Connecting LEDs in series is very efficient. You can light several LEDs - with the same amount of energy it takes to light just one. The reason is simple. With one LED - the series resistor wastes most of the energy - in the form of heat. But when you connect several LEDs in series - the energy that would have been wasted in the resistor - is turned into light instead.

The more LEDs you connect in series - the smaller the voltage drop across the resistor. And the smaller the voltage drop across the resistor - the smaller the amount of power it has to dissipate - i.e. waste.

Fake Alarm - Circuit No.1

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 Fake Alarm No.1

CIRCUIT DESCRIPTION With a 12v supply - you can flash up to 10 LEDs. With a 9v supply - the maximum is six. With a 6v supply - the maximum is four. The reason for the differences - is the voltage drop across the LEDs. *********************** Each LED has a forward voltage drop of about 2v across it. So 2 LEDs will have a 4v drop and 3 LEDs will have a 6v drop etc. When five LEDs are connected in series - the drop is about 10v. *********************** The power that drives the LEDs comes from pins 10 & 11 of the IC. When a Cmos output pin goes high - it's connected internally to the positive line. And current flows from the output pin - through Rx and the LEDs - to ground. The amount of current is regulated by the value of Rx. *********************** To glow at maximum brilliance - LEDs usually require a current of about 20mA. But the modern clear types will glow brightly at less than half that current. So we can have an effective display without coming anywhere close to overloading the Cmos outputs. *********************** When a Cmos output pin is high - it's connected internally to the positive line. It's not connected directly to the positive line. It's 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 a Cmos gate as having an internal 200 ohm resistor (Ri) in series with its output pin. This resistor is also in series with Rx and the LEDs. And it will reduce the current flowing through the LEDs. *********************** The following calculations are for a 12-volt supply. But the same approach can be used for lower supply voltages. *********************** The total voltage - across Ri, Rx and the LEDs - is 12v. A single LED has 2v across it. Therefore - the remaining 10v must be across Ri & Rx. And the current flowing through the LED - is the same as the current flowing through Ri & Rx. That is: V ÷ R = 10 ÷ (200R + 1k) = 8mA. *********************** Two LEDs have 4v across them - and the remaining 8v is across Ri & Rx. So the current flowing through the two LEDs is:- 8 ÷ (200R + 820R) = 8mA. *********************** Three LEDs have 6v across them - and the remaining 6v is across Ri & Rx. So the current flowing through the three LEDs is:- 6 ÷ (200R + 680R) = 7mA. *********************** Four LEDs have 8v across them - and the remaining 4v is across Ri & Rx. So the current flowing through the four LEDs is:- 4 ÷ (200R + 330R) = 7mA. *********************** Five LEDs have 10v across them - and the remaining 2v is across Ri & Rx. So the current flowing through the five LEDs is:- 2 ÷ (200R + 100R) = 7mA. *********************** With this sort of application - where the diodes are NOT being used to provide illumination - this level of current is usually perfectly adequate. *********************** THE OSCILLATOR *********************** Gates 1 & 2 are connected to form a simple oscillator. Its output - at pin 3 - is constantly switching from high to low - and vice versa. Pin 3 is connected to the inputs of gates 3 & 4 - and it switches the gates on and off. The speed at which it does so - is set by the values of C1 and R1. *********************** Note that each gate has its two inputs connected together. In this configuration - the gates are being used as simple inverters. In other words - you don't have to use the specified gates. Any type of Cmos inverter will do. However - the Schmidt-Trigger versions will flash the LEDs slightly more slowly. *********************** HOW THE OSCILLATOR WORKS *********************** 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 C1 & R1. *********************** Note that C1 & R1 are connected in series between pins 3 & 4. That is - pin 4 is connected to the resistor. The resistor is connected to the capacitor. And the capacitor is connected to pin3. *********************** Note also that pins 5 & 6 are connected to the junction of C1 & R1. The voltage at this junction controls the voltage on pins 5 & 6. And the voltage on pins 5 & 6 switches gate 2 on and off. *********************** Let's say we begin with pin 4 high and pin 3 low. This means that the junction of C1 & R1 represents the positive side of the capacitor. As pin 4 charges the capacitor - the voltage at the junction of C1 & R1 rises. When it reaches just over half of the supply voltage - it takes pins 5 & 6 high. *********************** When pins 5 & 6 go high - pin 4 goes low. Pin 4 takes pins 1 & 2 low - so pin 3 goes high. *********************** Note that the polarity of the charge across C1 & R1 has reversed. Pin 4 is low - and pin 3 is high. C1 starts to charge once more. But this time in the opposite direction. *********************** Since pin 3 is now high - and pin 4 is low - the junction of C1 & R1 represents the negative side of the capacitor. As the capacitor charges - the voltage on this junction will fall. This means that the voltage on pins 5 & 6 will fall. *********************** When it falls to just below half the supply voltage - the inputs change from high to low. And the output of gate 2 will go high. Pin 4 will take pins 1 & 2 high. So pin 3 will go low. *********************** Note that the polarity of the voltage across C1 & R1 had changed once more. It's back to where we started. Pin 3 is low - pin 4 is high - and the junction of C1 & R1 once again represents the positive side of the capacitor. In other words - the first cycle is complete. And the second cycle is about th start. *********************** The oscillator runs because the polarity of the charge on C1 is constantly reversing. As a result - the junction of C1 & R1 keeps taking pins 5 & 6 back and forth between high and low. And since pins 5 & 6 are constantly changing back and forth between high and low - the polarity of the charge on C1 keeps reversing. It's a sort of controlled feedback - where the frequency generated is fixed by the values of C1 & R1. ***********************

