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I am working with a CLICK PLC with the C0-08NA 120VAC input module.

According to the documentation, the C0-08NA's equivalent input circuit looks like this: enter image description here

A device I need to interface with is powered by 120VAC but uses diodes at various points in its "logic" to rectify the signal. I need to read one of these signals with this PLC, to determine if it's Off or On.

As a test, if I place a 1N4003 diode in series with a 120VAC input and connect it to one of the C-08NA's input pins, the input light flashes very briefly once, then stays off until the next time I touch the wire to the input. This obviously prevents me from detecting an "on" condition from this voltage level.

I assume this is because the rectified AC would only be about 60V instead of the 80V required to turn on the input, per the specifications.

What is a simple and safe way to convert this signal into a level that's readable by this PLC? Note that the PLC also has 24VDC inputs available if that helps. Commercially-available DIN-mount components are preferred.

FYI I tried actuating a 120VAC relay with this rectified signal and it just buzzes but won't close.

Thanks for any advice.

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    \$\begingroup\$ Try your test again, but connect a relay coil (or appropriately spec'ed resistor) across the 'Input' and 'Com' terminals - just like there would be in your actual application. I suspect you'll see a difference due to the fact that that the internal capacitor connected between 'Input' and the rest of the circuit will now discharge every cycle (I think the R in parallel with it is quite large and doesn't do that job quickly enough). \$\endgroup\$ – brhans Oct 26 '17 at 18:18
  • \$\begingroup\$ @brhans You are exactly correct: using a relay coil as a resistor (about 4.5k Ohms) in parallel with the input, the input stays on. (The relay buzzes of course, but that's inconsequential.) Should I add such a resistor to the final application, or are you saying this load is implied and is already present in the source device? Also, is it possible adding such a load to the device being monitored could affect the signal negatively in relation to its original purpose or damage the device? (i.e. too much current draw for the device's original design, etc) \$\endgroup\$ – Ryan Griggs Oct 26 '17 at 18:34
  • \$\begingroup\$ Is there some reason you have a half wave? The Opto can work on both polarities but must be low impedance source(<100 ohms ) Using a diode defeats this as it is open cct half wave so this blocks current half the time and the series cap blocks the input DC voltage reducing the current again. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 26 '17 at 18:41
  • \$\begingroup\$ @TonyStewart.EEsince'75 as stated in the original question, the device I'm trying to monitor rectifies the signal at various points. The point I need to measure is a rectified signal. \$\endgroup\$ – Ryan Griggs Oct 26 '17 at 18:42
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    \$\begingroup\$ I think that the load is already present as the relay coil in your other device, so this test with the resistor just simulated that and you shouldn't need to add it in the final application. \$\endgroup\$ – brhans Oct 26 '17 at 18:56
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The datasheet gives us a clue:

enter image description here

Figure 1. Note that the input current and impedance change with frequency.

Since both input current and impedance change with frequency it means that there's a reactive component to the input and that must be the first capacitor. It is doing most of the voltage drop and the parallel resistor is there to provide a discharge path.

  • \$ Z_{50} = \frac {V}{I} = \frac {100}{8.5m} = 11.7 \ \mathrm{k\Omega} \$.
  • \$ Z_{60} = \frac {V}{I} = \frac {100}{10m} = 10 \ \mathrm{k\Omega} \$.

These numbers are somewhat at odds with the input impedance figures but they're the right order of magnitude and the ratios are correct.

From \$ Z_C = \frac {1}{2 \pi fC} \$ we would expect that if the capacitor formed the bulk of the input impedance then the ratio of impedances to be \$ \frac {Z_{50}}{Z_{60}} = \frac {2 \pi 60C}{2 \pi 50C} = \frac {6}{5} = 1.2 \$. It actually works out at \$ \frac {11.7}{10} = 1.17 \$ which is close enough. We can ignore the other series resistor.

We can estimate the value as \$ C = \frac {1}{2 \pi fZ} = \frac {1}{2 \pi 60 \cdot 10k} = 0.83 \ \mathrm{\mu F} \$.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2. Equivalent circuit.

