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I have a leaf blower that I want to equip with a speed control (continuous, not just 2 or 3 speeds). Here's what I know about the leaf blower and its motor (photos follow):

  • Make & model: Black & Decker LB700
  • 120 VAC, corded, not grounded
  • motor current: 7 amps [per the manufacturer]
  • Single speed
  • Blower type: centrifugal, turbo
  • Shaft is parallel to air inlet, 90° to air outlet
  • Motor has no specification plate, only these markings:
    • 76-30-7A-L
    • 20 03 24
  • Motor has two wires:
    • black goes to power neutral
    • red goes to on/off switch, then to power hot
  • 2 brushes (not brushless)
  • 24 commutator segments
  • 12 slots in rotor

I'm less certain about this info:

  • Appears to have 2 sets of stator (external) windings
  • Appears to have no permanent magnets
  • Rotor appears to have 6 sets of windings

I don't see anything that looks like control circuitry, not even a capacitor

Can anyone identify what kind of motor this is, so I know what kind of speed control I need?

The first photo shows all the electrical components, including the power connector and the on/off switch. That's all there is to this leaf blower, except the plastic shell.

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1 Answer 1

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It's a series wound (universal) type motor. Some speed control can be accomplished with series resistance, but PWM is better. A phase modulated (TRIAC) control might work, but it should be designed for motors, and not for lamp dimming.

The pictures pretty clearly show the AC input wires going through the field and armature (through the brushes), so unless there are hidden connections, it has to be series wound. Also, equipment such as this (blowers, chain saws, string trimmers, circular saws, drills, and the like), are almost always series wound. Such motors have very high starting torque and are speed limited, at a given voltage, only by load and other physical factors.

In a series wound motor, the applied voltage causes a current through the stator (field) and the rotor (armature) that causes a magnetic field proportional to that current, which is ultimately limited by the resistance of the windings and saturation of the laminations. As the motor spins, it creates a back-EMF (BEMF) that counters the applied voltage and causes less current to be drawn, and eventually draws only as much power as needed to drive the load along with resistive and magnetic losses.

Shunt wound motors have a fixed magnetic field in the stator, based on applied voltage, so speed is more constant. The speed of the motor can be adjusted by varying current in the field, and reduced current (field weakening) can result in an increase of speed, because the motor must spin faster to generate the required BEMF to stabilize.

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  • \$\begingroup\$ +1 It may be useful if you briefly explain how you know it is a series wound motor and not a parallel wound one (which has different characteristics). eg explain from the wire connection sequence that the current flows through rotor and then through field (or vice versa). A comment why this is speed controllable and the limitations are beyond what was asked for but would also improve an already adequate answer. \$\endgroup\$
    – Russell McMahon
    Aug 17, 2022 at 9:22
  • \$\begingroup\$ @RussellMcMahon The pictures pretty clearly show the AC input wires going through the field and armature (through the brushes), so unless there are hidden connections, it has to be series wound. Also, equipment such as this (blowers, chain saws, string trimmers, circular saws, drills, and the like), are almost always series wound. Such motors have very high starting torque and are speed limited, at a given voltage, only by load and other physical factors. Shunt wound motors have a fixed magnetic field so speed is more constant. \$\endgroup\$
    – PStechPaul
    Aug 18, 2022 at 1:57
  • \$\begingroup\$ PStechPaul - You and I know about series motors - many will not. My comments were aimed at making your already OK answer better for people who would benefit from more than just a statement of fact. The information in your comment would be a valuable addition to your answer. \$\endgroup\$
    – Russell McMahon
    Aug 18, 2022 at 2:19

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