Thanks so much for the detailed description in this post and elsewhere.Richards wrote: ↑Tue Feb 23, 2021 3:50 pm It's always best to ask. The manual that I have from Teknic also shows a 1K resistor, so I thought that someone just had a typo when I saw 10K.
I assume that the manual was written for all users of ClearPath servos, so in actuality, it becomes the responsibility of the user to run some tests such that any resistors used keep the current within the specs of whatever he is using. Page 7 of the Acorn user's manual tells us that the Acorn can handle up to 50mA. A 1K resistor would allow 24mA at 24V, so a 1K resistor can be used with the Acorn. A 10K resistor would allow 2.4mA which is below the average current of the Acorn's open collector spec, so that also fits within the specs. As noted below, the ClearPath can still pull more than 2.4mA when the pull-up resistor is mounted in parallel.
The placement of the resistor, as shown in the Teknic manual, creates a parallel circuit. When resistors vary greatly in value, most of the current will flow through the path with least resistance. If there is less resistance through the ClearPath circuitry than there is through the pull-up resistor, then the pull-up resistor acts more like an open circuit than a resistor. If there is less resistance through the pull-up resistor than the ClearPath circuitry, then the ClearPath might not get enough current to function; however, Teknic lists 1K, so they know that a 1K resistor will work in parallel with their ClearPath motors.
The real question is what actually happens with a 1K resistor and a 10K resistor? An oscilloscope shows the answer. The higher the resistance, the greater the "shark fin" or "shark tooth". If I saw a pronounced "shark fin" on my 'scope with a particular length of cable and a 10K resistor, I would try using a 4.7K resistor, or a 3.3K, or a 2.4K. We're looking for results. In a lab setting, we might be able to say that a 10K resistor is best. In the shop, a 10K resistor might be the best or another value might produce a more ideal waveform.
Thanks,
Clay