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EP4 frequency
Hi,
I am new to this site and new to encoders so bear with me. I am using the EP4-100-079-D-H-T-B single-ended encoder on a 200 RPM motor, and I am not getting the frequencies I expected. I hook one of the output channels to an oscilloscope and get a clear square wave except the frequency is about 1.5 kHz when I am running the motor at about 20 RPM when I excepted the frequency to be about 33 Hz. The encoder is hooked up to a 5V supply as specified. Am I getting the correct output or is something causing the frequency to be high? If that is the excepted frequency for 20 RPM, why is so much more than I excepted? Thank you for any help. Ryan |
Re: EP4 frequency
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Most encoders don't create one pulse per revolution, but rather offer a pulse every portion of a revolution, sometimes 128 or 256. This is referred to as Pulses Per Revolution (PPR). Another unit used is the CPR, or Cycles Per Revolution - almost always four pulses. That's why you're seeing so many Hz. Just divide it by your encoder's PPR to get the encoder disc's RPS (Revolutions Per Second). You can then obviously multiply it by 60 to get the RPM. The E4P encoder comes with PPRs between 400 and 1400, and CPRs between 100 and 360. I don't know your particular model's spec. |
Re: EP4 frequency
It is 100 CPR (400 PPR) I believe. But if I am only running the motor at about 20 RPM, so the expected frequency should be about 33 Hz for just the one channel I believe.
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Re: EP4 frequency
400 Pulses/rev * 200 Rev/minute = 8000 pulses/min, divide by 60 (seconds per minute) and you get 1333 pulses per second (Hz).
Make sense now? |
Re: EP4 frequency
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1) he said 20 rpm, not 200 2) he said he put the scope on one channel only. so that would be 200 PPR on one channel, not 400. |
Re: EP4 frequency
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If those numbers are different, it will be easy to sub in the correct values and see what should happen. |
Re: EP4 frequency
The "100" in the part number EP4-100-079-D-H-T-B means 100 cycles per rev. In this context, a "cycle" means: - a rising edge on Channel A, followed by - a rising edge on Channel B, followed by - a falling edge on Channel A, followed by - a falling edge on Channel B If only one channel is being displayed on the scope, there will be 100 rising and falling edges per rev, which is 100 cycles per rev. So the math looks like this: 20 rev/min * (1/60) min/sec * 100 cycles/rev = 33 Hz ... which is what the OP calculated in post #1. |
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