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26-12-2014 #1
The formula you need to recall is power=voltage*current. I said in the previous post that the output power fro the motor driver must be roughly equal to the input power. In reality there are losses in the driver, but if we assume they're small then we can equate the input and output power to get ..
Now do you see why the input current, ., is lower than the output?
Here's an example of the same concept with a different stepper driver - in this case a 2m542 driver with a 3Nm motor from CNC4you connected to a lab PSU:
Current is on the left, voltage on the right.
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29-12-2014 #2
Thanks for taking the time to wire up a stepper & drive to a lab power supply Jonathan. I thought that the Power In = Power Out (V*I[in] = V*I[out]) only applied to transformers and voltage doubler circuits, given the stepper is acting as a chopper circuit, and no voltage change is taking place [albeit the stepper coil is an inductor].
separate AM882 question ->
seems you can get 3 variants of the AM882 drive, the AM882 (which has a datasheet from Leadshine), an AM882H, which seems to be above to take AC + DC and at a slightly higher voltage, and a AM882-DK (which is apparently a dedicated drive for an engraving machine).
The AM882H seems to have a fan as well, anyone got any experience of these (as I can't find an English language PDF) are they the same as the AM882, but with a heatsink mounted fan, and has a listed supply voltage of 18-70 VAC & 24-100VDC, with the supply terminals listed as "AC"
seems the AM882H can be had for ~£48
Any comments / experience would be very much appreciated.
Thanks again for your time,
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