FIXED!   HELP with measuring differential current output of linear encoder - Page 1
Forum section: electronics / projects

author: robrenz
author_rank: Super Contributor
timestamp: December 27, 2012, 02:43:08 pm
content: I have a linear encoder that has differential current outputs. It's output is supposed to be 7-16µA pp sine wave with a 1k ohm load. I have rebuilt the head and need to adjust the output levels. of each channel ( channel A, channel B, and the reference mark channel. The reference mark is 2-7µA pp with a 100k ohm load. It is powered by 5V. Each channel has its own plus and minus (differential) connection.Can anyone tell me how I properly measure the signal of the channel?I think I have all equipment necessary I just don't know how to properly terminate these differential leads to get a valid measurement. I have tried some things but I don't get a properly shaped sine signal and the pots that adjust the signal level seem to have no effect with the connections I have tried.Thanks in advance for any help

author: SeanB
author_rank: Super Contributor
timestamp: December 27, 2012, 04:04:23 pm
content: Does it expect a voltage level of 2.5V as a reference, and then a 1K resistor between that and the output? If you have that then simply read the voltage across the resistor, or use an opamp to amplify it a little and use it.

author: robrenz
author_rank: Super Contributor
timestamp: December 27, 2012, 04:57:45 pm
content: Thanks for your help SeanB,Scale output specs are attached, it is the Analog version. My DRO does its own subdivision of the sine wave I can pick the divisor from 2 to 100 to get the resolution I want from the scale. (raw scale signal period is a separate entry)I don't really understand what a differential current signal is, differential voltage I understand. Maybe both the plus and minus signal individually go to the ground of the 5V system through their own 1k resistor and then you take a differential voltage measurement between the two resistors? That would be the current differential wouldn't it?I could crack open the DRO and try to identify how it is terminated as it comes in from the connector but I thought I would try to get some help first.Thanks again, robrenz

author: SeanB
author_rank: Super Contributor
timestamp: December 27, 2012, 05:18:13 pm
content: Looks like the A and B channels are referenced to a half rail voltage as the current is both positive and negative. Thus with a 1k resistor to a 2V5 reference ( resistive divider on Vcc with an opamp buffer afterwards) you will get the 2 sine waves per output across a 1k resistor of around 8mV peak. You will need an opamp per output of around 200 gain to get an output of between 700mV and 2V swing per channel, though ideally you need a differential amplifier per channel, though a single ended one will work for just using the output to show steps of 90 degrees. The reference mark uses the same method, just with a 100k load so it can swing almost to the supply rails.

author: robrenz
author_rank: Super Contributor
timestamp: December 27, 2012, 10:00:46 pm
content: Its fixedHere is the rest of the story. One of my linear scales died on the hardinge lathe. I tried a spare read head that I had bought ages ago but it didn't work so I stole a head off the Z scale on my Bridgeport to get the lathe running again. Now I try to see why the spare read head didn't work. I hook it to the scope and sure enough its a TTL square wave output head that my analog input DRO is not happy with. The bad head from the lathe I disassemble to get the analog current output adjustment circuit board out of it. No small feat as these heads are potted in silicone. Xylene does a great job in turning the silicone to jello but it has to be carefully picked out of the circuitry under the microscope. Same thing on the spare TTL head because I want to use its good slider assembly combined with the salvaged analog board.I carefully take microscope pictures of all the wiring on the boards and do a quick check of the wiring on the slider assembly and it looks identical. I carefully strip and solder all those tiny #38 stranded wires to the analog circuit board and also redo all the cable wiring to the board. Now I try to adjust the current levels and all I get is a distorted sine wave that does not change amplitude with the current adjust pots. I try all of SeanB's suggestions to no avail. Now I double check the wiring on the slider comparing the analog to the digital. 2 wires are reversed on the damaged analog slider. I make the wire switch on the slider from the spare head. Put a 1k ohm resistor between A+ and A- and measure the voltage across the resistor with a Tek 7A22 differential amplifier in a 7603 and magic, perfect sine wave and the pots now adjust the current level. Set the A and B channel to 16mVpp and the ref channel to .6Vpp and sure enough the DRO is happy with it now.Thanks SeanB for your help. It helped me not give up on this repair. %-B I think the plus and minus current comes from a IR sensor pair per channel that are back to back and have their own individual grating phasing that makes one full on when the other is full off and vise versa. That is what generates the bidrectional current. and that current thru the 1k generates a 1mV/µA voltage drop. Same on the ref signal except the 100k ohm generates a 0.1V/µA voltage drop. The current adjust pots alter the individual led outputs which obviously alters the sensor current.

author: SeanB
author_rank: Super Contributor
timestamp: December 28, 2012, 04:24:17 am
content: Put the photos up, I have never seen the inside of one of these units. Nice to see you got it working, the TTL probably can be made to work if you divide the output down to a 10mV level and then apply it to the resolver. Good to see that you were able to unpot the unit.

author: robrenz
author_rank: Super Contributor
timestamp: December 28, 2012, 01:04:46 pm
content: I didn't take any pictures of the finished head because all I wanted to do was get it working but per your request I took some pictures of the TTL board and the old head/slider pieces. The analog board had only 3 resistors and 3 pots. the TTL board has one chip that does all the work to do a 1X, 5X, and 10X square wave output of the quadrature sine waves from the sensors. The shading masks are between the sensors and glued to the grating with clear optical cement. I believe they are positioned at assembly to balance the +/- current output. Fortunately I didn't have to do that since I used the good slider from the TTL head. The slider snaps onto a 1.5mm ball on the end of that spring steel wire. The wire forces the slider against the actual glass scale face and edge. The slider grating has a thin bearing strip on each edge that gives a small gap between the main scale chrome on glass and the slider grating chrome on glass. The gratings don't actually touch each other.

author: SeanB
author_rank: Super Contributor
timestamp: December 28, 2012, 02:42:40 pm
content: Very interesting. Always wanted to rip one apart and see how it works, only thing I have that puzzles me still is a old inkjet resolver, that is a straight wire and a sensor encapsulated around it.