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I have a Labview program which sends 4-20ma current signal through a ni usb device to control the speed of a motor by controlling a sew eurodrive MOVIDRIVE ® MDX60B / 61B. Here is the data sheet of MOVIDRIVE:https://download.sew-eurodrive.com/download/pdf/16837614.pdf. PAGE 60 shows the layout of signal terminals. On the actual device, the input terminals are just two normal terminals,nothing special.

Originally, this motor is controlled by manually pushing some physical buttons on a device. The device is just some kind of signal generator that generats 4-20ma current signal.

The MOVIDRIVE ® MDX60B / 61B is set to accept 4-20ma current signal as control input.

So how can I switch between the two control signals (1.from labview, 2. from the device) so that there won't be any gap between the switch? I want to accomplish this because if the control signal drops to 0 at some point, the driver may get damaged. Since these equipment (both the movidrive and the motor) is really expensive and important, I can't risk damaging them by placing a switch in the circuit and just switch between the signals. Any advice is much appreciated!

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  • \$\begingroup\$ Welcome to EE.SE. You'll need to expand your question. How does the original system control speed by "pushing buttons"? What is the control output from LabView (note capitals)? Is it analog or PWM? There's a schematic button on the editor toolbar. \$\endgroup\$ – Transistor Feb 5 '18 at 8:26
  • \$\begingroup\$ Welcome to the site. Please quickly realise that this is not a free design house, homework-answering service or an on-line technical encyclopedia, copied out to you on demand. People will help you take the next step if your question shows that you've done as much as you possibly could on your own - which your post doesn't, I'm afraid. Please revise your question showing your work and findings so far, in considerable detail. Or delete the question if Internet searches give you your answer anyway. Again, a warm welcome to the site. \$\endgroup\$ – TonyM Feb 5 '18 at 9:16
  • \$\begingroup\$ @Transistor, good advice but unfortunately the schematic editor's not available to a newcomer, they haven't enough reputation points. You can advise them to post a scan of a diagram, though. \$\endgroup\$ – TonyM Feb 5 '18 at 9:18
  • \$\begingroup\$ I edited the original post. Thank you for the tips on making myself more clear. Is there anything else I should add? \$\endgroup\$ – hiphip Feb 5 '18 at 9:20
  • \$\begingroup\$ A link to the drive datasheet and a page reference to the analog input circuit would be nice. I think your question boils down to, "how can I switch between two 4 - 20 mA control signals without dropping below the lower of the two during switching." Is this right? \$\endgroup\$ – Transistor Feb 5 '18 at 9:24
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The problem circuit with addition of C1.

  • Without C1 the input will drop to zero for an instant while the switch is toggled.
  • The addition of C1 will help but will have to be a largish value on account of the low impedance of the 500 Ω input shunt resistor. This will result in a lag in the system response which may not be acceptable.

schematic

simulate this circuit

Figure 2. Conversion to voltage control makes the solution quite simple.

The SEW drive can be configured for 0 - 10 V input control. This will, quite likely, have a high input impedance of > 10 kΩ. By configuring the drive for 0 - 10 V control and adding external shunt resistors you can convert the 4 - 20 mA signals to 0 - 10 V.

With this arrangement C1 will track the selected current source. During the inverval when the switch is thrown C1 just has to maintain the voltage to the high impedance input of the drive. When the switch contact makes again C1 will very quickly settle to the new control voltage due to the low impedance of the 500 Ω resistor now placed in parallel.

Work out your time constants from the usual RC formula.

RS sell a 500 Ω, 0.1% resistor which seems to be designed for this purpose.

enter image description here

Figure 3. The SEW drive supports voltage control input from -10 - 0 - +10 V giving direct forward / reverse control, if required, from a voltage source.


I'm not sure how real your problem is. If you have a non-zero deceleration rate programmed in your drive I would think that you have some protection already.

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  • \$\begingroup\$ If this answer works then please accept the answer to indicate that it is correct. \$\endgroup\$ – Transistor Feb 8 '18 at 10:32
  • \$\begingroup\$ oh I tried to upvote but the system says I can't do it because I don't have enough reputation, I didn't realize there is also an acceptance button. I just did that! \$\endgroup\$ – hiphip Feb 9 '18 at 11:10
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Effectively you want to maintain a current signal during switchover. Inductors are to current signals what capacitors are to voltage signals, so an inductor and a path to ground ought to do it:

schematic

simulate this circuit – Schematic created using CircuitLab

When the switch disconnects briefly, the inductor maintains the current into R1 for a little while. D1 provides the return path in this situation.

Unfortunately with a 500Ohm burden you'd need a very large inductor in order to get ride-through in the order of milliseconds. Check your SEW's input circuit - if it uses a smaller burden you can get away with a smaller inductor.

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  • \$\begingroup\$ that's really helpful. thank you! unfortunately I can't upvote your answer because I don't have enough reputation \$\endgroup\$ – hiphip Feb 6 '18 at 9:29

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