[Kal22378]

I am new to the cnc world and was wondering instead of using a voltage divider circuit to drop the voltage from 12v to 5v would it be possible to use a buck converter to bring the voltage down to the necesary 5v for my BOB?

[Toinvd]

If you mean the supply voltage + 5Volt of the BOB, then the answer is YES. A buck converter will do it even better then a voltage divider.

[Kal22378]

Yes. My proximity sensors run on 12v and need to get the voltage down to 5v correct? I am using a DB25 1205 BOB.

[Kal22378]

Let me rephrase this...sorry. I am trying to step the voltage down from the 12v my proximity sensors output to 5v using a buck converter. My BOB is powered with 5v already.

[rodw]

Personally, I would use a 12v capable optocoupler to switch your existing 5V power to your BOB input. I think a Buck power supply is overkill and it may add some delay to the signal.

[Toinvd]

So it's not the supply. Don't use the buck converter. Have a look into the manual and see if the inputs are optocoupler. If so, they might even accept the 12 V or just put a serial resistor ( about 10 Kohm will do )
If it's a direct input on a IC then you can use a voltage divider 8K2 and 5K6 will do the job.

[Kal22378]

The inputs and outputs are all optoisolated

[Kal22378]

Attachment 392718

- - - Updated - - -

This is the board I have

[davidloomes]

The output of an NPN NO sensor is just a switch to ground. It doesn't have any output voltage of its own, so it doesn't need to be stepped down.

[john-100]

If your proximity sensors have true open collector outputs with no pull up resistors or LED indicators
connected between the positive supply and output

the sensor will act like a N/O switch between the output and ground terminals like this -

Attachment 392734


your break out board will be some thing like this -

Attachment 392736

( NOTE - the photos found on line did not make it clear if the 12 IC inputs from the printer port have the 10K resistors connected as pull up or pull down resistors - I have shown them as pull down resistors )

the 5 inputs from your E-stop switch and proximity sensors etc
are connected to a 330 ohm resistor and LED that is part of an opto-isolator
when a switch or proximity switch grounds the input

approximately 10mA will flow from the BOB's +5V supply .though the resistor & LED and the switch / sensor to ground and the light from the LED switches on the NPN photo transistor connected to the printer port input

the board has two 5V power inputs
one for the PC side of the BOB and one for the isolated
input / output connections

some versions of the board has two jumpers that enables the use of one 5V supply but this will by pass the galvanic isolation that is needed if you have problems with ground loops for instance

John

[Kal22378]

John....would you mind if I private message you?

[john-100]

OK with me if you want to PM me
the only thing is as I don't use any 3rd party to store pictures
I have not found a way to add photos to private messages

John

[john-100]

I have downloaded the data for your NBN8-12GM50-E2-V1 sensors
Attachment 392964

this confirms they have N/O NPN outputs
Attachment 392966

which can be wired in parallel and connected to a BOB input like this
Attachment 392968
(provided the outputs are open collector with no pull resistor up to the +12V supply)

John

[Kal22378]

Ok....so i can just run the sensor Black wire directly to my BOB with no resistor?

[john-100]

I was going to say connect the sensor output directly to the BOB input
until I read the NBN8-12GM50-E2-V1 data sheet again

the maximum reverse voltage for the LED inside the PC817 opto-isolator used on the breakout board is 6V

I am not 100% sure but I assume the 0.5mA leakage current referred to on the sensor data sheet
is from the positive supply to the output
due to a component / circuit to protect the NPN output transistor from the back emf when it switches off an inductive load like a relay
Attachment 392976

if I am correct, the leakage could damage the opto-isolator
(as the cathode will be taken to +12V while the anode is at +5V reverse biasing the LED by 7V )

what voltage do you measure between the negative supply and the output when the sensor is switched off ? (I am assuming you have a multimeter )

if its a volt or two then you can connect the output directly to the BOB
but if its around +10 to +12V
I would add a diode with the cathode connected to the sensor output and anode to the BOB input

any silicon diode like a 1N4148 or 1N4001 to 1N4007 will do
(the 1N4001 is a 1A diode and has thicker wires that may be easier to connect to)

Attachment 392978

the problem with adding a resistor instead of the diode is
you need a low value resistor to ensure the NPN photo transistor inside the opto-isolator is turned hard on
but you will need a high value resistor to limit the reverse current through the LED when it breaks down

John

[Kal22378]

Ok. I put a multimeter on the sensors output and ground. When no metal near the sensor it reads 11.70v. When i move metal to the sensor the meter reads .26v. I had some 1n4148 diodes and connected one up to the output as your drawing shows. I have .04v when no metal is near and .02 when the sensor activates.

[john-100]

"When no metal near the sensor it reads 11.70v. When i move metal to the sensor the meter reads 0.26v."
makes sense if there is a high impedance between the sensor + supply and output

"0 04v when no metal is near and .02 when the sensor activates"
. looks like you have either a faulty component or a short circuit

what do you have between the BOB's ground and an input with no sensors connected ?
I would expect it to be about 3.2V and 5V


try using two diodes like this

Attachment 393070

john

[Kal22378]

Ok John. When I took the readings on my multimeter the sensor was not hooked up to the BOB. I hooked one sensor to the input of the BOB following your diagram. When sensor is off I am getting 4.14v at the input. When the sensor is on I am getting 1.13v.

[john-100]

that's more like I expected

this should sum up the results



John

[Kal22378]

Should I put an external pull up resistor like your diagram suggests?