# Two Simple Thermistor-Controlled Relay Circuits

Posted on Feb 4, 2014

You can use a Cmos 4001 or a Cmos 4011 in this circuit. They have four - two-input - gates. In each case - the two inputs are joined together. So the gates are being used as simple inverters. = This means that when the two inputs are high - the output will be low. And when the two inputs are low - the output will be high. = The output from the cir

cuit is at pin 10. When pin 10 goes low - it`s connected internally to the negative line. So it connects the base of Q1 to ground - through R3. This allows base current to flow through the transistor`s emitter-base junction. And R3 limits the size of that current - to about 5mA. = When current flows through the emitter-base junction of Q1 - it switches the transistor on. The transistor connects the negative side of the relay coil to ground - and the relay energizes. = When pin 10 goes high again - it`s connected internally to the positive line. So it connects the base of Q1 to the positive line - through R3. This cuts off the base current`s path to ground. = Without its base current - the transistor switches off. This disconnects the negative side of the relay coil from ground - and the relay de-energizes. = In other words - every time pin 10 goes low - the relay will energize. And every time pin 10 goes high - the relay will de- energize. Thus - the state of the relay is controlled by the state of pin 10. = The state of pin 10 is controlled by the voltage on pins 5 & 6. = The gates are being used as simple inverters. So - if pins 5 & 6 are high - pin 4 will be low. Pin 4 takes pins 1 & 2 low - so pin 3 will be high. Pin 3 takes pins 12 & 13 high - so pin 11 is low. Pin 11 takes pins 8 & 9 low - so pin 10 is high. = In other words - while pins 5 & 6 remain high - the relay will remain de-energized. The opposite is also...

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