Help me understand this photogate (Robotics)

Thread Starter

mcardoso

Joined May 19, 2020
152
Hi all,

I am working on getting an older industrial robot working again as a hobby. Getting close to being done but I need to get all the homing sensors working.

3 out of the 4 joints use an Omron EE-SV3 "Photomicrosensor (Transmissive)" with a fiberglass "code wheel" to trigger it. I'm pretty sure I have the EE-SV3-B model. Here is the link the the datasheet: https://omronfs.omron.com/en_US/ecb/products/pdf/en-ee_sv3.pdf

Here is a schematic of the system. On the far left is the servo interface module I need to work with. It uses sinking inputs with an input impedance of 7500 ohms. The numbered terminals just to the right of that are terminal blocks that I am wiring to and where I have connected the 24V power supply. To the right of those terminals is the signal cable that connects to the robot itself. Inside the robot, between the dotted lines in the middle of the page is a PCB. The PCB has 3 components that interact with this portion of the circuit. One is NF1, a ceramic noise suppression filter, and R4, a 1200 ohm current limiting resistor. Finally there is HL4 which is a LED that illuminates when the sensor is on to help troubleshoot robot operation (notice the direction of the LED). Further right from there, a cable harness travels through the entire robot arm and ends up near the end where the optical gate sensor is located on one of the pulley assemblies.


1600866766858.png

Can you help me understand how to hook up this sensor to my servo interface module? The big issue is that the 1756-M02AE servo interface module expects a sourced 24V to be applied to it and sink it to 0V, while the sensor here is set up to sink current to 0V. This means I cannot read this sensor.

A similar circuit was used on a different axis on this robot, but instead of the optical gate, it used an inductive proximity sensor. I was able to add a 2000 ohm pullup resistor at the servo module and pull the voltage high enough for the card to read it properly. I just don't understand these optical sensors quite well enough to say that this is the correct way to hook them up as well.

1600867481526.png

I need to get the input at the far left "Home0/1" above 17.0V for a logical 1 and below 8.5V for a logical 0. The second sketch (with the proximity sensor) gave me 1.9V OFF and 19.4V ON. Not shown is the 7500 ohm resistor internal to the card between terminals 16 and 14 on the left.

I'm sort of stuck keeping the robot wired how it is unless there is absolutely no way to add external components to make it work.

Thanks,

Mike
 

Thread Starter

mcardoso

Joined May 19, 2020
152
I'm going to reply to my own thread with a few thoughts. I'm trying to understand the datasheet for this sensor. Let me know if any of these conclusions are incorrect.

1) This sensor can be powered by any voltage as long an an appropriate current limiting resistor is used. In my case 24V is applied and a 1200 ohm resistor is added in series. This gives 20mA of forward current on the Emitter and a forward voltage at 25*C of 1.2V (used the second graph on page 2 of the PDF "Forward Current vs. Forward Voltage Characteristics (Typical)"). This is nicely within the operating specifications.

2) When the sensor is not blocked, the Collector will pass a certain amount of light current. At 20mA forward current (found in #1), the Collector will pass 6mA of light current (IL). I found this from the 3rd graph on page 2 of the PDF "Light Current vs. Forward Current Characteristics (Typical)". This makes me a little unsure since it doesn't seem to account for the aperture sizes of the different models. My model says it should pass between 0.5 and 14mA.

3) Now I get to the point where I am unsure of how to connect this. If I make the assumption I can use the same 2000 ohm external resistor as I did in the case of the proximity sensor then I end up with a voltage divider at the input card. The upper resistor is 2000 ohms and the lower is 7500 ohms. This gives 18.9V or so at the middle. The optical gate must sink enough current to drop this below 8.5V to bring the input to a logical 0. I'm not sure how to know if the 6mA of light current are sufficient to do so.
 

Tonyr1084

Joined Sep 24, 2015
5,039
Are you sure about this?
1600870787667.png
I don't see how the Opto can fire in this config.

[edit] What or where is the sensor in question? What is the Opto supposed to be switching on and off?
 

Thread Starter

mcardoso

Joined May 19, 2020
152
Are you sure about this?
View attachment 217817
I don't see how the Opto can fire in this config.

[edit] What or where is the sensor in question? What is the Opto supposed to be switching on and off?
That's a weird component, but I think it is just a high frequency power harmonics filter. For DC voltages, you can pretend it isn't there.

