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I was salvaging a motor / gearbox from a breast pump and noticed something I didn't understand at all. The control panel consists of a 6 position switch (5 speeds and off) and then a series of diodes. No resistors, no capacitors, and no 555 timer. It doesn't seem to be a PWM situation, so what am I not understanding about diodes that let them act as a speed controller? I thought their only function was to limit power flow in a single direction.

I also got a couple 1/2 horsepower DC permanent magnet motors for a go kart project and wondering if this information could be useful at all on a larger motor.

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  • \$\begingroup\$ Possible that the switch adjusts how many diodes are in series. Each diode would drop the voltage by around 0.5-1V, so it could be used for a rather odd and crude speed controller. Can you post a pic of the back of the board and tell us what the input voltage is from the DC adapter? \$\endgroup\$
    – Phil C
    Commented Apr 22, 2018 at 0:59
  • \$\begingroup\$ Thanks Phil, I didn't realize that you would get a voltage drop through a diode, that may be just what is happening. Would that create the same inefficiency that would exist with resistors in generating heat? I could just use a PWM, but more than anything I am trying to understand what I was looking at. The supply is a 12V 2A source. \$\endgroup\$
    – Alan
    Commented Apr 22, 2018 at 1:19
  • \$\begingroup\$ draw the schematic ... it may help you understand the circuit \$\endgroup\$
    – jsotola
    Commented Apr 22, 2018 at 3:14
  • \$\begingroup\$ @Alan The most common application of standard diodes is to control the direction of current flow, but that's not all they're good for and there's no such thing as a component without some level of voltage loss and heat dissipation. Check out this article if you want to get a better introduction to how diodes work. \$\endgroup\$
    – Phil C
    Commented Apr 22, 2018 at 4:43

2 Answers 2

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It's a very cheap way to create a speed controller that has a predictable voltage drop without the same level of current limiting created by a resistor based voltage divider.

The switch puts a variable number of diodes in series to create a combined voltage drop.

The larger diodes (D1-D5) seem to read "1N593" from what I can make out. If that's correct, they would be zener diodes in the forward biased mode. That family of zeners have a 15-39V zener voltage and a 1.5V forward voltage. They're using the 1.5V forward voltage to limit the motor input.

The smaller diode (D6) is probably a standard rectifying diode. Hopefully one with a relatively high reverse voltage, because they don't seem to have anything else to deal with back EMF when that motor stops. This would add another 0.6V voltage drop.

If that's all correct, you should expect to see about 10V on high and about 4.5V on low with 1.5V increments between. Regardless of exact numbers, that's the basic theory of operation.

It's not very efficient and is only useful for small increments, so I would not be trying to adapt this for larger motors.

Here's a simulated version of the circuit (based on my assumptions for exact parts used and with some equivalent components for the rotary switch and motor)

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  • \$\begingroup\$ So are they actually 2 zeners in series inside then? \$\endgroup\$
    – Henry Crun
    Commented Apr 22, 2018 at 3:43
  • \$\begingroup\$ @HenryCrun I'm sorry, I don't understand the question. This board has as many as 5 or 6 zeners in series (assuming I'm right about it having any). If you're asking about the zeners themselves, none of the datasheet I've seen have shown an equivalent circuit diagram showing 2 diodes in series to explain the voltage drops. It's possible they are made of 2 (or more) PN junctions in series, but I'd think it far more likely that it is a single PN junction with a specific doping method to create the characteristics. I'm not going to claim to be a zener diode physics expert though. \$\endgroup\$
    – Phil C
    Commented Apr 22, 2018 at 4:02
  • \$\begingroup\$ Can't find much information, only a few seem to give Vf, BZX79C 0.9V@10mA (cf. 1n4148=0.7), 1N593X 1.2V@200mA (cf 1n4001=0.8V). Nothing about it varying with Vz. Guess its just an effect of the doping levels. \$\endgroup\$
    – Henry Crun
    Commented Apr 22, 2018 at 10:17
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A series of diodes will create a series of stepped voltage drops. The voltage drops will not change proportionally with current as the voltage drops across resistors would. That means that there will be less speed variation with changing load. That is an advantage over series resistors. Using diodes in this way still causes heat and inefficiency similarly to series resistors. If the motor requires a maximum of 2 amps and the voltage drop is 1 volt per diode, six diodes would dissipate 12 watts and the motor would dissipate another two watts or so in winding resistance, friction etc. leaving about 10 watts converted to mechanical energy with an efficiency of about 42%. That is a pretty low efficiency, but only 14 watts dissipated as heat.

The same technique could be used with a larger motor, but the energy lost as heat would increase proportionally. With a 1/2 Hp motor, a similar setup would use 888 watts and dissipate more than 500 watts of power as heat to provide 373 watts of mechanical power.

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  • \$\begingroup\$ Thanks for your excellent response Charles, I marked Phil as correct as the diagram and explanation helped with the issue, though yours gave great insight on what would happen as it scaled up. \$\endgroup\$
    – Alan
    Commented Apr 23, 2018 at 20:43

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