Two Simple Thermistor-Controlled Relay Circuits

Circuit No.1 - Description

Click Here For A Photograph Of The Prototype.

Circuit Diagram For A
Temperature Controlled Relay

 
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 circuit 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 true.

                 ============

If pins 5 & 6 are low - pin 4 will be high.

Pin 4 takes pins 1 & 2 high - so pin 3 will be low.

Pin 3 takes pins 12 & 13 low - so pin 11 is high.

Pin 11 takes pins 8 & 9 high - so pin 10 is low.

                 ============

In other words - while pins 5 & 6 remain low - the relay will remain energized.

                 ============

 THE ROLE OF THE THERMISTOR

                 ============

When the temperature is high - we want the relay to be energized.

So we want pins 5 & 6 to be low.

And when the temperature is low - we want the relay to be de-energized.

So we want pins 5 & 6 to be high

                 ============

A cmos input pin is high - when it's over half the supply voltage.

And a cmos input pin is low - when it's below half the supply voltage.

I've used a 12-volt supply in the diagram.

This means that - while the temperature is high - we need pins 5 & 6 to be below 6-volts.

And - while the temperature is low - we need pins 5 & 6 to be above 6-volts.

                 ============

We can control the voltage on pins 5 & 6 with a potential divider.

A potential divider is no more than a number of resistors connected in series.

A portion of the supply voltage appears across each resistor.

And the size of that portion depends on the value of the resistor concerned.

                 ============

For example - if we connect two 10k resistors in series across our supply - there will be a drop of 6v across each one.

In other words - the 12-volt supply potential will be divided into two equal portions - of 6-volts each.

                 ============

But the resistors don't have to be the same value.

When resistors with different values are connected in series - the voltage across each resistor - is in proportion to its value.

For example - if you connect a 10k resistor and a 15k resistor in series across the supply - two-fifths of the 12v will appear across the 10k resistor - and the remaining three-fifths will be across the 15k resistor. 

In other words - there'll be 4v8 across the 10k resistor - and 7v2 across the 15k resistor.

                 ============

Back to pins 5 & 6 and the thermistor.

R1 & R2 form a potential divider with the thermistor.

R2 only plays a minor role - and you can ignore it for the present.

So our potential divider is made up of R1 and the thermistor.

And the voltage across the thermistor determines the state of pins 5 & 6.

                 ============

When it's warm - the thermistor has a relatively low resistance when compared with R1.

Consequently - most of the 12v is across R1.

And only a relatively small portion of the 12v is across the thermistor.

That is - the voltage across the thermistor is less than 6 volts

In other words - pins 5 & 6 are low.

So pin 10 is low - and the relay is energized. 

                 ============

But - as it gets colder - the value of the thermistor increases.

And more and more of the supply voltage is dropped across it.

So there's less and less voltage remaining across R1.

That is - the voltage across R1 falls.

And the voltage across the thermistor rises.

Consequently - the voltage on pins 5 & 6 rises.

                 ============

Eventually - depending on the temperature and the setting of R1 - the voltage across the thermistor rises above 6-volts.

That is - the voltage on pins 5 & 6 rises above 6-volts.

In other words - pins 5 & 6 go high. 

So pin 10 goes high - and the relay de-energizes. 

                 ============

While pins 5 & 6 remain high - pin 10 will remain high.

So the relay will remain de-energized.

And it won't energize again until the temperature rises.

Then - the value of the thermistor will fall - and pins 5 & 6 will go low once more.

This will take pin 10 low - and the relay will energize.

                 ============

               THE ROLE OF R2

                 ============

At very high temperatures - the value of the thermistor will fall to a few hundred ohms or less.

If R1 were accidentally adjusted to zero ohms - there's a chance that excess current might destroy the thermistor.

The purpose of R2 is to limit the maximum possible current - through the thermistor - to about 10mA.

So it doesn't matter how hot it gets - or where you set R1. 