The sensor in question is the diagram at the far right of the page. It is an optical gate with a built in LED emitter and a phototransistor output. It is physically measuring a slotted disc (which passes between the arms of this gate) on the robot joints to give a repeatable "home" position that the robot can measure.

1600871581314.png

My goal is to hook up the sinking output of this sensor to my Allen Bradley 1756-M02AE Analog Servo Interface Module. https://literature.rockwellautomation.com/idc/groups/literature/documents/in/1756-in071_-en-p.pdf

You don't need to worry about this module, other than it has a 24V logic sinking input with 7500 ohm impedance. This module controls the robot servo drives, but needs this sensor hooked up to home the motors against.
 

Thread Starter

mcardoso

Joined May 19, 2020
152
I did some math (probably wrong) to calculate the largest external pullup resistor needed to give the logic low voltage (8.5V) assuming the sensor forward current of 6mA (discussed above). This yielded 3088 ohms. If I then turn OFF the sensor (0mA) I see that the new voltage divider (3088 ohms on top, 7500 ohms on the bottom) gives 17.0V in the High state.

This means I am sitting perfectly on top of my noise margins of the input - no good. I think the only way to solve this is to increase the light current of the sensor beyond 6mA.

If I again go back to the assumption that I want to use the 2000 ohm external resistor, I can solve for the needed light current from the sensor.

With the desired high voltage of 19V and low voltage of 6V and external resistor 2000 ohms, I need 8.2mA of light current out of this sensor.

No idea how to do that, lol
 

Tonyr1084

Joined Sep 24, 2015
5,039
The reason why I asked if this is correct is because the sensor LED is connected to ground on both pins. NF1 (or NFI) being grounded will not pass any current through the sensor to light the LED. Unless 24V is coming from somewhere (not shown) it isn't going to light.

Now, if I assume it's not there then I'm also assuming it's not grounded. If it were truly not there, not grounded, just a pass through jumper then you would have 19mA going through the sensor. You probably don't need that much current. But for the sake of being sure it has sufficient input - I wouldn't go below 15mA. I know we're talking milliamps, but that's where I'd go with it.

So, with the above assumptions, what I'm reading is pin 15 (1756-Mozae) is supposed to see a 24V source? The unknown resistor makes this difficult to understand. Pin 13 is grounded. Between pins 13 and 15 is the input for the 1756 as a signal. So what are you using to using to limit current to LED (HL4)? I'm guessing the unknown resistor is for that purpose? If so, the LED could be limiting your input at pin 15 to just around 4 volts (depending on Vf's of HL4 & EE-SV3-B). Remember, the resistor (?), HL4 and the PN Junction of the sensor transistor form a voltage divider.
 

Tonyr1084

Joined Sep 24, 2015
5,039
I did some math (probably wrong) to calculate the largest external pullup resistor needed to give the logic low voltage (8.5V) assuming the sensor forward current of 6mA (discussed above). This yielded 3088 ohms. If I then turn OFF the sensor (0mA) I see that the new voltage divider (3088 ohms on top, 7500 ohms on the bottom) gives 17.0V in the High state.
I was assuming 7.5KΩ was the input resistance between pins 13 & 15 of the 1756 board. If it's a hard 7.5KΩ then it acts in part as a voltage divider. But we're not sure that's the case. If we assume pin 15 is an input going to a gate of some sort, then its current would be fairly low. If it's a MOSFET type gate then all it needs is to see a voltage. The 7.5K might be a pull-down to turn the gate off when there is no voltage present. If it's a BJT then it shouldn't be 7.5K. Really, I am unsure of the MOZAE board.

Assuming R? is for pull-up AND pin 15 is an input, you're going to have only around 4V (depending on the Vf of HL4 and the PN junction of EE-SV3-B). That might be your problem. Without knowing more about the required input signal at pin 15 I'd be guessing.
 

Thread Starter

mcardoso

Joined May 19, 2020
152
Tony, thanks so much for chatting with me about this!

The reason why I asked if this is correct is because the sensor LED is connected to ground on both pins. NF1 (or NFI) being grounded will not pass any current through the sensor to light the LED. Unless 24V is coming from somewhere (not shown) it isn't going to light.
Well, for starters, it would help if I drew the symbol correctly. I was missing the capacitor part of that filter element. The symbol should look like this:

1600874222855.png

Where the 24V source is connected to the left terminal, the 24V load (photogate) is connected to the right terminal, and the bottom terminal goes to 0V (DC Common). The current passes through uninhibited.

