The first circuit is described in complete detail. The remaining two circuits are only slightly different from the first - and their descriptions deal only with those differences. Therefore - to understand how the later circuits work - you should begin with Doorbell No.1

The circuit uses a Cmos 4001. It has four two-input NOR gates. Gates 1 & 2 form a simple oscillator. The frequency of that oscillator is set by C2 & R3. *********************** The output from the oscillator is at pin 3. When the oscillator is running - pin 3 keeps changing back and forward between high and low. *********************** Pin 3 is connected to the inputs of gates 3 & 4. When pin 3 goes high - it takes the inputs high. And when pin 3 goes low - it takes the inputs low. *********************** Gates 3 & 4 each have their two inputs connected together. In this configuration - the gates are being used as simple inverters. When the inputs are high - the outputs are low. When the inputs are low - the outputs are high. *********************** Every time pin 3 takes the inputs of gates 3 & 4 low - pins 10 & 11 go high. When pins 10 & 11 go high - they're connected internally to the positive line - and supply current to the LEDs - through Rx. *********************** So - when the oscillator is running - gates 3 & 4 are switching on and off. And pins 10 & 11 are switching the LEDs on and off. In other words - the LEDs connected to pins 10 & 11 - are flashing. *********************** Each LED has a forward voltage drop of about 2v across it. You can think of the LED as a sort of 2v zener diode. When they are connected in series - 2 LEDs will have a 4v drop. And 3 LEDs will have a 6v drop etc. *********************** The power that drives the LEDs comes from pins 10 & 11 of the IC. To glow at maximum brilliance - LEDs usually require a current of about 20mA. This is a bit too much to draw from Cmos output pins. *********************** A good rule of thumb is that the maximum current should not exceed 1mA per supply volt. Modern clear types of LED will glow brightly at these reduced levels of current. So we can have an effective display without coming anywhere close to overloading the Cmos outputs. *********************** The purpose of Rx is to limit the current the LEDs draw from the output pins. However - Rx is not the only resistance in the circuit. *********************** When pins 10 & 11 are high - they're connected internally to the positive line. The output pins are not connected directly to the positive line. They're connected through a metal-oxide transistor. And the transistor offers some resistance to the passage of current. Typically - it's the equivalent of about 200 ohms. *********************** So you can think of each gate as having an internal 200 ohm resistor - "Ri" - in series with its output pin. Ri is also in series with Rx and the LEDs. And it will reduce the current flowing from the output. *********************** The total voltage across Ri, Rx and the LEDs is 9v. A single LED has 2v across it. Therefore, the remaining 7v must be across Ri & Rx. And the current flowing through the LED is the same as the current flowing through the two resistors. That is: - V ÷ (Ri + Rx) = 7 ÷ (200R + 560R) = 9mA. *********************** Two LEDs have 4v across them - and the remaining 5v is across the two resistors. So the current flowing through the two LEDs is:- 5 ÷ (200R + 330R) = 9mA. *********************** Three LEDs have 6v across them - and the remaining 3v is across the two resistors. So the current flowing through the three LEDs is:- 3 ÷ (200R + 150R) = 9mA. *********************** HOW THE OSCILLATOR WORKS *********************** So far - we have only described what happens when the oscillator is running. But what is an oscillator - and what makes it run? *********************** The essence of an oscillator is that its output is constantly switching back and forth from one state to another. The output from our oscillator is at pin 3. Pin 3 changes back and forth - from high to low - because gate 2 keeps switching gate 1 on and off. The speed at which gate 2 switches gate 1 on and off - is controlled by R3 and C2. *********************** In standby mode R2 holds pin 6 high. This forces the gate 2 output to remain low. See the truth table. While at least one input is high - the output will be low. *********************** Because the gate 2 output is prevented from changing its state - it cannot change the state of gate 1. The result is that pin 4 holds pins 1 & 2 low - so the output of gate 1 is high. Pin 3 holds the inputs of gates 3 & 4 high - so pins 10 & 11 are low. In other words - both groups of LEDs are switched off. *********************** Note that C2 & R3 are connected in series between pins 3 and 4. Note also that pin 5 is connected to the junction between the resistor and capacitor. It's the voltage at this junction that controls the voltage on pin 5. And it's the voltage on pin 5 that switches gate 2 on and off. *********************** In standby mode - pin 3 is high - and pin 4 is low. So - in effect - pin 5 is connected to the negative plate of the capacitor. Current from pin 3 charges the capacitor - and the voltage at the junction of C2 & R3 falls to zero. Since the voltage at the junction of C2 & R3 is zero - pin 5 is low. *********************** When the push switch is operated - two things happen. Firstly - the switch connects the negative lead of the buzzer to ground. This causes the buzzer to sound. It will go on sounding until the button is released. *********************** The second job of the push switch is to discharge C1 through R1 & D2. This takes pin 6 low. It will remain low until C1 charges again - through R2. This takes about 30-seconds. *********************** During this time - pin 4 is no longer forced to remain high. So the oscillator is no longer disabled. In other words - the oscillator will start to run - and the LEDs will flash. *********************** You'll recall that in standby mode - pin 5 is already low. When pin 6 goes low - both of the gate 2 inputs are low. So the output at pin 4 will go high. *********************** Pin 4 takes pins 1 & 2 high - so the gate 1 output will go low. Note that the polarity of the voltage across C2 & R3 has changed. Pin 3 is now low - and pin 4 is high. C2 begins to charge once more - but this time in the opposite direction. *********************** Since pin 3 is now low - and pin 4 is high - the junction of C2 & R3 represents the positive side of the capacitor. As the capacitor charges - the voltage on this junction will rise. This means that the voltage on pin 5 will rise. *********************** When it reaches just over half the supply voltage - about 5-volts - the input changes from low to high. Now that one of its inputs is high - the output of gate 2 will go low. Pin 4 will then take pins 1 & 2 low. So pin 3 will go high. *********************** Note that the polarity of the voltage across C2 & R3 had changed once more. Now pin 3 is high and pin 4 is low. Since pin 3 is high - the junction of C2 & R3 once again represents the negative side of the capacitor. *********************** As the capacitor charges - the voltage on this junction will fall. This means that the voltage on pin 5 will fall. When it reaches just under half the supply voltage - about 4-volts - the input changes from high to low. The output of gate 2 will once again go high. It will take pins 1 & 2 high - and pin 3 will go low. *********************** For about 30-seconds - until C1 charges through R2 - the oscillator will continue to run. The oscillator runs because the polarity of the charge on C2 is constantly reversing. As a result - the junction of C2 & R3 keeps taking pin 5 back and forth between high and low. And since pin 5 is constantly changing back and forth between high and low - the polarity of the charge on C2 keeps reversing. It's a sort of controlled feedback - where the frequency generated is fixed by the values of C2 & R3. Until C1 charges through R2 - every time pin 3 takes pins 8, 9, 12 & 13 low - pins 10 & 11 will go high - and switch the LEDs on. *********************** Some bells and buzzers can be a source of high reverse voltage spikes. These spikes will destroy Cmos ICs. D1 is included to short-circuit the spikes at source - before they can do any damage. *********************** D2 creates a one-way path. It allows the push switch to discharge C1. But it prevents current through the buzzer from charging C1 - through R1. In other words - the time it takes for C1 to charge - is set by R2 alone. *********************** R2 & R3 can both be variable resistors. They can be used to vary both the speed of the flashes - and the length of time the LEDs will go on flashing. *********************** R1 does two jobs. It limits the C1 peak discharge current through D2 and the push switch. It also protects D2 and the push switch from a possible short circuit. This could arise if a variable R2 were accidentally adjusted to zero ohms. ***********************

