What Size encoder is the perfect fit?

[slimneill]

What Size encoder is the perfect fit?
1. The Viper 200F Driver has a step rate of 300Khz. And a Max encoder frequency of 625Khz.
Isn't your maximum speed limited by the lower of either the input and output count rates? In this case the 300Khz output?
2. When using quadrature, is the 4x multiplier a result of counting 4 times as many points on the encoder wheel, or is it done in the driver after receiving the single count form the wheel? I.E. will quadrature multiply a 1000 line encoder and make it appear to be a 4000 line encoder (not in quadrature) and reduce the max encoder/motor speed by a factor of 4?
3. I also read that Mach3 has a max 100Khz output, and that parallel port is somewhere in the neighbourhood of 40Khz? Will something else be the bottle neck in the chain and reduce my motor speeds if I go for a higher line count encoder?
Am I reading too much into these different factors in determining the corrector line count for an encoder?

My aim is to have machine accuracy to 0.5/1000". I read that you want to have your encoder accuracy to be 5x more precise then your machine accuracy. I.E. 0.1/1000"
The encoder will be mounted directly to the servo motor shaft, there is a 5:1 pulley gear reduction to the ball screw which was a pitch of 1.847 inch/rev. The motor has a max RPM of 3000RPM.
The fastest speed i would want this machine to go would be 600ipm, doing the math that means the motor will only ever reach a max speed of 1624 RPM.

1000 Cycle/Rev Encoder in Quadrature
1000 x 4 =4000 pulse/rev (assuming 4x as many pulses, which could be the wrong assumption)
4000 pulse/rev x 1624 rpm / 60 sec/min = 108,266 pulse/sec
=108.26 KHz input to the driver

2000 Cycle/Rev Encoder in Quadrature
2000 x 4 =8000 pulse/rev (assuming 4x as many pulses, which could be the wrong assumption)
8000 pulse/rev x 1624 rpm / 60 sec/min = 216,533 pulse/sec
=216 KHz input to the driver

2500 Cycle/Rev Encoder in Quadrature
2500 x 4 =10,000 pulse/rev (assuming 4x as many pulses, which could be the wrong assumption)
10,000 pulse/rev x 1624 rpm / 60 sec/min = 270,666 pulse/sec
=270 KHz input to the driver

I was looking at the E6 encoder from US-Digital and they have different wheel sizes for different applications, US Digital | Products » E6 Optical Kit Encoder has anyone used these encoders? Are they a good fit? Do they work well?

Encoder Resolution
Ball Screw Pitch = 1.847 inch/rev with 5:1 reduction between motor and ball screw: 1.847 / 5 = 0.3694"
1 Rev or motor = 0.3694" of ball screw travel

1000 Cycle/Rev Encoder in Quadrature
1000 x 4 =4000 pulse/rev (assumes 4x pulses are physical and not created electronically in the processor)
0.3694"/4000 pulse = 0.0000923 inch/pulse

2000 Cycle/Rev Encoder in Quadrature
2000 x 4 =8000 pulse/rev (assumes 4x pulses are physical and not created electronically in the processor)
0.3694"/8000 pulse = 0.000046175 inch/pulse

2500 Cycle/Rev Encoder in Quadrature
2500 x 4 =10000 pulse/rev (assumes 4x pulses are physical and not created electronically in the processor)
0.3694"/10000 pulse = 0.00003694 inch/pulse

Does my Math make sense? Am i off by a factor of 4?
Am I splitting hairs, does it not make that much of a difference?
Thanks for all of your expert help, it is GREATLY appreciated.

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[Larken]

What Size encoder is the perfect fit?
1. The Viper 200F Driver has a step rate of 300Khz. And a Max encoder frequency of 625Khz.

To run at 625Khz, you need to have a very noise free twisted pair setup.


Isn't your maximum speed limited by the lower of either the input and output count rates? In this case the 300Khz output?

Depends on the step multipler. With a 10X step multiplier , 65Khz step would give 650Khz encoder.


2. When using quadrature, is the 4x multiplier a result of counting 4 times as many points on the encoder wheel, or is it done in the driver after receiving the single count form the wheel? I.E. will quadrature multiply a 1000 line encoder and make it appear to be a 4000 line encoder (not in quadrature) and reduce the max encoder/motor speed by a factor of 4?

I recommend a 4X step multipler so 4x quad cnts are just used to increase positioning accurace, not as resolution


3. I also read that Mach3 has a max 100Khz output, and that parallel port is somewhere in the neighbourhood of 40Khz? Will something else be the bottle neck in the chain and reduce my motor speeds if I go for a higher line count encoder?
Am I reading too much into these different factors in determining the corrector line count for an encoder?

The PP can easily run 100 Khz. With a good terminated BOB. (All BOB's aren't made the same )


My aim is to have machine accuracy to 0.5/1000". I read that you want to have your encoder accuracy to be 5x more precise then your machine accuracy. I.E. 0.1/1000"
The encoder will be mounted directly to the servo motor shaft, there is a 5:1 pulley gear reduction to the ball screw which was a pitch of 1.847 inch/rev. The motor has a max RPM of 3000RPM.
The fastest speed i would want this machine to go would be 600ipm, doing the math that means the motor will only ever reach a max speed of 1624 RPM.

Thats fast for a 5tpi screw.

The rest... ive got a headache !
Larry K

[Al_The_Man]

W

2. When using quadrature, is the 4x multiplier a result of counting 4 times as many points on the encoder wheel, or is it done in the driver after receiving the single count form the wheel? I.E. will quadrature multiply a 1000 line encoder and make it appear to be a 4000 line encoder (not in quadrature) and reduce the max encoder/motor speed by a factor of 4?
.

Very simple, a quadrature encoder, so named because the two basic pulses are 90° apart (in quadrature).
Whether the receiving end design decides to chose to use the basic encoder counts of 1 leading edge (x1) or two leading edges ( X2) or both leading and trailing edges of both pulses ( x4) is up to the designer of the drive or controller.
For example, there are some motion cards out there which offer the option of all three resolutions.
IOW, the multiplier increases the encoder resolution.
Al.

[Larken]

You didn't mention what type of machine you have. Is it a machineing center retrofit , a cnc router or a desktop hobby mill.

If you are using a 5-1 belt reduction, you probably only need a 500 line differential encoder. This is also the best deal at western digital, since the 500 line is 10$ cheaper than the 1000 line.

Larry K

[slimneill]

It's a large 4 x 8 CNC router for wood working, aluminum and light steel.

High accuracy is the goal, not high speed.

I already ordered a 2000 line encoder from US-Digital. Will see if it works. I

have to order 3 more for the other motors eventually.

[Al_The_Man]

It's a large 4 x 8 CNC router for wood working, aluminum and light steel.
High accuracy is the goal, not high speed.
I already ordered a 2000 line encoder from US-Digital.

2000 Cycle/Rev Encoder in Quadrature
2000 x 4 =8000 pulse/rev (assumes 4x pulses are physical and not created electronically in the processor)
0.3694"/8000 pulse = 0.000046175 inch/pulse
.

0.000046175" will be your least input increment, do you need that resolution for a wood working router?
Al.