Help creating a circuit to detect 22ohms and turning on a switch

[parapants]

Hello,

I have a CAD table and I have a cutting laser. I've mounted the laser to the table, but the two talk a different language. The table wants the laser to fire when it changes a line from 1 ohm to 21 ohms (+/-1 ohm). The laser wants to fire when it receive a 5v DC signal. I'm trying to make on a breadboard a circuit that will trip a switch when it detects the 21 ohms on the line.

So, to make sure I've explained this correctly, the breadboard has 2 sets of inputs (a 5v dc power supply, and 2 wires from the CAD table which have a normal 1 ohm reading). The breadboard has one output, which is two wires used to send the 5v dc signal to the laser. When the two wires from CAD table indicate 21 ohms, then the switch sends the 5v dc current to the laser.

I've tried use the CAD table's output wires as the switch, but it wasn't designed to work that way, and all I get are random voltages. The only consistent results from the CAD table is the ohms shown on the tables indicator wires.

Thank you for your help.

 

[Grossel]

Hi.

It's done by connecting your variable resistor in series to a fixed resistor between ground and a voltage. Then you'll get a variable voltage output.
Next: Make an opamp window detector (is on when voltage is between a max/min value) and you've done it

 

[alec_t]

What voltages are present at the two ends of your resistor? Is it part of a current-loop?

 

[circuitspecialists]

You can also use a Schmitt Trigger IC and a variable resistor. Form a voltage divider out of the variable resistor and the 21 ohm resistance so that it "trips" when the resistance changes from 1 to 21 ohms.

 

[parapants]

What voltages are present at the two ends of your resistor? Is it part of a current-loop?

Hello,

There is no current in the loop, only a resistance. Right now I'm going to try using a Arduino board, and a portion of the previous post to control the laser.

Thank you both for your help.

 

[alec_t]

only a resistance.

That is true only in the unlikely case that the resistor is completely isolated from the remainder of the circuitry in the CAD table. Can you confirm that is the case? If it is not isolated there is a risk that the solutions proposed above could feed back harmful voltage or current to the CAD table.

 

[WTP Pepper]

If there is no current in the loop then you cannot measure resistance unless you have a camera reading the colour code of the resistor in place

Seriously though, can you let the forum know if either side of the resistor is grounded and whatever is inside the box is measuring current to ground and not to some arbituary mid point voltage. Measure the voltage on each end of a resistor under both conditions with respect to ground and post back.