An LDR (Light Dependent Resistor) lets you set the level of darkness at which this repeating timer will stop working.

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Circuit Description

Click Here For A Photograph Of The Prototype.

Circuit Diagram For
A Light Controlled
Repeating Timer

You can use a Cmos 4011 or a Cmos 4093 in this circuit.

Each has four - two-input - NAND gates.

Only three of the four gates are used - gate 3 is spare.

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Gates 1 & 2 are wired to form an astable oscillator.

For a detailed account of how the oscillator works - click the link below the diagram.

Basically - Gates 1 & 2 take turns at switching each other on and off.

The speed at which they do this is controlled by the value of C1 and the four resistors.
 
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The output from the oscillator is at pin 3.

While the oscillator is running - pin 3 changes back and forth between high and low.

As it does so - it takes the inputs of gate 4 high and low.

And this switches pin 11 on and off. 

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When pin 11 switches on - it's connected internally to the positive line.

And it supplies base current to Q1 - through R5.

This switches the transistor on.

The transistor connects the negative side of the relay coil to ground.

And the relay energizes.

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When pin 11 switches off again - it's connected internally to the negative line.

This means that it can no longer supply the transistor with base current.

So the transistor switches off.

And the relay de-energizes.

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   THE ROLE OF THE LDR

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The oscillator output is at pin 3.

To control the relay - pin 3 must be free to change back and forth between high and low.

If we prevent pin 3 from changing its state - it can't switch gate 4 on and off.

And - if pin 11 doesn't switch on and off - the transistor can't energize and de-energize the relay.

In other words - by prevent pin 3 from changing state - we can inhibit the relay output.

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We can prevent pin 3 from changing state - by taking pin 1 low.

While at least one of the gate 1 inputs is low - the output will remain high.

See the Truth Table.

In other words - while pin 1 remains low - pin 3 is forced to remain high.

And - since pin 3 can no longer switch gate 4 on and off - the relay output is disabled.

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We only want to disable the relay during the hours of darkness.

During the daytime - we want pin 1 to be high.

And after dark - we want pin 1 to be low.

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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 - during the daytime - we need pin 1 to be above 6-volts.

And - after dark - we need pin 1 to be below 6-volts.

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We can control the voltage on pin 1 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.

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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.

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

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

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

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Back to pin 1 and the LDR.

R6 & R7 form a potential divider with the LDR.

R6 only plays a minor role - and you can ignore it for the present.

So our potential divider is made up of R7 and the LDR.

And the voltage across R7 determines the state of pin 1.

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In the daylight - the LDR has a relatively low resistance when compared with R7.

Consequently - most of the 12v is across R7.

In other words - pin 1 is high.

So pin 3 is free to change its state - and the relay will energize and de-energize. 

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But - as it gets darker - the value of the LDR increases.

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

So there's less and less voltage remaining across R7.

That is the voltage across R7 falls.

And consequently - the voltage on pin 1 falls.

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Eventually - depending on the level of darkness and the setting of R7 - the voltage across the LDR rises above 6-volts.

This leaves less than 6-volts across R7.

That is - the voltage on pin 1 is less than 6-volts.

In other words - pin 1 is low. 

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While pin 1 remains low - pin 3 can't change its state.

So the relay is prevented from energizing and de-energizing.

It won't start up again until the light level rises.

Then - the value of the LDR will fall, and pin 1 will go high once more.

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In very bright light - the value of the LDR will fall to a few hundred ohms.

If R7 were accidentally adjusted to zero ohms - there's a chance that excess current might destroy the LDR.

R6 limits the maximum possible current - through the LDR - to about 10mA.

So it doesn't matter how bright it gets - or where you set R7. 

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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 - C2 & C3 are like a short circuit.

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

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        THE ROLE OF D4

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Changes in light levels can happen slowly.

So pin 1 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 it's supposed to be day or night.

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D4 helps to reduce the length of this period of indecision. 

While pin 3 has any slight tendency to go low - pin 11 will tend to go high.

As it does so - it will tend to take pin 1 high - through D4.

And a high pin 1 will help pin 3 to complete its journey low.

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It's a sort of positive feedback.

If pin 3 is still trying to go low - D4  will give it the push it needs.

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D4 also insures that the relay will remain energized for the whole of the final preset time period.

Without D4 - as it gets darker - the voltage on pin 1 will fall.

Half-way through the output period - pin 1 might become low.

This would take pin 3 high - and switch gate 4 off.

In other words - the relay might drop out before its time was up.

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D4 allows pin 11 to override the light sensor.

It holds pin 1 high for the whole of the final output period.

So the length of the final output period is determined in the usual way - by R1, R2 & C1.

And it's not cut short by the falling light level.

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We can't simply link pins 1 & 11.

We have to use a diode.

The diode creates a one-way path.

It allows a high pin 11 to take pin 1 high.

But it prevents a low pin 11 from taking pin 1 low.

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     THE ALTERNATIVE C1	

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While the oscillator is running - the polarity of the charge on C1 is constantly reversing.

Sometimes pin 3 is high and pins 5 & 6 are low.

And sometimes pin 3 is low and pins 5 & 6 are high.

