You can use AC coils on DC circuits providing you apply enough DC voltage to draw the same amount of current as when operating on AC.
Here's an alternative to rewinding AC coils so that they can operate on DC circuits. It's a relatively simple procedure requiring some calculations and some bench testing. Basically, you apply a DC test voltage to the coil until you draw the same magnitude of operating (holding) current as when the coil operates on AC. Then, applying this DC voltage, you insert a fixed but adjustable resistance to obtain the desired dropout characteristics. This simple field conversion takes only an hour or two. A note of caution: You should use a DC bridge that is vastly oversized so that it can withstand the heat generated and any transient overvoltages that may occur. Generally, a 1000PRV bridge can be used on a 120V circuit, but it should be rated at [+ or -]4 times the rated current of the coil.
What it's about
A holding coil, or solenoid, is a current operated device. It doesn't care what voltage (AC or DC) is impressed on it as long as the voltage level does not exceed the voltage rating of its magnet wire insulation. As such, you can apply a DC voltage to an AC coil. To get the AC coil to work on a DC system requires a sufficient DC voltage be impressed on the coil so that the same amount of current is drawn as when the coil is operated on AC.
Test ranges. This DC voltage usually will be within [+ or -]30% of the AC voltage. Also, the DC resistance usually will be within the same range of the AC resistance and the DC pick-up voltage usually will be within this same range of the AC pick-up voltage. The DC drop-out voltage usually will be within [+ or -]10% of the AC drop-out voltage.
Will there be any benefits associated with the DC operation? Yes, you can expect coil life to be extended three-fold due" to the [+ or -]70% power reduction. In addition, you'll have clean, solid make-and- break operation with no contact chatter, hum, or buzz.
There are four steps involved in this procedure.
Step 1: Measurements. First, measure the AC current drawn by the coil at rated AC voltage. Next, measure the AC pickup and drop-out voltages of the coil. Finally, measure its DC resistance.
Step 2: Calculations. First, calculate the coil's AC impedance ([Z.sub.AC]) by dividing the coil's AC voltage ([E.sub.AC]) by its AC current draw ([I.sub.AC]), or
[Z.sub.AC] = [E.sub.AC]/[I.sub.AC]. (equation 1)
Second, calculate the coil's AC power ([P.sub.AC]) by multiplying its AC voltage times its AC current draw, or
[P.sub.AC] = [E.sub.AC] x [I.sub.AC]. (equation 2)
Step 3: Branch tests. Gradually apply a DC voltage to the coil until you see the same magnitude of current obtained from the Step 1 measurement (coil operated on AC). Then, measure the DC pick-up and drop-out voltages.
Step 4: Calculations again. First, calculate the coil's DC resistance ([R.sub.DC]) by using the following equation and verify that it's almost equal to the measured DC resistance obtained in Step 1.
[R.sub.DC] = [E.sub.DC]/[I.sub.AC] (equation 3)
Second, calculate the coil's DC power (watts) ([P.sub.DC]) using the following equation and verify that it's [+ or -]30% of the coil's AC power ([P.sub.AC]) obtained in the Step 2 calculation.
[P.sub.DC] = [E.sub.DC] x [I.sub.DC] (equation 4)
Finally, calculate the power factor (PF) by dividing the coil's DC power ([P.sub.DC]) by its AC power ([P.sub.AC]), or
PF = [P.sub.DC]/[P.sub.AC] = W/VA (equation 5)
Sample conversion problem
Suppose a signal relay with its coil rated at 120VAC. The measured DC coil resistance is 2000 ohms and the AC power is 2VA. The relays contacts are 10A, coin-silver.
Measure the coil's AC current draw or we can calculate it, or do both as a check. By using equation 2 and solving for AC current draw ([I.sub.AC]), we have the following.
[I.sub.AC] = [P.sub.AC]/[E.sub.AC]
= 2VA/120V = 0.017A
We then verify that this is very close to what is measured.
Now, calculate the coil's AC impedance ([Z.sub.AC]), using equation 1, as follows.
[Z.sub.AC] = [E.sub.AC]/[I.sub.AC].
= 120V/0.017A = 7058.9 ohms
Next, we calculate the coil's DC voltage by using equation 3 and solving for [E.sub.DC] as follows.
[E.sub.DC] = [R.sub.DC] x [I.sub.AC]
= 2000 x 0.017A = 34V
As we said earlier, this can be measured through bench testing. Nevertheless, the calculation provides a good check that the DC voltage falls within the prescribed range.
Next, we calculate the DC power ([P.sub.DC]) by using equation 4 as follows.
Remember our original premise for using AC coils on DC circuits: apply a DC test voltage to the coil until you draw the same magnitude of holding current as when the coil operates on AC. In other words, [I.sub.AC] equals Inc. Thus, you can insert the calculated AC holding current value in place of [I.sub.DC] in equation 4.
[P.sub.DC] = [E.sub.DC] x [I.sub.DC]
= 34V x 0.017A
This is 28.9% of the AC power (2VA).
Finally, we calculate the coil power factor (PF} using equation 5 as follows.
PF = [P.sub.DC]/ [P.sub.AC]
= 0.578W / 2VA
= 0.289 or 28.9% PF
Roger D. Hoestenbach is Senior Consulting Engineer, Paragon Engineering Services, Inc., Houston, Tex.