Field Supply Dropping Resistor

Running a field coil from a supply higher than its rated voltage? A series resistor takes up the difference. Give the supply voltage, the field resistance and the rated field power; the plate sizes the resistor and its wattage.

Parameters

Supply voltage

V_s

Field coil resistance

R_f

Rated field power

P_f

Typical domestic speakers: 4–12 WW

Fig. 12Series dropping resistor feeding a field coil.

Dropping resistor

Exact resistor

R_s

1.92 kΩ

Nearest E121.8 kΩ

Resistor power

P_R

6.08 W

Rating (×2)15 W

Field voltage

V_f

141 V

Field current

I_f

56.6 mA

Actual field power8.45 W

Efficiency

\eta

58.1 %

☞ Use 1.8 kΩ (E12) rated 15 W or more. 41.9 % of supply power is wasted as heat in the resistor.

Field power vs. series resistance

How it works

Theory · assumptions · formulas

From power to current

A field coil is rated by its dissipation, the figure that, for the designer's chosen gauge, gives the intended ampere-turns. From $P_f = I^2 R_f$ :

I_f = \sqrt{\frac{P_f}{R_f}}, \qquad V_f = \sqrt{P_f R_f}

Sizing the resistor

R_s = \frac{V_s - V_f}{I_f}, \qquad P_R = I_f^2 R_s

The calculator picks the nearest E12 value and recomputes the actual current, since a few percent of resistance error changes the power by roughly twice that percentage: $\tfrac{\Delta P_f}{P_f} \approx -2\,\tfrac{\Delta R_s}{R_s+R_f}$ .

Efficiency

The resistor wastes $\eta = V_f/V_s$ of the supply power, a heavy penalty when dropping from a high HT rail. If the supply is far above the field voltage, consider a switch-mode converter, or rewinding the field for the available voltage (Plate IV).

Self-regulationAs the coil warms its resistance rises; with a series resistor the current falls less than it would on a fixed voltage, so the field power stays slightly steadier. Rate resistors at least twice their dissipation and mount them away from the speaker cone.