That's why C1 needs to be non-polarised.

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A regular electrolytic capacitor will conduct DC current when it's biased in the wrong direction.

It could act like a resistor connecting pins 5 & 6 to pin 3.

So when pin 3 goes low - it will hold pins 5 & 6 low.

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

Pin 4 will hold pin 2 high - causing pin 3 to remain low.

In other words - the oscillator will not run - pin 3 will not change its state - and pin 11 will not switch on and off.

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Connecting two regular capacitors back-to-back will simulate a non-polarised capacitor.

Since one of the capacitors is always correctly biased - no DC current can flow through the pair.

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When a number of capacitors are connected in series - their combined capacitance is reduced.

The formula for capacitors connected in series - is similar to the formula for resistors connected in parallel.

1/Ct = 1/Ca + 1/Cb + 1/Cc .....

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Where just two capacitors are connected in series - the formula may be expressed as : -

     (The Product ÷ The Sum)
                        Or 
        (Ca x Cb) ÷ (Ca + Cb)

So two 1000uF capacitors - connected in series - have a combined capacity of : -

       (1000 x 1000) ÷ 2000 
                = 500uF

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When - as here - both capacitors have the same value - there's a simpler calculation.

Their combined capacitance is equal to half the value of one of the capacitors.

So two 1000uF capacitors - connected in series - have a combined capacity of : - 

         1000 ÷ 2 = 500uF.

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     THE ROLES OF D1 & D2

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D1 and D2 create one-way paths.

They allow independent control of the rate at which C1 charges and discharges.

R1 & R2 control the length of time required - for a high pin 4 - to take pins 5 & 6 high.

And R3 & R4 control the length of time required - for a low pin 4 - to take pins 5 & 6 low again.

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While pins 5 & 6 are high - pin 4 is low.

Pin 4 takes pin 2 low - so pin 3 is high.

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

While pin 11 is low - Q1 is switched off - and the relay is de-energized.

It will remain de-energized until C1 discharges into pin 4 - through R4, R3 & D1.

Thus - the length of time the relay remains de-energized - is controlled by R3 & R4.

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Similarly - while pins 5 & 6 are low - pin 4 is high.

Pin 4 takes pin 2 high - so pin 3 is low.

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

While pin 11 is high - Q1 is switched on - and the relay is energized.

It will remain energized until C1 is charged - by pin 4 - through R1, R2 & D2.

Thus - the length of time the relay remains energized - is controlled by R1 & R2.

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R1 and R4 are not strictly necessary to the operation of the circuit.

They set an upper limit on the frequency of the oscillator.

Their presence means that the oscillator will remain stable - even if R2 and R3 are adjusted to zero ohms.

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You can change the values of R1 and R4 to suit your own requirements.

You can choose values that take you close to your target times.

Then use lower value presets - for fine adjustment only.

If you don't need too fine an adjustment - you can rely solely on R1 and R4 to give you the time periods you require.

That way - you can leave out the pots altogether.

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I used an SPCO/SPDT 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.
 
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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. 

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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.

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The inputs of spare gates should always be connected to something.

If they're left free - the gate can switch rapidly and randomly.

And this causes unnecessary current consumption.

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The easiest solution here - was to leave the copper strips intact.

Pin 9 will change its state - along with pins 5 & 6.

But - because pin 8 is held permanently low - the output of gate 3 is forced to remain high  (See the Truth Table).

Since the gate 3 output isn't changing - it isn't wasting energy.

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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 D3 - 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.

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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.
 
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Learn More About How An Astable Oscillator Works

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Construction Guide

Click here if you're new to constructing stripboard projects.

The prototype of the daylight timer was built using only the Stripboard Layout as a guide. So - if you reproduce the layout - you will have a working timer. Details of how to Test Your Finished Circuit Board are also provided.

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 22 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 .

Repeating Timer No.3
Pattern For Cutting The Tracks

Repeating Timer No.3
Pattern For Cutting The Tracks

Actual Size

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 Nine Wire Links - the three presets - and the four 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.

Recurring Timer No.3- Make A

Then fit the four diodes and the transistor. Pay particular attention to the orientation of the diodes. See the Photograph Of The Prototype. Note that D3 is facing upwards. All the diodes are shown lying flat on the board - but D2 is actually mounted standing upright.

Next, fit the capacitors - the relay - the IC socket - and the LDR. If you are using a non-polarised capacitor connect it across pins 3 & 6 of the IC. If you are using two regular capacitors - pay particular attention to the orientation of each.

Repeat Timer No.3 - Make B

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 5 solder bridges.

Finish off by inserting the Cmos IC into the socket. Pin 1 of the IC should be in the top left-hand corner. Check that all 14 pins have entered the socket. Sometimes - instead of entering the socket - a pin will curl up under the IC.

You Are Now Ready To Test Your Finished Circuit Board.

General Information

Click Here For A Photograph Of The Prototype.

Repeat Timer No.3 Schematic	The Rest of Ron's Circuits	Write To Ron	More Free-to-Use Circuits	Circuit Exchange International

Repeat Timer No.3 - Support Material

Construction Guide

Actual Size

General Information