Thread: AM882 Alarm Circuit
I'd just like to quickly double check that my proposed Leadshine AM882 driver alarm circuit for is ok.
I looked at the manual last night and I've worried myself that I'm misunderstood it. The typical configuration diagram in the Leadshine manual shows a 2k ohm resistor on the ALM+ feed:
I think this resistor is simply there as a current limiter - if it was absent when the alarm went off you'd have a short to ground. Since I have the relay coil in the circuit I think this will have the same effect but I'm not 100% sure (as you can probably tell I'm new to electronics).
Reading the Wikipedia page on open collectors it looks like I have the coil in an unusual place. I think you would normally place the coil before the alarms (e.g. in the positive leg of the circuit) is that right?
My electronics is a little rusty so wait for confirmation but I think you might damage the driver doing this.
The 2k resistor is there to pull the line high when the phototransistor is off. When it's on, the phototransistor pulls the voltage down to near ground potential.
Without knowing the spec for the phototransistor and relay I can't tell whether the phototransistor is robust enough to cope with the loads a relay would impose. With a 2K resistor there will be around 12 milliamps flowing when the photodiode is conducting. I think the relay would be much more than this and there will also be voltage spikes from the inductance of the coil.
Again, wait for confirmation but I don't think you need a relay - just use the leadshine circuit.
Cheers, some additional info. The manual isn't completely clear for the alarm connectors but I would read it as it can handle 30V at 100mA:
The relay coil requires 24V @ 20mA to activate.
I could use the Leadshine circuit but that would have it running at the 5V the BOB supplies and I was hoping to get the voltage a bit higher to help avoid interference.
I don't have a resistor and it works fine, the relay has a built in diode across the coil to prevent back emf, as you say the relay is usually connected before the alarms and that's how mine is wired.
You can also use the ProTuner software to make the alarms normally closed and put them in series but I've just let mine as normally open.
also there's this
Last edited by EddyCurrent; 28-03-2014 at 12:26 PM.Spelling mistakes are not intentional, I only seem to see them some time after I've posted
Sounds like it will be fine then, you definitely don't need the 2k resistor which would limit the current to a maximum of 12ma - too little to activate the relay.
Brilliant, thanks guys.
I'm sure that there is something wrong with that Leadshine circuit. The "R" for the alarm circuit should not be 0 for the 5V case. That would short out the 5V supply when the alarm went off, followed by the demise of the opto-isolator microseconds later. What you should use as a pull-up will depend on what is in the block on the left. Personally, I would put the 24V relay where R sits, connect alarm outputs in parallel, and corresponding terminals to 24V ground, but it's 6 of one and half-a-dozen of the other and the other suggestions would work fine.
Last edited by Neale; 28-03-2014 at 12:55 PM.
Not sure why you want the relay, its one more thing to go wrong. The FAULT input on a PMDX is already pulled up internally so just wire all ALM+ to Fault and ALM- to GND on J13. The ALM outputs are already opto-isolated so the additional isolation of the relay is superfluous.
The relay is there so that I can run the majority of the alarm circuit at 24V rather than the 5V the fault input on the PMDX is pulled up to, my understanding was that would provide more resilience to noise.
On the PMDX the fault and e-stop inputs are sampled together which to my mind means double the risk of noise causing a false e-stop (the manual doesn't say but I'm assuming they are OR'ed together).
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