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Hello. I'm in the process of wiring up the two NPN NC proximity sensors to the c10 BOB. I've read since they are NC that they should be wired in series. Is that correct? I've also read that I require a resistor to drop the voltage from the +12v to around +6v for the c10. I've read I should use a 10k ohm resistor but I guess that would change after its wired in series? A diagram would be most helpful.
All help is appreciated. Thanks.
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it would be easier with N/C switches with a PNP open collector output
what is the part number of the proximity sensors you have ? and do you have a link to the data sheet ??
can they work when powered by the C10 BOB's 5V supply supply ??
if they have to be powered by a 12 V supply you need to limit the voltage that's applied to the C10 input to 0 to +5V
(the absolute limit for input of the 74ACT245 IC used for the C10 is -0.6V to +5.6V)
John
I don't have the data sheet but this is the part number LJ12A3-4-Z/Ax your basic Chinese npn switch. I'll be using a 12v power supply to power the switches, not the c10 BOB. Forgive me but I'm a bit of a newbie.
Thanks
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my first thought is to connect the proximity sensors via a diode "OR" gate like this -
Attachment 388362
the 4.7V zenner protects the C10 input
John
Can I wire it in parallel? Or run each one individually to c10 inputs?
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assuming you have use one input for an E-stop
you could have 3 proximity switches and a touch probe Individually connected to the remaining 4 inputs
Attachment 388414
you can only wire normally open switches in parallel
using normally closed switches the control would only see a change of state when all switches open
John
PS
the circuit in post 4 is a way of connecting 2 proximity switches per axis
and possibly all proximity switches to 1 input
What I did for my switches was buy an adjustable charger adapter. You just need to hook ground back to board ground. Then you can adjust the voltage for your sensor to work properly. I think mine took 9 something after voltage drop. This way no resistors.
The #4 diagram will do two npn nc switches in series per axis correct? That is what I would like to do. I have six switches in total. Two per axis.
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John,
Very awesome solution, this will help anyone that has NPN N/C proximity switches.
Greatly appreciated!
Jeff...
Quote:
Originally Posted by john-100
in post 4 the proximity switches are not in series
the 1N4148 diodes effectively put all the proximity switches in parallel when they are open but
isolates any closed switch and prevents it grounding the C10's input
for example with 4 proximity switches
Attachment 388434
John
I think I'll do this diagram. Couple questions about it. There are many different 1n4148 diode values, does it matter which one I get? I need 1k ohm resistors? Ive read to use 10k ohm resistor to drop voltage down required for c10. Thanks for the help.Quote:
Originally Posted by john-100
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the 1N4148 is just a small wire ended silicon diode and is a popular diode in this type of logic circuit
http://uk.farnell.com/multicomp/1n41...-35/dp/9565124
(just check any suffix the component supplier adds to indicate the 1N4148 is supplied in a bulk pack so you don't order 40000 instead or 4 !!! or its a surface mount version with out any wires 1N4148-W )
but you could use a 1N4001 1A rectifier if that is whats you happen have to hand
if you had just one proximity switch connected to the C10 input you could calculate what resistance you need to have between the switch positive supply and output
but once you have 2 switches or more the voltage will depend on how many switches are open - the 4.7V zenner diode simplifies the solution
the maximum resistance between the proximity switch +12V and its output works out as 6400 ohms in order to have 5V dropped across the C10's 4K7 pull down resistor
John
PS
if using a UK (or european) supplier the BZX85C4V7 is an alternative to 1N4732 zener diode
Thanks for all the great info. I'll buy some 1k ohm resistors.
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Sorry bit I'm still a bit confused. Is 6400 ohm resistor used for one switch and 1000 ohm resistor for using two switches like in the diagram? Appreciate all the help John.
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If I remember correctly
depending on the tolerance of the 74ACT245 IC a logic high can be up to +3.8V
with a 10K resistor across the switches 12V & output terminals
the volt drop across the C10's 4K7 pull down resistor will be 3.837V which may not always be enough
depending on the tolerance of the components
so your looking to replace the 10K resistor with a 8.2 k resistor at the most
the 1K resistor I chose is probably the lowest value you want to add to 10 switches (for 5 axis)
with 1N4148 diodes in series with each switch output and the 1W 4,7V zener across the C10 BOB's input
as long as the C10's input is protected by a 1W 4.7V zener
you can us any resistor value from 1K to 8.2K
John
PS
it was just out of curiosity after re reading post 1
I calculated that for a single switch that
a 6400 ohm resistor across the proximity switch +12V & output terminals
would result in 5V being dropped across the 4700 pull down resistor when the switch opens
note-
for multiple switches you need to connect the 1W 4.7V zener diode across the C10 BOB's input
Would it matter if I use 1/4 watt or 1 watt resistors?Quote:
Originally Posted by john-100
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You can use either
a 1K resistor across a 12V supply would only dissipate 0.144 W
John