I have built a circuit which inverts the signal of a ULN2003A, it is in turn connected to a relay. It is possible for the ULN2003A output to be disconnected from the external circuit as it is remote from the driver. I have tried different methods to try and make it that the circuit defaults to an off state, making the relay go to NC, yet retain normal operation.
I have been trying to work out he circuit for the whole day, to no avail. I have been trying to keep the circuit as simple as possible in that I want to use transistors.
My current circuit:
simulate this circuit – Schematic created using CircuitLab
When the driver is active it switches the relay to NC.
The way in which it is supposed to operate is that while the driver is active the relay is contacts are NC, when the driver is OFF it should activate. Then the important part, if the driver line is removed it should then default to NC.
The last part is what has been troublesome, I need help in figuring out the circuit to achieve that.
I have also noted 0.6V on the ULN2003A pins when its is attached to the circuit, or not.
Best Answer
What you want to do is not possible. The datasheet shows the ULN2003A to have open collector outputs. That means it only switches between open and actively driven low. Open is the same as disconnected. Your circuit therefore can't tell the difference between open and disconnected.
To make this work, you need the off state to be when the ULN is open, and active when it is pulling low. That way, disconnected will cause off, not active.
The ULN does have some diodes connected to the output in addition to the open collector. With more circuitry and some cleverness you could detect the presence of one of the diodes. However, that's quite a kludge around a fundamentally bad architecture. It is better to fix the real problem instead of kludging around it.