Diagram of Testing Circuit

So I am tinkering with making a circuit that will allow a motor to generate power to charge a battery. I have this, what I presumed to be an, AC induction motor I salvaged from an old Christmas decoration.

After removing the motor from its casing, I have learned it is a synchronous motor.

This one with the ratings of 120VAC 3.8W 4.2/5 RMP

It can output upwards of 200 Volts AC in short circuit (or DC using a rectifier bridge) at ~ 6.7mA. I could only get the amperage reading from a short circuit through the rectifier. Could be my $7 multimeter not doing well with AC or my ignorance on reading AC amperage.

Strangely enough (at least to me), no matter how I worked it: the amperage would stay at a consistent cap of ~ 6.7mA. In my tinkerings I figured out there is, at least almost, a straight line which shows that the resistance given to the circuit will output a maximum voltage I can get from the motor itself.

The diagram posted is a test circuit to gather data points on this.

I'm wondering if there is anyone with an idea as to what is causing this phenomena?

Here is a chart and graph of the voltages across the entire circuit (from either end of the rectifier bridge), given the different values of R1.

Graph of Data Points

Definitely some good answers. Not sure which is the best answer. I appreciate all of the input, and will select the best answer once I get back from work and have time to do some more testing, and take apart the motor to see what's really going on inside.

To clarify: the end game is to maximize on incoming voltage so I can reduce the voltage later in the circuit and bump up the amperage to charge a battery somewhat efficiently. Also to understand why there seems to be this constant 6.5mA coming from this motor.

I may just go back to my research and choose a best answer for now. If I run into anything interesting down the road, I'll make sure to post again.

  • \$\begingroup\$ The problem is the motor windings are too thin to be able to increase current when you decrease voltage. How about not decreasing voltage and use a transformer to knock it down to the voltage you want? Or a buck converter... \$\endgroup\$ Jun 19 '17 at 1:03
  • \$\begingroup\$ The idea is low amperage high voltage coming from the motor; which is then converted to low voltage high amperage later on in the circuit. \$\endgroup\$ Jun 19 '17 at 4:50
  • \$\begingroup\$ Not quite right. Read the answers below about maximum power transfer and then look up the theory (which is easy enough). You want maximum power - not maximum voltage. \$\endgroup\$
    – Transistor
    Jun 19 '17 at 6:51
  • \$\begingroup\$ If the voltage is increasing and the amperage is remaining constant, doesn't that mean the wattage is also increasing? I will read further, as I've been wrong before haha. \$\endgroup\$ Jun 20 '17 at 19:21
  • \$\begingroup\$ Re, "It can output upwards of 200 Volts AC in short circuit." I bet you meant to say, open circuit. \$\endgroup\$ Jun 22 '17 at 21:12

Great job on the experiment!

The motor that you are using as a generator has a high internal impedance largely due to the windings and the internal magnetics structure. You can think of this as a resistor inside of the motor that is in series with its output. Of course this is not a real resistor, just a way of modeling it.

You didn't mention it, but when you are experimenting consider that the load placed on your motor acting as the generator may cause the driving motor to speed up or slow down. This will of course affect the results of your experiment.

In electronics, we know that maximum power will be transferred to a load when the load impedance matches the source impedance. You may find it interesting to add a column to your chart that shows power in the load (= V 2/R) to see if you can find the point of maximum power transfer. You will need to extend your experiment with higher values of resistance most likely.

Once you have determined the maximum power you can obtain from your generator, you can then see if it is suitable to power your target device. If it does have enough power, then the most likely the solution will require a buck regulator to efficiently step down the higher voltage.

Keep up the good work.


Over the range you measured, the generator is largely acting like a current source.

To first approximation, the generator can be modeled as a voltage source in series with a resistance. The voltage is directly proportional to rotation speed, and the resistance is reasonably fixed.

You say you are getting 200 V open circuit voltage, and about 6.6 mA short circuit current. Assuming that the generator is still spinning at the same speed when shorted as when open, the internal resistance of the generator is (200 V)/(6.6 mA) = 30 kΩ. This value will be skewed if the generator actually slowed down when the output was shorted. Here is the simplified model of your generator and rectifier diodes:

If the above is correct, then you will get largely constant current for loads significantly less than 30 kΩ. At 30 kΩ, you should get half the short circuit current at half the open circuit voltage. That is the point where the generator produces maximum power. At significantly above 30 kΩ load, the generator will look largely like a voltage source of 200 V.

  • \$\begingroup\$ I actually mistyped when I said it pulls 6.7 mA openly. Only when I have a load such as a resistor in tow will it read 6.7 mA (6.8 now that I've removed the motor from it's casing with mains lead wire). I also measure the resistance of the motor itself: which is 2.25KΩ. \$\endgroup\$ Jun 19 '17 at 4:48

An induction motor would not generate very much voltage into a rectifier circuit. The rotor might have a small amount of residual magnetisim to allow it to generate a little voltage acting as a permanent-magnet synchronous generator.

If the motor is generating more than 50 volts, it may be a permanent-magnet synchronous motor like a clock or timer motor. It could also be a permanent-magnet DC motor with a commutator, but that would be unusual for a small motor at that voltage level.

When using a salvaged motor, it is very helpful to find all of the information marked on the motor and on the product from which it was removed. If the product contains other electrical components, that are connected to the motor, it is important to have those components and be able to re-connect them after they and the motor are removed. It is also helpful to know about any other use of electrical power in the product.

Before attempting to use a motor as a generator, the motor should be tested as a motor. It is best to test it as it was originally used. Determine the voltage, current, power and speed with the original load and with no load. Measure the DC resistance.

Detailed pictures and dimensions can be helpful.


We learned in the motor class at UCLA that every motor is also a generator. Syncronous motors will generate and buck the power draw according to the phase relationship between the applied voltage and the motor position when it is under load. When the load is negative (somebody is turning the crank and trying to make the motor go faster), the power draw becomes negative. That is how a synchronous motor becomes a generator. You regulate the output by regulating the mechanical power applied to the shaft.

The whole thing is an exercise similar to financial book keeping: account for all of the energy.

I don't think Tesla had DC in mind when he invented AC motors.


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