SCOPE Soft Front Panel

[nyborn]

Hi DFGRAY:

It has been more than a week that I haven't heard from you back. I would like to ask.

Could you please write me back what is going on?

[DFGray]

In general, when you have both analog and digital circuitry, you use separate ground paths for both.  This is because digital circuits tend to have fast, transient current spikes, inducing changes on the ground path, while analog circuits tend to be slower with fewer current spikes.  Keeping the ground paths connected at only one point helps prevent digital noise from getting into your analog system and analog noise from getting into your digital system.  In your case, the analog ground is the ground of the scope.  The digital ground is the ground pin on the 7805 on your circuit board.  Your circuit has the digital and analog sections totally isolated from each other.  Using the 6250 for monitoring and control will join the analog and digital grounds through the 6250.  This may or may not cause problems, depending on the magnitudes of the current and voltage through the UUT.  If you do need isolation, use differential mode for the analog input to the 6250 and you should be good.  Connect a 10k resistor from the AI- input of the 6250 to the digital ground of the 6250 to finish the isolation (this is the bias resistor).

Please learn what your digital circuit does or you will continue to have problems with the interface.  All your digital inputs run through latches (the paired 74HC00 gates).  This locks the output to a high or low, depending on which gate sees a transition.  Inputs 3, 4 and 5 are further modified to produce an output pulse (about 0.5sec?) using the 78LS123.  Yes, you can produce the necessary signals on two lines with a single button press, but it will require you to write a program to do it.  If you have LabVIEW, you can use the DAQ assistant to help.  You can program the 6250 with C, C++, CVI, and LabWindows as well.  I can't help you much here, since I have not programmed DAQ in over five years (my recent experience has been with the NI-SCOPE API).

Take home message - learn how your circuit operates, then try to interface to it.  If you are going to write a program, it would probably be just as easy to replace your whole digital front end, once you know how it works.

[nyborn]

Hi DFGRAY:

I have already taken time to understand the circuitry work. But your explainations and guildlines made me get perpective on how electronics work. Thanks.

I have a perpective question that want you to verify. According to your previous mail, you said in my case, the analog ground is the ground of the scope.  The digital ground is the ground pin on the 7805 on your circuit board.  Your circuit has the digital and analog sections totally isolated from each other.

If the test fixture is out of BNC cable, just as normal. The analog ground will be the negative terminal of the power supply. Is my perpective right?

The scond question is why the pulse voltage is about 5V which is greater than the voltage measured from AI+ and AI-. If you look at the circuitry, the analog ground is the ground of the scope (NI 5112). All the voltage and current now flow to this hard groud. The voltage in between AI+ and AI- is about 0.004V something like that a small value. Why I pressed the "bark" button, the motor vibrate and close the hamlin relay which is near the negative terminal of the UUT. You can see there is a analog ground under the hamlin relay. I think at this moment the analog is not that ground, it is the BNC hard ground, all the voltage and current will not go to that ground, it will go to the hard ground, am I right? The right side of the UUT is like an open ciruit? Am I right? and why the voltage become more than 5V. Could you explain here? I don't understand.

The last qustion is about ni Scope Multi Record.vi file and it is very important. I hope you can reply to me early. Otherwise I got stuck when I need to generate the waveform in the program. I followed guildline and setup the hardware with 10K bias resistor. Then I immediately used SFP to test the hardware. Attached is the diagram that I tested. You can see, I passed all the hardware setup test in SFP. But when I used the ni Scope Mulit Record.vi file which is preinstalled with the NI 5112 software. If I don't connect the digital IO from my test fixture, the waveform looks correct. But if I put digital lines and bias resistor for ni Scope Multi Record.vi file, the waveform parameters and time domain changed. Why? I have no clue. Would you please help me out here.

[DFGray]

Your scope problem is pretty easy to fix.  You have not set the trigger level correctly.  The trigger level on the VI is set to 0V.  It is essentially triggering on noise.  Set it to about 1V and you should be fine.  This is an easy error to make.  The autosetup on the SFP automatically sets a "good" trigger level.

Your electronics questions can be answered if you look at what power supplies are connect to what lines.  The relays totally isolate the UUT.  The motor is connected to the digital power supply, not the UUT one.  The schematic is too low resolution and fuzzy to figure out the voltage on the UUT from the information I have.  Note that, according to the schematic,  the motor does not close the relays.  What does?  Look at the schematic and you will see.

[nyborn]

Hi DFGRAY:

I am so sorry. Attached is better schematic that scanned in pdf file. I am so sorry that made you hard to see.

Back to the first questions of electronics, that means the analog ground is the BR1 of the power supply in the schematic out of BNC cable connected to the fixture. That is what you meant? Do I get your meaning right? I would like you to verify.

