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I am building a homemade ebike using a treadmill motor. I know this motor is not the best for ebike application, but I am trying to work with what I have and see how far I can take it. I am wondering what battery capacity I need to maximize the motor's efficiency.

The motor is a permanent magnet brushed DC. It it is rated for 1977 Watts at 130 VDC (15.2 A) and 1119 W cont. duty at 95 VDC (11.7 A). Obviously, for ebike application I am not giving it 130 V but 60 VDC. Its velocity constant is 65.

I have built a substantial gear reduction to improve torque (8.3 times reduction) and still propel the bike to a decent speed.

My question is, since this motor is capable of producing 1977 Watts of power, that theoretically means to get max power at 60 V I will need around 30 Amperes from the battery. However would this be too much for the motor as it is designed for 15.2 amps? The voltage is half of what it is designed to use so can I go up to 30 Amperes without frying the coils? I am new to electricity so I don't really know if excessive amps will fry the coils if the power output is still the same.

If not then limiting the battery to 15 Amps is totally fine, I am trying not to destroy my motor. Thanks

Davide Andrea
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MechEboss
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    _”30 Ah without frying the coils”_ Battery capacity has practically zero impact on your motor, only range. – winny Dec 05 '22 at 17:19
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    You are confusing current (A) with capacity (Ah). – Justme Dec 05 '22 at 17:22
  • Sorry I should have been more clear. What I want to know is if the motor is under full load connected to a 30 Ah battery, how much current will flow through the coils – MechEboss Dec 05 '22 at 17:32
  • Pondering the current to aim for is a good idea, but unless running full throttle, that's the task of the motor controller to, well, control, no matter the capacity of the battery. `can I go up to 30 Amperes without frying the coils?` No. Max current & torque are largely independent of speed&voltage, (more than) halving the voltage will do the same to power. Oddly, the numbers presented seem to indicate about 100% efficiency: your mileage *will* vary. – greybeard Dec 05 '22 at 17:58
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    @AhmadInsanally You still are asking the same thing. Ah does not matter. Ah is a measure of capacity. No matter if 30 Ah or 3000 Ah capacity, the current to your battery is equal. How much current (A) there is depends on voltage (V). – Justme Dec 05 '22 at 18:06
  • @Justme got it thank you for clarifying – MechEboss Dec 05 '22 at 18:12
  • @greybeard yes these are just ballpark numbers, since I am still building the bike I have no idea what the actual current draws will be or how effective the gear ratio is. Since I am aiming to keep this bike homemade and low tech, would a rheostat throttle work as substitute for a motor controller (assuming it is rated to the safe current for the motor)? – MechEboss Dec 05 '22 at 18:18
  • > What I want to know is if the motor is under full load connected to a 30 Ah battery, how much current will flow through the coils -- It depends on what you mean by "full load". Stalled? Top speed unloaded? Top speed on level ground? Top speed uphill? What is the rolling resistance of the bike? How steep is the hill? How much do you weigh? As you see, it's not answerable, neither by us nor by you. You'll have to try. – Davide Andrea Dec 05 '22 at 22:16

2 Answers2

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At 60V the motor will produce less than 1/4th the power as it would at 130V. That’s still more than 400W, which is more than the Class 1 limit of 250W.

How to figure? The power output is going to be roughly the ratio of the voltages squared times the ‘known’ max power output at the rated voltage:

  • (60V/130V)^2 * 1977W = 421W

You will not 'fry' the motor by reducing the voltage. Its max (lock-rotor) torque will be reduced (limited by coil resistance) and its max power output will also be reduced.

At any rate the motor should be able to handle that operating condition easily as long as it has adequate cooling.

More about motor power here: https://www.electrically4u.com/voltage-and-power-equation-of-a-dc-motor/

hacktastical
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  • At half the voltage, a motor will produce slightly less than half the power: at half speed, but it can deliver about the same torque. – greybeard Dec 05 '22 at 22:22
  • Assume the motor has an approximately constant ohmic resistance. That determines the amount of current that can be forced through it. – hacktastical Dec 05 '22 at 23:47
  • `ohmic resistance [determines] the amount of current that can be forced through it.` at stand-still. In regular operation, back-EMF "gets" most of the voltage. Optimum efficiency is at the geometric mean between that current and running without load. – greybeard Dec 05 '22 at 23:54
  • Yes, I was speaking to lock-rotor condition. As you probably know, the **max power output** (not efficiency -OP isn't asking about that) is when the resistive IR drop voltage is about equal to the back-EMF voltage. That is, an impedance *Z* approx. equal to *2R*. Halve the voltage, you get a max-power point at half the RPM, but *Z* = *2R* still holds. Therefore current is also halved, and power is one-fourth. So, max power out is in proportion to the square of the voltage, not linear as for max torque or max RPM. – hacktastical Dec 06 '22 at 23:54
  • @hacktastical thank you so much for the link and for the theoretical insights. I am going to run through the calculations and compare with practical testing to measure actual RPMS under load, torque produced/current draws etc. And yes cooling is definitely something I am considering. it will have a air scoop vent facing the front and integrated fan, as well as heat sensor which opens the circuit when it gets too hot. Hopefully these measures will prolong motor life – MechEboss Dec 07 '22 at 04:25
  • `not efficiency -OP isn't asking about that`? `battery` - the OP is not about max. power. As you probably know, max. power overheats most motors in very short time. – greybeard Dec 07 '22 at 06:54
  • (Then again, with *electric bike*, I have been thinking *permanent magnet*. With a treadmill motor, I'd expect a shunt field coil. Probably not with twice the field necessary for nominal torque/power: Torque *will* be reduced at not even half the voltage. Even with sufficient field, the torque at max. efficiency will be below .7 that at nominal voltage.) – greybeard Dec 07 '22 at 08:12
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If it produces 1977 W at 130 V and 1119 W at 90 V, then there's no way it's going to produce 1977 W at only 60 V. So that's not even an issue.

Simon B
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  • That's not really how it works. The motor might very well produce 2000 Watts at 60 V if the load demands it and the battery can supply enough amps. However, it won't do it for long without burning up. – user57037 Dec 07 '22 at 02:58
  • @mkeith yea I am going to be very careful with current flow. Fuses in every part of the circuit and controller which limits the maximum amps to what the motor would normally use when it was powered with rectified mains voltage. I think this motor will be limited to max draw of 15 amps which is still more than enough power for this application. – MechEboss Dec 07 '22 at 04:11