If you are connecting to an INPUT logic of your DEVICE, you need to add a resistor from the UNL2003A to a 12V supply (1k to 10k is typical and depends on the speed, input capacitance and impendence of your DEVICE. Then, connect your DEVICE to the node between resistor and ULN2003A. As said above, the output will be inverted. If an inverted signal does not work for you, run it through an open block on the UNL2003A to invert it again (resistor to +12 on each is needed).I have a project. To get an output of +12V for High or 0V for Low, the maximum current is 25mA. As tm4C123 can output 5V or 3.3V and small current. I choose ULN2003. Could anyone have such experiences on the wiring please?
The don't think so, see the "input" on far left (into ULN2003A)? It looks like everything is flowing left to right. He just doesn't show output or action of DEVICE.... and remember he is using right to left on his diagram.
That is not how I read it. Since the tm4C123 cannot (does not) need 12v, I assume the tm4C123 is to the LEFT of the ULN2003A and driving the input of ULN2003A. Therefore, circuit would be (is) drawn in standard left to right flow.Yea, but his text says something else.
by Duane Benson
by Duane Benson
by Duane Benson