Fake Alarm - Circuit No.2

The last circuit will flash up to two groups of 5 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.

CIRCUIT DESCRIPTION This circuit is almost exactly the same as the first - so you should begin by reading the description of Fake Alarm No.1 The only difference here is that the two LEDs - or two groups of LEDs - will flash alternately. *********************** Note that pin 3 is no longer connected to the inputs of gate 3. They are connected instead to the gate 4 output. Along with flashing it's own group of LEDs - pin 11 also inverts the output from the oscillator. *********************** When pin 11 takes pins 8 & 9 high - pin 10 will go low. And - when pin 11 takes pins 8 & 9 low - pin 10 will go high. Thus - the state of pin 10 will always be opposite to that of pin 11. In other words - the two pins will flash their LEDs alternately. ***********************

Warning Signal - Circuit No.3

This circuit is like the previous one - except that it will flash up to 15 groups of 5 LEDs per output. The LEDs will also flash at a faster rate - and they will glow at their full brilliance. Unlike the more discrete "alarm fitted" indicators - this circuit is designed to get your attention.

CIRCUIT DESCRIPTION This circuit is almost exactly the same as the first two - so begin by reading the descriptions of Fake Alarm Nos. 1 & 2 The difference here - is that this circuit allows you to increase both the size and the brightness of the display. Instead of drawing the LED current from the IC - this circuit draws it directly from the power supply. So you can light more LEDs - and drive them at their full brilliance. *********************** When the Cmos outputs go high - they supply a small base current to the transistors. This switches the transistors on. The transistors connect the negative side of the diodes to ground. And the battery supplies current to the LEDs - through the series resistors. *********************** Maximum brightness is typically achieved with a current of about 20mA. The values of Rx were chosen to keep the current - through the LEDs - just below this figure. *********************** The calculations here are similar to those described in Fake Alarm No.1 With a 12v supply - there will be 10v across the five LEDs - and 2v across the resistors. So the current through the LEDs will be about 2 ÷ 120R = 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 also. *********************** The frequency of the oscillator - the speed at which pins 10 & 11 go high and low - is set by the values of C1 & R1. Here - the value of R1 is roughly halved. So the frequency of the oscillator is roughly doubled. In other words - the LEDs will flash about twice as fast. *********************** THE TRANSISTORS *********************** Each 2k7 resistor supplies a base-current of about 4mA to its transistor. E.g. 12v ÷ (Ri + R2) = 4mA 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 ÷ 4 = 15. 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. 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 ÷ 4 = 75. The BC337 easily satisfies both requirements. ***********************

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