Now, if I assume it's not there then I'm also assuming it's not grounded. If it were truly not there, not grounded, just a pass through jumper then you would have 19mA going through the sensor. You probably don't need that much current. But for the sake of being sure it has sufficient input - I wouldn't go below 15mA. I know we're talking milliamps, but that's where I'd go with it.
Just curious where you got 19ma? I see 24V/1200ohm = 20mA. The component datasheet for the photogate specifies a working forward current of 30mA with an absolute maximum of 50mA. Not sure why Epson (robot Mfg.) decided on the 1200 ohm resistor built into the robot. I could change that out, but I'd rather keep the robot original if I can.

So, with the above assumptions, what I'm reading is pin 15 (1756-Mozae) is supposed to see a 24V source? The unknown resistor makes this difficult to understand. Pin 13 is grounded. Between pins 13 and 15 is the input for the 1756 as a signal. So what are you using to using to limit current to LED (HL4)? I'm guessing the unknown resistor is for that purpose? If so, the LED could be limiting your input at pin 15 to just around 4 volts (depending on Vf's of HL4 & EE-SV3-B). Remember, the resistor (?), HL4 and the PN Junction of the sensor transistor form a voltage divider.
Pin 15 of the 1756-M02AE module is the home sensor input. A short snippet from the manual shows what they expect most people to do with it...

1600874832903.png

The unknown resistor is the external one I am trying to figure out if I can add to make this work. It is not currently part of the circuit and things do not work.

Pin 13 is ground.

HL4 is a small indicating LED which is installed in the rear of the robot in a small window. I believe the original robot controller built by Epson must've had sourcing inputs which pulled up to 24V. When this sensor was triggered, the sensor would have sunk current to ground, pulling down the input close to 0V. This current would have flowed from the input point, through HL4, through the photogate to ground. This current flow would illuminate HL4.

I was assuming 7.5KΩ was the input resistance between pins 13 & 15 of the 1756 board. If it's a hard 7.5KΩ then it acts in part as a voltage divider. But we're not sure that's the case. If we assume pin 15 is an input going to a gate of some sort, then its current would be fairly low. If it's a MOSFET type gate then all it needs is to see a voltage. The 7.5K might be a pull-down to turn the gate off when there is no voltage present. If it's a BJT then it shouldn't be 7.5K. Really, I am unsure of the MOZAE board.
Its pretty close to that. Not sure about the internal circuitry unfortunately. Only what is published in the manual.

1600875511270.png

Assuming R? is for pull-up AND pin 15 is an input, you're going to have only around 4V (depending on the Vf of HL4 and the PN junction of EE-SV3-B). That might be your problem. Without knowing more about the required input signal at pin 15 I'd be guessing.
I need >17.0V when the photogate is blocked by the target (aka, transistor OFF) and <8.5V when the photogate is unobstructed (aka. transistor ON)

Without the Resistor "?" in the circuit, the input sits at 0V since the internal resistor to the input card AND the photogate are pulling it to 0V.
 

Thread Starter

mcardoso

Joined May 19, 2020
152
Here is a simplified version of the circuit (as it exists currently) with all terminals and extraneous components removed.

1600876116000.png

I think it is easy to see now that the INPUT will never have anything other than 0V since it has no connection. My hope is that I can find a value of R1 in the second image such that the INPUT has voltage >17.0 in the ON state and <8.5V in the OFF state.

1600876738207.png
 

Tonyr1084

Joined Sep 24, 2015
5,039
Just curious where you got 19ma? I see 24V/1200ohm = 20mA.
(24V - 1.2Vf) ÷ 1200Ω = 0.0198 [close enough to 20mA]. Don't forget to subtract the forward voltage of the sensor LED.
BUT add in resistance from NF1 (not expecting to see much resistance) ~ 50Ω changes things up.
(24V - 1.2Vf) ÷ (1200Ω + 50Ω) = 0.0192A. I'd call that 19mA.
Granted, this is nit-picky, the forward voltage of the sensor can not be ignored. In this case it's extremely low. Typical LED's run between 2 to almost 4Vf. Failure to account for their values, you'd end up with less current than you expected to see.

As far as the Opto (photo gate), what is triggering it? is the 24V present as a high signal and not present as a low signal?

Give me some time to digest this and I'll see what I can bang out for you.
 