Doorbell Circuit No.2

The last circuit will flash up to two groups of 3 LEDs in tandem. This circuit will flash the two groups alternately. The alternate flashing creates the illusion of movement - and makes the display more eye-catching. Note that - although I've drawn the two groups of LEDs side by side - the individual LEDs can be mounted in any pattern you like

CIRCUIT DESCRIPTION This circuit is almost exactly the same as the first - so you should begin by reading the description of Doorbell No.1 *********************** The only difference here is the connections to the gate 3 inputs. In the previous circuit - pins 8 & 9 are connected to pins 12 & 13. That is - pins 8 & 9 are connected to the inputs of gate 4. So - when gate 4 switches on - gate 3 will switch on also. And when gate 4 switches off - so too will gate 3. In other words - the two groups of LEDs flash at the same time. *********************** In this circuit - pin 8 is connected to the gate 4 output. As well as lighting its own LEDs - pin 11 is inverting the output from the oscillator. While the oscillator is taking pins 12 & 13 low - the gate 4 output is taking pin 8 high - and vice versa. *********************** So the inputs of gates 3 & 4 - and consequently their outputs - are always the opposite of one another. When pin 11 is high - pin 10 is low. When pin 11 is low - pin 10 is high. In other words - the two groups of LEDs flash alternately. *********************** In this circuit - R2 is doing two separate jobs. When it charges C1 - it both disables the oscillator - and it switches the gate three output off. *********************** When the oscillator stops - pin 3 is high - pins 12 &13 are high and pin 11 is low. Therefore - the LEDs connected to pin 11 are switched off. *********************** But the two groups of LEDs flash alternately. So - when the pin 11 group is switched off - the pin 10 group will be switched on. And this means that in standby mode - half of the LEDs would be lighting. Apart from the possible annoyance - this would have a serious effect on battery life. *********************** R2 not only disables the oscillator - it also switches the gate 3 output off. When R2 takes pin 9 high - it forces pin 10 to go low. In other words - it switches the pin 10 group of LEDs off. With all the LEDs off - the standby current is virtually zero - and battery life is excellent. ***********************

Doorbell Circuit No.3

The main difference between this circuit and the last one - is the addition of the two transistor switches. The switches will each flash up to 15 groups of 3 LEDs. And - because they are getting power directly from the battery - the LEDs will glow at their full brilliance.

CIRCUIT DESCRIPTION This circuit is almost exactly the same as the first two - so begin by reading the descriptions of Doorbells Nos. 1 & 2 *********************** The only difference here is that this circuit makes more current available. The current comes directly from the power supply - and not from the IC. This allows you to flash more LEDs. It also allows you to drive them at their full brilliance. *********************** When the output pin goes high - it supplies a small base current to the transistor switch. This will switch the transistor on. The transistor will connect the negative side of the LEDs to ground. And the battery will supply current to the LEDs - through the series resistors. *********************** Maximum brightness is typically achieved with a current of about 20mA. The value of the resistors was chosen to keep the current through the LEDs just below 20mA. *********************** The calculations here are similar to those described in Circuit No.1 With a 9v supply - there will be 6v across the three LEDs - and 3v across the resistors. So the current through the LEDs will be about 3 ÷ 180R = 17mA. With fewer LEDs - the voltage across the resistors is higher. So - for any given current - the value of the resistors has to be higher. *********************** Each 2k7 resistor supplies a base-current of about 3mA to its transistor. And each group of LEDs takes less than 20mA from the supply. So with three groups of LEDs - there's a total current of about 60mA flowing through each transistor. *********************** This means that the DC gain (hfe) of the transistor needs to be at least 60 ÷ 3 = 20. It also means that the transistor must be able to pass a collector current (Ic) of at least 60mA. This is well within the capability of most - if not all - small NPN transistors. *********************** I used BC337s because they're cheap and widely available. They'll also allow you to increase substantially - the number of LEDs in your display. *********************** The BC337 has an Ic(max) of 500mA. And it can have a DC gain (hfe) of anything from 100 to 800. *********************** This means that you can easily connect up to 15 groups of LEDs to each transistor. If each group requires less than 20mA - the total current through the transistor is below 15 x 20 = 300mA. And the gain required is less than 300 ÷ 3 = 100. The BC337 easily satisfies both requirements. ***********************

Changing The Supply Voltage

I've drawn the circuits with a 9-volt supply. However - they can be adapted to a 6-volt or a 12-volt supply. This may involve changing the number of LEDs in the groups - and the value of Rx will need to be re-calculated. The following example is for Circuit No.3.

The LEDs have a forward voltage drop of about 2v each. With the 9-volt supply and three LEDs - you have 6-volts across the LEDs and 3-volts across the resistor. The current flowing through the three LEDs is the same as the current flowing through the resistor. And the current flowing through the resistor is V ÷ R. That is: -

3v ÷ 180Ω = 17mA.
With the 6-volt supply you should use groups of two LEDs. Then there's a total of 4-volts across the LEDs and 2-volts across the resistor. Change the value of the resistor to 120 ohms - and the current flowing through the two LEDs is: -

2v ÷ 120Ω = 17mA.
With the 12-volt supply you can use groups of five LEDs. Then there's a total of 10-volts across the LEDs and 2-volts across the resistor. Again - with a 120 ohm resistor - the current flowing through the five LEDs is: -

2v ÷ 120Ω = 17mA.