## 2017-06-04 Making A NC Switch Look NO

I’ve been thinking about this for awhile.  Let’s say I have a switch, for example a thermal switch that only has contacts that are NC (normally closed) until a certain temperature is reached.  At that point, it changes to open, interrupting the circuit.  But the input to my controller senses NO (normally open) switch contacts that close when a certain temperature is reached.  So how can I change the switch so that it will work with the controller?

We will assume that there are no other ways to make it work, such as changing the controller or the switch themselves.  And the controller puts a current of 10 milliamps DC through the contacts when they close, and the controller outputs 5 volts DC when contacts are open.  Also, we want to avoid using a circuit that requires external power, such as a relay.

I considered using a single transistor and resistor as an inverter.  The collector and emitter would be connected to the controller with the correct polarity.  The switch would be connected to the emitter and base.  And to turn the transistor on when the switch contacts are open, a resistor must be between the collector and base.  And this resistor must have high resistance, enough so that the controller will not sense it as being closed (the voltage is not high enough), but its resistance must be low enough so that the transistor is turned on with a low voltage.

Let’s say this resistor is 10k.  When the thermal switch is closed, the transistor is not conducting and the 10k resistor will be connected directly across the input of the controller.  With 5 volts across the 10k resistor, there will be 1/2 milliamp flowing.  But the controller would have a current limiting resistor in series with the 5V.  If the current limiting resistor is 5V / 10 mA or 500 ohms, then about 500ohm / .0005A or .25 volt would be dropped across the current limiting resistor, giving us about 4.75 volts at the controller input.

When the thermal switch is open, the 10k will be in series with the base – emitter junction across the input.  So 5V – 0.6V will be across the resistor.  The current would be 0.44 mA.  But this is flowing through the transistor, and turns on the transistor.  If the gain of the transistor is 100, the current flowing would be 44 mA, but obviously the maximum is only 10 mA, so more voltage is dropped across the input’s resistor, and the voltage at the input drops until it reaches equilibrium, at wich time the input will have a low voltage that the input thinks that its switch is closed.

Right now, I’m thinking that it would be easy to just connect a few resistors, a transistor and a switch together and try this.  I’d use a pot for the 10k to see what the optimum value is.