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I'm attempting to wire a Yaskawa Sigma-V Series AC servo drive to a Rosetta CNC, and I came across this symbol, a triangle that has one line in, one line out, and what seems to be a return line connected to what looks to be a small circle touching the triangle.

image of the electronic symbol described above

Here's another photo, zoomed out a bit.

the same picture zoomed out to show the full circuit, including what seems to be an optocoupler

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1 Answer 1

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It's a buffer with differential output. The pin with the circle is the inverted output. The circle means the state of that pin is the opposite of the main output. When the output is high, the pin with the circle is low.

When you apply a high level to the input of the buffer, the main output of the buffer goes high and the negative output goes low. This will drive the current through the LED. When you set the input of the buffer to low, the main output goes low and the negative output goes high. This will turn off the LED.

For this specific circuit (driving an LED), you can get a similar result by disconnecting the negative output and connecting the Cathode of the LED to the GND of the circuit; assuming the same GND is used for the buffer and the LED.

It is safer to use differential output rather than using the same GND. When a shared ground is used between devices, the current going through the ground wire can create a voltage difference between the nodes (because of the wire resistance). If we connect the Cathode of the LED to the ground and then because of the current of the rest of the circuit, the ground on the buffer side goes higher, there is a risk that that voltage might turn the LED on unintentionally. The opposite is possible too; a high signal might not be able to turn the LED on. By using the differential output, you can guarantee that regardless of the current in the rest of the circuit, the LED only follows the command on the buffer input.

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