So, with BNC connected to NI 5112, the analog ground will be the hard ground of 5112, not the ground of BR1 in the schematic in this case, am I right?

The third questions is why get the voltage like 5.02V which is greater than 0.004V in R1 in BNC cable. Let me tell you the voltage supplied to UUT is 6V. The 2N3904 amplifier and the shkrelay input is close to the hamlin relay. If you look at the bark push button, the shkrely from logic gate is connected to the shkrelay of the hamlin relay. Why get about 5.02V at NI5112 for short pulse waveform.  But the test push button do not have shkrelay wire connected to each other and it will still get the same short pulse waveform about 5.02V at NI5112 from the BNC cable. I don't understand here. Would you please explain to me?

[nyborn]

One last question I forgot is the attached graph of the about 0.5 offset adding R1 resistor. Is this result from the graph acceptable with digital ground connected and AI- connected to 10k resistor and BNC cable with R1? Can I make the offset back to 0V or is there any way to do it or it is acceptable, there is no reason to make thing complicated? I am not quite sure. I am afraid someone in my job might challenge this question. I tried to test a hardware setup without connected to digital I/O in CB68LP, just only BNC cable with R1 and AI+ and AI- connected to r1. The graph is normal which have no offset starting from 0V. I thought it should be the potential difference of digital I/O ground which is different than the analog BNC hard ground with R1 connected that caused offset about 0.5V (ground loop). Without the R1 connected, just BNC with digital I/O lines to cb68 board, the signal had no offset which is starting 0V. Am I thinking this thought right? if not, would you please correct me? I need to make sure everthing is correct. So, what do you about that 0.5V offset? Attached is another graph for you to see the result that you might see last time

[nyborn]

One last question I forgot is the attached graph of the about 0.5 offset adding R1 resistor. Is this result from the graph acceptable with digital ground connected and AI- connected to 10k resistor and BNC cable with R1? Can I make the offset back to 0V or is there any way to do it or it is acceptable, there is no reason to make thing complicated? I am not quite sure. I am afraid someone in my job might challenge this question. I tried to test a hardware setup without connected to digital I/O in CB68LP, just only BNC cable with R1 and AI+ and AI- connected to r1. The graph is normal which have no offset starting from 0V. I thought it should be the potential difference of digital I/O ground which is different than the analog BNC hard ground with R1 connected that caused offset about 0.5V (ground loop). Without the R1 connected, just BNC with digital I/O lines to cb68 board, the signal had no offset which is starting 0V. Am I thinking this thought right? if not, would you please correct me? I need to make sure everthing is correct. So, what do you about that 0.5V offset? Attached is another graph for you to see the result that you might see last time

[nyborn]

Hi DFGRAY:

May be the third question is very diifficult to answer because I did not provide you enough information. I found that the 5.02 V is represented a shock pulse waveform after pressing "bark" push button to see if the waveform looks good on NI 5112 for motor vibration. That is why the actual voltage would be from the positive terminal to the negative terminal of the hamlim realy switch ground. I have a question if you connected with a BNC hard ground. Would the ground of the hamlim relay switch would be the hard ground of the BNC.

2) The offset voltage in R1 which is about 0.5V. According to previous mail, you said the BNC orginial has 0 impendance without R1. Wiith R1, the BNC will have some impendance. In fact, according to ohm law, the impendance has small effect. We don't care, is that right in this cace. I would like to verify with you. I don't understand how the offset come. Is it becasue eventhough we don't the R1 impendance, but there will be some voltage from R1 to Ni5112 ground. That is why is cause 0.5 offset. The ground right now is different than floating souce ground in the analog test fixture. Am I thinking it correctly. I would like to verify with you too.

3) You said I should use 100 ohm for R1 which is correct value, but when I tried to program in Labview, I use AI+ and AI- to measure the voltage in R1, the voltage varied from 0.02ma, 0.03ma, 0.04ma unsteady. If I use the 18kohm instead of 100ohm for R1. The program would not jump the voltage value a lot. It will be very steady to decrese from 1.1ma to 0.034ma and then stay there.

It make the operator in my manufacturing area easy to see. This value 0.034ma is corrosponding to my customer specification. Could I use the 18kohm in this real situation?

4) I don't understand why I need to use the value of 10kohm for bias resistor. Is that because this is the general purpose according to the ground tutorial. I would like to make sure with you.When AI-  in CB68LP board is connected to DGND with 10kohm resistor the analog ground and digital ground are completely sepearted. Do you mean these two ground are in the CB-68LP board. I would like make sure. Also, the digital ground is completely separated with the analog ground in the test fixture without any digital I/O connected or BNC connected thing like that. How can identify the difference between the analog ground and digital ground in the test fixture. Could you tell me? because I really don't get it and understand about it.

At last, I hope you can verify with me these 4 important question, Because without asking you, I could not verify and I will get stuck.