC pick-up and drop-out voltages.

Step 4: Do some more calculations. First, calculate the coil's DC resistance (R subscript DC) by using the following equation:

R subscript DC = E subscript DC x I subscript AC (equation 3).

Then, calculate the coil's DC power, in watts, (P subscript DC) using the following equation:

P subscript DC = E subscript DC x I subscript AC (equation 4).

Verify that the power is 530% of the coil's AC power (P subscript AC) calculated in Step 2.

You may wonder why we're using AC current in equation 3 and equation 4. Our original premise for using AC coils on DC circuits: Apply a DC test voltage until you draw the same magnitude of holding current as when the coil operates on AC. In other words, I subscript AC equals I subscript DC. So, we can insert the measured AC holding current value in these two equations.

Finally, calculate the power factor (PF) by dividing the coil's DC power (P subscript DC) by its AC power (P subscript AC), or

PF4P subscript DC3P subscript AC (equation 5).

Let's try a sample conversion. Suppose you have a signal relay with a coil rated at 120VAC. You measure the DC resistance (2000 ohms) and note the nameplate says AC power is 2VA. The relay contacts are 10A, coin-silver.

Step 1. You can measure the AC current draw (I subscript AC), calculate it using the above data, or do it both ways as a check. By using equation 2, but solving for current draw, you have:

I subscript AC = P subscript AC/E subscript AC = 2VA/120V = 0.017A.

Now, verify that this is close to what you're measuring.

Step 2. Calculate the coil's AC impedance (Z subscript AC) by using equation 1:

Z subscript AC = E subscript AC/I subscript AC = 120V/0.017A = 7059 ohms.

You don't have to calculate the coil's AC power because it's on the nameplate (2VA).

Step 3. Do some bench testing to find the DC voltage that generates the same magnitude of current as when the coil operates on AC. You can also calculate the coil's DC voltage by using equation 3, but solving for E subscript DC:

E subscript DC4R subscript DC= = I subscript AC = 2000 ohms x 0.017A = 34V.

By measuring and doing the calculation, you get a good sense that the DC voltage falls within the prescribed range.

Step 4. You measured the DC coil resistance (2000 ohms, remember?) so you already have this information. But, you need to calculate the coil's DC power (P subscript DC) by using equation 4:

P subscript DC = E subscript DC x I subscript AC = 34V x 0.017A = 0.578W.

This is almost 29% of the AC power (2VA), which is within the prescribed range.

Finally, calculate the coil power factor (PF) by using equation 5:

PF = P subscript DC/P subscript AC = 0.578W/2VA = 0.289 or 28.9%.

That's all there is to it! Thanks to Roger D. Hoestenbach, former Senior Consulting Engineer, Paragon Engineering Services, Inc., Houston, who provided the details for this unique application and procedure.

TAGS: Design