Osilloscope to check encoder

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buffalonymann

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I'm trying to examine the pulse of a rotary encoder with my scope - BEI encoder with A+ A- B+ B- Z+ Z- PWR and Gnd . I connect my probe to A+ and the clip on GND but am not seeing the 5 volt pulse. Anybody help me with this?
 
Are you connecting A- to the supply common and reading from A- to A+. You would also need to connect a DC +5 volt supply to the supply terminal (I think its D for + and F for common but I don't know what model you are using).
 
BEI xh25d-ss-2500-abzc-28v/v-em

The encoder is connected to a MINT Drive and also they are taking a signal from GND and A+ to a red lion controller to display linear ft. That's why I am on GND and A+
 
Are you connecting A- to the supply common and reading from A- to A+. You would also need to connect a DC +5 volt supply to the supply terminal (I think its D for + and F for common but I don't know what model you are using).

The MINT drive is powering the encoder with 5 VDC - I removed the 2 signal wires from the red lion controller and attached my scope to those, but do not see the pulses
 
181115-1526 EST

buffalonymann:

A quick look at the datasheet and I conclude that power input is between common and V+. This would also mean that the output signals are referened to common. All electronics are probably floating from case which is the ground pin.

With 5 V between V+ and common you should see a voltage between 0 and 5 on all of the signal lines relative to common.

A is one channel B is another and these are phase shifted relative to each other. Call it a 90 degree shift. The Z channel is probably a once per revolution, and likely in phase with A channel.

A+ and A- are a line driver output. When A+ is high, meaning towards 5 V, A- is low, (meaning toward 0 V, not negative). These switch on the next half cycle. The Bs switch on a 1/4 cycle relative to As.

Is there input current flow to the 5+ terminal? Is it large enough to indicate that the LED is consuming power? With 5 V power applied what are the various signal voltages relative to common with the encoder in a fixed position? Does very slow motion of the shaft produce any change in signal voltages?

.
 
LD - Sinking input required

LD - Sinking input required

Try a lower impedance termination in parallel with your scope input.
 
181115-2327 EST

ELA:

The connection labeling implies differential line drivers. Thus, each output can be considered a totem pole output between common and 5+. These would require no pull-up or pull-down resistors.

Some static voltage measurements will help us understand what may be happening.

.
 
181115-1526 EST

buffalonymann:

A quick look at the datasheet and I conclude that power input is between common and V+. This would also mean that the output signals are referened to common. All electronics are probably floating from case which is the ground pin.

With 5 V between V+ and common you should see a voltage between 0 and 5 on all of the signal lines relative to common.

A is one channel B is another and these are phase shifted relative to each other. Call it a 90 degree shift. The Z channel is probably a once per revolution, and likely in phase with A channel.

A+ and A- are a line driver output. When A+ is high, meaning towards 5 V, A- is low, (meaning toward 0 V, not negative). These switch on the next half cycle. The Bs switch on a 1/4 cycle relative to As.

Is there input current flow to the 5+ terminal? Is it large enough to indicate that the LED is consuming power? With 5 V power applied what are the various signal voltages relative to common with the encoder in a fixed position? Does very slow motion of the shaft produce any change in signal voltages?

.

Gar,

My understanding is that when channel A+ is energized, it is 5VDC with respect to GND or common and A- is -5VDC with respect to comm. Am I mistaken?

I have little experience using a scope, so I'm not certain what settings I should be using. Trigger default is rising, (tried rising pulse but no difference)
 
181115-1526 EST

buffalonymann:

A quick look at the datasheet and I conclude that power input is between common and V+. This would also mean that the output signals are referened to common. All electronics are probably floating from case which is the ground pin.

With 5 V between V+ and common you should see a voltage between 0 and 5 on all of the signal lines relative to common.

A is one channel B is another and these are phase shifted relative to each other. Call it a 90 degree shift. The Z channel is probably a once per revolution, and likely in phase with A channel.

A+ and A- are a line driver output. When A+ is high, meaning towards 5 V, A- is low, (meaning toward 0 V, not negative). These switch on the next half cycle. The Bs switch on a 1/4 cycle relative to As.

Is there input current flow to the 5+ terminal? Is it large enough to indicate that the LED is consuming power? With 5 V power applied what are the various signal voltages relative to common with the encoder in a fixed position? Does very slow motion of the shaft produce any change in signal voltages?

.

When I put my fluke across A+ and Comm I see a steady 2.9 volts. I run the speed of the encoder up and down to almost nothing and do not see any fluctuations in the voltage

I could take snapshots of what I see on the screen and attach if there is a way to do it
 
181115-0850 EST

buffalonymann:

I would not expect +/-5 output. You have a 5 V supply between common and +5. Unless there was something in the encoder to produce -5 there would be no -5 output.

If you totally stop the encoder, absolutely no motion, then what are the voltages from each of A+, A-. B+, B-, Z+, Z-, to common.

.
 
181116-0906 EST

buffalonymann:

In a totem pole output you have two output transistors. One connects to the + supply, and the other to common. When one is on the other is off. This provides a low driving point impedance in both output states.

Your 2.9 V looks like an average that would occur during rotation.

.
 
181115-0850 EST

buffalonymann:

I would not expect +/-5 output. You have a 5 V supply between common and +5. Unless there was something in the encoder to produce -5 there would be no -5 output.

If you totally stop the encoder, absolutely no motion, then what are the voltages from each of A+, A-. B+, B-, Z+, Z-, to common.

.

The wiring is not color coded - need a few minutes to pull the connector from the servo drive and wring out the pairs
 
181115-0850 EST

buffalonymann:

I would not expect +/-5 output. You have a 5 V supply between common and +5. Unless there was something in the encoder to produce -5 there would be no -5 output.

If you totally stop the encoder, absolutely no motion, then what are the voltages from each of A+, A-. B+, B-, Z+, Z-, to common.

.

Encoder stopped I have 5v on A+ A- and 2.5 on B+ B- they're using only channels A and B as a master encoder to a servo drive
 
181116-1035 EST

buffalonymann:

Were your measurements relative to common?

Relative to common, when fully stopped and stable in position, I would expect A+ relative to common to be somewhere near +5 or near 0. Near might mean +3.5 to +5, or 0 to 1.5+ .

Same would be true for B or Z. The receiving detector will look at the difference between A+ and A-. If slightly above a 0 difference that is one binary state, and if slightly below 0 difference that is the opposite binary state. This is called a balanced line system and has good immunity to common mode noise.

If both A+ and A- are +5 relative to common when the encoder is stable, then there is an internal problem. And with B+ and B- both at 2.5 V relative to common there is a different kind of problem.

If you monitor A+ relative to common and rotate the encoder is there any voltage change. I might expect somewhere near 2.5 V. Same check on A- relative to common.

I will be gone until late afternoon. There should be others that can help you.

.
 
I'm trying to examine the pulse of a rotary encoder with my scope - BEI encoder with A+ A- B+ B- Z+ Z- PWR and Gnd . I connect my probe to A+ and the clip on GND but am not seeing the 5 volt pulse. Anybody help me with this?
Bear in mind that the clip is probably grounded so whatever terminal you clip it to.........
 
181119-0807 EST

Logically it is a complimentary signal, but that does not mean it is negative. When one goes high the other goes low.

Basically the individual lines are TTL totem pole outputs. Thus, somewhere between 0 and 5 V.

.
 
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