[keen]

Hi Guys. I recently had my oldest 1100 spindle operation fault out, and so did some study, then a low cost repair.

Thought I would share the journey in this video. Please post any information you have on the subject.

Cheers Cliff

[Fenichel]

Using a higher-wattage resistor than the original will be fine.

Here is a thought experiment: Leaving aside the weird things that happen at high frequencies, it's not misleading to imagine that one 75-ohm resistor is made of 1 foot of wire whose resistance is 75 ohms/foot, while another is made of 3 feet of wire whose resistance is 25 ohms/foot. At a given applied voltage, they'll carry the same current, and draw the same I^2*R power, dissipated as heat. The power will be dissipated over a bigger area in the second case, assuming that the longer wire is not just cramped into a tight coil. The pertinent failure mode is a matter of inability to get rid of heat as fast as it's produced, so the second resistor has a better chance of surviving a given intensity of service.

Does this help?

[TurboStep]

Please post any information you have on the subject.

Hi Cliff
I assume when you said you got "a zero reading" across the resistor that you meant "no reading" rather than zero ohms?
Although the RS resistor is rated at 300 W this is only when mounted on a "reference" heatsink of 5870cm^2. You have the VFD and resistor mounted on the smaller Tormach supplied sub-plate so you will actually have a lower power capability – but the original resistor was also operating under the same conditions. As you're probably not continuously starting and stopping the spindle the overall power loss, averaged over time, will likely be much less. I'd be more concerned about only mounting the resistor just along one side, the contact between the mounting face of the resistor and the plate may be reduced and the thermal conductivity will likely be limited.
That said, probably the most important parameter in this case is not the average power rating but the short-term overload capability. All (mostl) the kinetic energy in the motor and spindle will be transferred to the resistor within around 2 seconds. I can't realistically estimate how much energy this will be, but I would expect the power dissipated during this short time to be considerably higher than the steady state rating of 300 resp. 200 W. For around 2 seconds your new resistor will tolerate around 8x the steady state rating which corresponds to around 2.4kW. I suspect this would be ok.
Step

[keen]

Using a higher-wattage resistor than the original will be fine.

Here is a thought experiment: Leaving aside the weird things that happen at high frequencies, it's not misleading to imagine that one 75-ohm resistor is made of 1 foot of wire whose resistance is 75 ohms/foot, while another is made of 3 feet of wire whose resistance is 25 ohms/foot. At a given applied voltage, they'll carry the same current, and draw the same I^2*R power, dissipated as heat. The power will be dissipated over a bigger area in the second case, assuming that the longer wire is not just cramped into a tight coil. The pertinent failure mode is a matter of inability to get rid of heat as fast as it's produced, so the second resistor has a better chance of surviving a given intensity of service.

Does this help?

Yes, thanks for your input! Cliff

[keen]

Hi Cliff
I assume when you said you got "a zero reading" across the resistor that you meant "no reading" rather than zero ohms?
Although the RS resistor is rated at 300 W this is only when mounted on a "reference" heatsink of 5870cm^2. You have the VFD and resistor mounted on the smaller Tormach supplied sub-plate so you will actually have a lower power capability – but the original resistor was also operating under the same conditions. As you're probably not continuously starting and stopping the spindle the overall power loss, averaged over time, will likely be much less. I'd be more concerned about only mounting the resistor just along one side, the contact between the mounting face of the resistor and the plate may be reduced and the thermal conductivity will likely be limited.
That said, probably the most important parameter in this case is not the average power rating but the short-term overload capability. All (mostl) the kinetic energy in the motor and spindle will be transferred to the resistor within around 2 seconds. I can't realistically estimate how much energy this will be, but I would expect the power dissipated during this short time to be considerably higher than the steady state rating of 300 resp. 200 W. For around 2 seconds your new resistor will tolerate around 8x the steady state rating which corresponds to around 2.4kW. I suspect this would be ok.
Step

Hi Step. Thanks for your input, always appreciated! Yes I meant no reading. I love when someone is more pedantic than me!
(but I understand your points )
Cliff