Note that in Figure 2 you now have half of 120 V AC on the left of the capacitor so effectively you only have 60 V AC on the input. This isn't enough to reliably turn on the opto-isolator.

schematic

simulate this circuit

Figure 3. Addition of R1 to allow C1 to discharge

As others have noted in the comments, without R1 D3 charges the left side of C1 and it has no discharge path so the next positive going half-cycle can't pump more current in. Adding R1 discharges C1 when the voltage falls and allows it to accept more current on the next half-cycle.


enter image description here

Figure 4. The low-pass filter on the logic input.

Note that on the logic side of the opto-isolator there is a low-pass RC filter. This is what is keeping your input "on". I reckon that this is also just on the edge of working with the reduced input voltage caused by half-wave rectification. You might not see the LED blink but you may find that the program glitches intermittently if the logic-side input falls below the the threshold for one scan or more.

You could be a long time debugging this if you weren't aware of the internals. It might be worth debouncing the input with a 200 ms off-delay timer in the program.


Possible work-arounds to increase reliability:

I've thought about this overnight and I don't think there is any "right" solution - but I often think that much of engineering is finding "the least worst" solution.

  • The first thought when faced with increasing the output of a half-wave rectified signal is to add a smoothing capacitor on the output (as one would on a power-supply). This won't work in this case as we would then be feeding DC into the AC input and the capacitor would block it.

schematic

simulate this circuit

Figure 5. Using a 110 V DC relay to buffer the signal.

  • A second option would be to use a relay. A 110 V DC relay such as Finder 46.52.9.110.5000T presents a nice high resistance load to the sensor. With this example a 24 ms time constant would average out the voltage enough to give 110 V DC. A 470 nF adequately rated capacitor would do this.

enter image description here

Figure 6. Coil voltage with filter capacitor on 60 Hz supply (for your convenience).

I think this is the simplest reliable option. The time delay added by C1 is short and repeatable. There are no heat-generating components other than the relay coil and everything is readily available. Make sure the capacitor peak voltage is > 200 V DC.

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  • \$\begingroup\$ Wow, thanks @Transistor for the detailed explanation. I understand now why the LED flashes on once and then goes off without the resistive load on the input. Is there an industry standard way of conditioning this type of signal so that it's appropriate for this type of input, or an industry standard way of converting it to 24VDC? I looked for DIN rail opto-isolators rated for similar input voltages but did not find anything suitable. Since this is an industrial application, I don't feel good about bodging a resistor across the input (safety and compliance concerns etc). \$\endgroup\$ – Ryan Griggs Oct 27 '17 at 2:00
  • \$\begingroup\$ What about a solid state relay with some sort of time delay to smooth the output? I'm not even sure a 120VAC SSR would properly switch on a half-wave input. Your thoughts appreciated. \$\endgroup\$ – Ryan Griggs Oct 27 '17 at 2:02
  • \$\begingroup\$ See the update. Thanks for accepting the answer. I wouldn't try feeding the input from an SSR. The input is too sensitive and any leakage through the SSR snubbers, etc., could play havoc. \$\endgroup\$ – Transistor Oct 27 '17 at 9:30
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A device I need to interface with is powered by 120VAC but uses diodes at various points in its "logic" to rectify the signal. I need to read one of these signals with this PLC, to determine if it's Off or On.

Well, if your "device" is ac powered, and it incorporates a power transformer, you can simply wire the plc input to the primary of the device's transformer. If it does not have a transformer, you can tap off the AC input signal using 1uf non-polarized caps of a non-isolated dc supply (if you need to isolate it, but I doubt it).

As a test, if I place a 1N4003 diode in series with a 120VAC input and connect it to one of the C-08NA's input pins, the input light flashes very briefly once, then stays off until the next time I touch the wire to the input. This obviously prevents me from detecting an "on" condition from this voltage level.

I assume this is because the rectified AC would only be about 60V instead of the 80V required to turn on the input, per the specifications.

What is a simple and safe way to convert this signal into a level that's readable by this PLC? Note that the PLC also has 24VDC inputs available if that helps. Commercially-available DIN-mount components are preferred.

FYI I tried actuating a 120VAC relay with this rectified signal and it just buzzes but won't close.

Well your devices are AC not DC so yes, its not going to work rectified to DC.

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