                 ============

        THE ROLES OF C1 & C2

                 ============

Cmos inputs react very fast.

The state of a cmos gate can be changed by high frequency interference.

This might be electro-magnetic waves picked up by the wiring - or the power lines.

To high frequencies - C1 & C2 are like a short circuit.

They conduct any stray signals to ground - before they can do any harm.

                 ============

               THE ROLE OF C3

                 ============

Changes in temperature can happen slowly.

So pins 5 & 6 can spend a significant time hovering around the 6-volt mark.

In this area - the status of the input is uncertain.

It's neither high nor low.

In other words - the circuit can't tell if the temperature is above or below the preset level.

                 ============

Although Cmos gates are digital devices -during this period of uncertainty - their outputs can behave in an analogue manner.

Instead of switching sharply from high to low - and vice-versa - the pin 10 voltage rises and falls slowly.

This may mean that your relay will energize and de-energize slowly.

And the uncertainty about its status can cause the relay coil to tend to oscillate.

In other words - the relay may rattle or buzz for several seconds.

C3 prevents the oscillation.

So the relay changes state cleanly - and without the noise.

                 ============


I used a single pole relay in the prototype - but you can use a multi-pole relay if it suits your application.

The circuit will work from a minimum of 5-volts up to a maximum of 15-volts. 

You just need to make sure that the coil of your relay will operate at the voltage you are using.

Also - the relay coil shouldn't draw more than about 50 mA.

Otherwise the transistor might be overloaded.
 
                 ============

With a 12-volt supply - I specified a minimum coil resistance of about 270 ohms. 

This gives a maximum current of 
    
          12 ÷ 270 = 45 mA.

45mA is well within the limits of the BC557. 

                 ============

There's nothing special about the BC557. 

Any small PNP transistor with a gain (hfe) greater than 100 and an Ic(max) of at least 100mA should do.

But remember that the pin configuration of your transistor may be different from that of the BC557.

                 ============

Coils of all sorts give off high reverse-voltage spikes that will destroy Cmos ICs. 

Always connect a diode across relay coils - and across any other device that might contain a coil. 

Note that the cathode of D1 - the side with the bar - is connected to the positive terminal of the coil. 

The small arrowhead indicates the direction in which the diode will conduct.

Thus, any reverse-voltage spikes will be short-circuited at source - before they can do any damage.

                 ============

Although static electricity can destroy Cmos ICs - they are extremely reliable in operation. 

Most modern versions will survive a fair degree of abuse.

But it's worth taking a couple of basic precautions. 

Avoid touching the pins - and don't overheat them with the soldering iron. 

A socket will reduce the chances of damaging the IC - and will make it easier to replace if necessary.

If you're using a socket - always switch-off the power before inserting or removing the IC.

Also - check that the IC is the right way round - and that all of the pins are correctly inserted into the socket. 

Sometimes - instead of entering the socket - a pin will curl up underneath the IC.
 
                 ============

Circuit No.2 - Description

Click Here For A Photograph Of The Prototype.

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 circuit is at pin 10.

When pin 10 goes high - it's connected internally to the positive line.

So it connects the base of Q1 to the positive line - through R3.

This allows current to flow through the transistor's base-emitter junction.

And R3 limits the size of that current - to about 5mA.

                 ============

When current flows through the base-emitter 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 low again - it's connected internally to the negative line.

So it connects the base of Q1 to ground - through R3.

This cuts off the transistor's base current.

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 high - the relay will energize.

And every time pin 10 goes low - 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 energized.

The opposite is also true.

                 ============

If pins 5 & 6 are low - pin 4 will be high.

Pin 4 takes pins 1 & 2 high - so pin 3 will be low.

Pin 3 takes pins 12 & 13 low - so pin 11 is high.

Pin 11 takes pins 8 & 9 high - so pin 10 is low.

                 ============

In other words - while pins 5 & 6 remain low - the relay will remain de-energized.

                 ============

 THE ROLE OF THE THERMISTOR

                 ============

When the temperature is low - we want the relay to be energized.