Thread Starter

mcardoso

Joined May 19, 2020
152
(24V - 1.2Vf) ÷ 1200Ω = 0.0198 [close enough to 20mA]. Don't forget to subtract the forward voltage of the sensor LED.
BUT add in resistance from NF1 (not expecting to see much resistance) ~ 50Ω changes things up.
(24V - 1.2Vf) ÷ (1200Ω + 50Ω) = 0.0192A. I'd call that 19mA.
Granted, this is nit-picky, the forward voltage of the sensor can not be ignored. In this case it's extremely low. Typical LED's run between 2 to almost 4Vf. Failure to account for their values, you'd end up with less current than you expected to see.

As far as the Opto (photo gate), what is triggering it? is the 24V present as a high signal and not present as a low signal?

Give me some time to digest this and I'll see what I can bang out for you.
Wow, totally did not account for the forward voltage! Thanks for catching that.

The Opto / photogate is always powered. There is a code wheel (flag) that is mechanically driven between the two arms of the gate as the robot moves. The servo control module would be looking for the ON to OFF transition of the input when trying to determine the home position.

So the sensor emitter is always powered, but the phototransistor on the other side will either be ON or OFF depending of the presence of the code wheel / flag sitting between the arms of the gate.

Here is a rough idea of what it is doing in the robot:

1600884460682.png

Thanks again so much for all the help.
 

Tonyr1084

Joined Sep 24, 2015
5,039
Is this a circuit that already exists? Somethings are not making sense to me. Not that I'm any kind of expert on this subject. It's likely others are holding back their comments to see what I come up with. Then they may critique what I've drawn.
 

Thread Starter

mcardoso

Joined May 19, 2020
152
Here's some thoughts.

I can easily add and modify the value of "R1"

I can, but would prefer not to have to modify the following:
-The Resistor "R4"
-The photo sensor itself
-The direction of the LED "HL4"

I cannot modify the following:
-Power supply voltage
-7500 ohm input pulldown resistor
 

Thread Starter

mcardoso

Joined May 19, 2020
152
Is this a circuit that already exists? Somethings are not making sense to me. Not that I'm any kind of expert on this subject. It's likely others are holding back their comments to see what I come up with. Then they may critique what I've drawn.
Yes, I bought a 1999 Seiko (Epson) SCARA robot. Bigger version of this guy.

1600885476387.png

It was pretty cheap but did not come with a controller. I am building my own from industrial automation components I have access to. I have all the motors working and the last step is getting it to read the homing sensors. I have 1/4 working, which was the inductive proximity senor mentioned earlier, however the other 3 axes use these photo gates.

I'm hoping to only add external circuitry to the robot, however if I must, I can modify components on the PCB inside the unit.

The circuit I have drawn is my reverse engineered sketch of the wiring inside the robot and the PCB.
 

Tonyr1084

Joined Sep 24, 2015
5,039
I just don't see how your schematic works. I may be wrong but I think you might need to approach this more in this way:
You have 23.4V (24V - 0.6Vf PN Junction)
You have 20.4V (respectively, depending on the Vf of HL4)
20.4V @ pin 15 when the beam is NOT blocked. Essentially 0V @ pin 15 when the beam IS blocked.
1600887009311.png
 
Last edited:

Thread Starter

mcardoso

Joined May 19, 2020
152
I just don't see how your schematic works. I may be wrong but I think you might need to approach this more in this way:
View attachment 217843
I see what you are doing here. You are swapping the mode of the sensor from a sink to a source. Doing so would require inversion of HL4 on the PLB, as well as rewiring the sensor cable.

This is the way I wish the robot was hooked up in the first place!

It is an easy solution for the sensor I am working on, but quite difficult for the remaining two which are deeply buried in the mechanics of the robot and will be difficult to remove for modification.

I'll keep this one in my pocket, but might explore other options first which require less modification of the robot cabling and PCB.

I'll try to include a case example of how the pullup resistor R1 almost solves the issue.
 

Tonyr1084

Joined Sep 24, 2015
5,039
Pin 15 is held high by R1 as long as the opto is blocked. As soon as the arm (or leg or whatever) moves away from "Home" pin 15 goes low. Is that what you want?
1600887839580.png
 

Tonyr1084

Joined Sep 24, 2015
5,039
OK. So make R1 a 1KΩ resistor. Use a 1 watt resistor as you could be seeing almost half an amp. (490mA) Can the opto handle that much current without blowing out?
 
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