Is it possible to take a regular dc motor, use it to spin some strong magnets over some coils, and get a higher voltage than what's being used to drive the motor? I was thinking; if you're able to create a series of coils around the magnet, wouldn't the output voltage eventually surpass the input voltage?
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\$\begingroup\$ See dynamotor. Very old technology. You can get any voltage you want, but the total power output will always be less than the power input. \$\endgroup\$– Dave TweedCommented Jun 4, 2019 at 23:27
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4 Answers
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Absolutely! That is the basis for motor-generators, which, before power semiconductors, were one of the few devices able to change DC voltage with comparatively little loss.
Of course, stepping up voltage comes with a concomitant reduction of current, so you don't gain any power. E.G. 12 volts at 30 amps is equivalent to 120 volts at 3 amps, 360 watts.
But you don't actually break even. If the motor has an efficiency of 95%, and the generator has the same efficiency (those are high figures), the combined efficiency is 90.25%, so with 360 watts in, you'd get only ~325 watts out. TANSTAAFL, as Heinlein summarized the first law of thermodynamics.
Or, maybe not! It's possible to get instantaneous power far exceeding the input power due to a flywheel effect. The motor and generator armatures store kinetic energy as they spin up, so that even with the motor disconnected, energy can be withdrawn until rotation ceases. Sadly, though, you still get less out than what you put in while spinning things up. Heinlein got it right.
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\$\begingroup\$ So just to clarify, even if I had a dc motor with strong magnets attached to the shaft, and maybe 15 coils in series around the magnets, the output would still be lower than the input? \$\endgroup\$ Commented Jun 4, 2019 at 23:35
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\$\begingroup\$ Again, you asked about voltage, and you can increase it. You just don't get more power. \$\endgroup\$ Commented Jun 4, 2019 at 23:37
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\$\begingroup\$ Ah ok, I think I understand now \$\endgroup\$ Commented Jun 4, 2019 at 23:39
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\$\begingroup\$ Motor-generator sets are still used in some circumstances to change voltage, frequency, provide isolation or the ability to ride out surges. E.g. acim.nidec.com/generators/kato-engineering/products/… \$\endgroup\$ Commented Jun 4, 2019 at 23:48
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\$\begingroup\$ And in Tokamaks and magnet labs where a flywheel stores power for each shot. The F Bitter Magnet Lab at MIT has a 40 tonne flywheel on each motor generator of its 25 Tesla magnets: henrykolm.weebly.com/mit-magnetic-lab-1961-82.html [Running the flywheels up to speed occasionally dimmed lights throughout Cambridge.] \$\endgroup\$ Commented Jun 5, 2019 at 16:01
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You can easily get more output voltage from one motor driving a generator, the same as using an electronic voltage boosting circuit or using a transformer for AC. But the power cannot increase then the output current will be a little less than the input current.
Are you thinking of Over Unity to feed the higher output voltage to power the original motor forever? It don't work.
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\$\begingroup\$ Well basically I'm thinking about creating a series of coils, maybe about 500 turns each, and testing how many of those are needed to get more voltage than what the input is. In theory it sounds plausible? More coils = more electricity? \$\endgroup\$ Commented Jun 4, 2019 at 23:29
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\$\begingroup\$ @CouchMango Don't confuse "voltage", "power", and "electricity". You could easily come up with an arrangement that provides a higher output voltage than the input voltage, but the output power must be less than the input power. The phrase "more electricity" really has no meaning. \$\endgroup\$ Commented Jun 4, 2019 at 23:34
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If using two different DC brush motors, one to drive the other. The output V*I=P, power out is still less than the input ( due to losses). But it works as a DC transformer with the sum of both motor impedances in series.
Most motors are rated in V/kRPM, Volts per 1000 RPM for no load, thus the sum of series impedance affects the output voltage by the load resistance.
But with no load a 10V/kRPM motor driving a different motor rate 50V/kRPM will produce 5x the output voltage again at no load yet draw ~10% of rated input current at a rated voltage just for the armature field coupling current and nothing out much like small AC transformers.
There will be commutation current spikes on the output, so a suitable storage cap is necessary to integrate the current.
This will likely give more satisfaction and better results than any DIY generator with massive rare earth magnets due to all the assembly issues of balance, gap leakage, winding/pole design, brush-armature design, etc,etc.
Don't exceed RPM of either motor rating unless you want to break something or reverse the motor with a polar cap on output.
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You will run into some mechanical problems with your arrangement mostly.
However, if you had a DC machine with a field winding instead of permanent magnet excitation, you can control the field strength.
- The stronger the field, the lower the speed for a given voltage
- The stronger the field, the higher the voltage for a given speed
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\$\begingroup\$ It does work, but not with the mechanical arrangement you think of. \$\endgroup\$– JankaCommented Jun 4, 2019 at 23:31