So we want pins 5 & 6 to be high.

And when the temperature is high - we want the relay to be de-energized.

So we want pins 5 & 6 to be low

                 ============

A cmos input pin is high - when it's over half the supply voltage.

And a cmos input pin is low - when it's below half the supply voltage.

I've used a 12-volt supply in the diagram.

This means that - while the temperature is high - we need pins 5 & 6 to be below 6-volts.

And - while the temperature is low - we need pins 5 & 6 to be above 6-volts.

                 ============

We can control the voltage on pins 5 & 6 with a potential divider.

A potential divider is no more than a number of resistors connected in series.

A portion of the supply voltage appears across each resistor.

And the size of that portion depends on the value of the resistor concerned.

                 ============

For example - if we connect two 10k resistors in series across our supply - there will be a drop of 6v across each one.

In other words - the 12-volt supply potential will be divided into two equal portions - of 6-volts each.

                 ============

But the resistors don't have to be the same value.

When resistors with different values are connected in series - the voltage across each resistor - is in proportion to its value.

For example - if you connect a 10k resistor and a 15k resistor in series across the supply - two-fifths of the 12v will appear across the 10k resistor - and the remaining three-fifths will be across the 15k resistor. 

In other words - there'll be 4v8 across the 10k resistor - and 7v2 across the 15k resistor.

                 ============

Back to pins 5 & 6 and the thermistor.

R1 & R2 form a potential divider with the thermistor.

R2 only plays a minor role - and you can ignore it for the present.

So our potential divider is made up of R1 and the thermistor.

And the voltage across the thermistor determines the state of pins 5 & 6.

                 ============

When it's warm - the thermistor has a relatively low resistance when compared with R1.

Consequently - most of the 12v is across R1.

And only a relatively small portion of the 12v is across the thermistor.

That is - the voltage across the thermistor is less than 6 volts

In other words - pins 5 & 6 are low.

So pin 10 is low - and the relay is de-energized. 

                 ============

But - as it gets colder - the value of the thermistor increases.

And more and more of the supply voltage is dropped across it.

So there's less and less voltage remaining across R1.

That is - the voltage across R1 falls.

And the voltage across the thermistor rises.

Consequently - the voltage on pins 5 & 6 rises.

                 ============

Eventually - depending on the temperature and the setting of R1 - the voltage across the thermistor rises above 6-volts.

That is - the voltage on pins 5 & 6 rises above 6-volts.

In other words - pins 5 & 6 go high. 

So pin 10 goes high - and the relay energizes. 

                 ============

While pins 5 & 6 remain high - pin 10 will remain high.

So the relay will remain energized.

And it won't de-energize again until the temperature rises.

Then - the value of the thermistor will fall - and pins 5 & 6 will go low once more.

This will take pin 10 low - and the relay will de-energize.

                 ============

               THE ROLE OF R2

                 ============

At very high temperatures - the value of the thermistor will fall to a few hundred ohms or less.

If R1 were accidentally adjusted to zero ohms - there's a chance that excess current might destroy the thermistor.

The purpose of R2 is to limit the maximum possible current - through the thermistor - to about 10mA.

So it doesn't matter how hot it gets - or where you set R1. 

                 ============

        THE ROLES OF C1 & C2

                 ============

Cmos inputs react very fast.

The state of a cmos gate can be changed by high frequency interference.

This might be electro-magnetic waves picked up by the wiring - or the power lines.

To high frequencies - C1 & C2 are like a short circuit.

They conduct any stray signals to ground - before they can do any harm.

                 ============

              THE ROLE OF C3

                 ============

Changes in temperature can happen slowly.

So pins 5 & 6 can spend a significant time hovering around the 6-volt mark.

In this area - the status of the input is uncertain.

It's neither high nor low.

In other words - the circuit can't tell if the temperature is above or below the preset level.

                 ============

Although Cmos gates are digital devices -during this period of uncertainty - their outputs can behave in an analogue manner.

Instead of switching sharply from high to low - and vice-versa - the pin 10 voltage rises and falls slowly.

This may mean that your relay will energize and de-energize slowly.

And the uncertainty about its status can cause the relay coil to tend to oscillate.

In other words - the relay may rattle or buzz for several seconds.

C3 prevents the oscillation.

So the relay changes state cleanly - and without the noise.

                 ============

I used a single pole relay in the prototype - but you can use a multi-pole relay if it suits your application.

The circuit will work from a minimum of 5-volts up to a maximum of 15-volts. 

You just need to make sure that the coil of your relay will operate at the voltage you are using.

Also - the relay coil shouldn't draw more than about 50 mA.

Otherwise the transistor might be overloaded.
 
                 ============

With a 12-volt supply - I specified a minimum coil resistance of about 270 ohms. 

This gives a maximum current of 
    
          12 ÷ 270 = 45 mA.

45mA is well within the limits of the  BC547. 

                 ============

There's nothing special about the BC547. 

Any small NPN transistor with a gain (hfe) greater than 100 and an Ic(max) of at least 100mA should do.

But remember that the pin configuration of your transistor may be different from that of the BC547.

                 ============

Coils of all sorts give off high reverse-voltage spikes that will destroy Cmos ICs. 

Always connect a diode across relay coils - and across any other device that might contain a coil. 

Note that the cathode of D1 - the side with the bar - is connected to the positive terminal of the coil. 

The small arrowhead indicates the direction in which the diode will conduct.

Thus, any reverse-voltage spikes will be short-circuited at source - before they can do any damage.

                 ============

Although static electricity can destroy Cmos ICs - they are extremely reliable in operation. 

Most modern versions will survive a fair degree of abuse.

But it's worth taking a couple of basic precautions. 

Avoid touching the pins - and don't overheat them with the soldering iron. 

A socket will reduce the chances of damaging the IC - and will make it easier to replace if necessary.

If you're using a socket - always switch-off the power before inserting or removing the IC.

Also - check that the IC is the right way round - and that all of the pins are correctly inserted into the socket. 

Sometimes - instead of entering the socket - a pin will curl up underneath the IC.
 
                 ============

The prototypes of both circuits were built using the same construction guide. The two layouts are virtually identical. The only difference between them is the type and orientation of the transistor.

The terminals are a good set of reference points. To fit them - you may need to enlarge the holes slightly. Then turn the board over and use a felt-tip pen to mark the 17 places where the tracks are to be cut. Before you cut the tracks - use the "actual size" drawing to Check That The Pattern is Correctly Marked .

When you're satisfied that the pattern is right - cut the tracks. If you don't have the proper track-cutting tool - then a 6 to 8mm drill-bit will do. Just use the drill-bit as a hand tool - there's no need for a drilling machine. Make sure that the copper strip is cut all the way through. Sometimes a small strand of copper remains at the side of the cut and this will cause malfunction. Use a magnifying glass - and backlight the board. It only takes the smallest strand of copper to cause a problem.

Next - fit the Seven Wire Links - the preset - and the two fixed resistors. For the links - I used bare copper wire on the component side of the board. Telephone cable is suitable - the single stranded variety used indoors to wire telephone sockets. Stretching the core slightly will straighten it - and also allow the insulation to slip off.

Next - fit the diode and the three capacitors. Pay particular attention to the orientation of both the diode and the electrolytic capacitor. See the Photograph Of The Prototype. Then - fit the relay - the IC socket - and the thermistor.

Turn the board over and examine the underside carefully - to make sure that there are no unwanted solder bridges or other connections between the tracks. If you backlight the board during the examination - it makes potential problem areas easier to spot. When you're satisfied that everything is in order - add the 6 solder bridges.

Temperature Controlled Relays - Support Material

Construction Guide

Actual Size

Fit The Transistor

Circuit No.1 - BC557 PNP

Circuit No.2 - BC547 NPN