How To Use A Cmos 4017 To Build A Sequential Timer

This circuit uses a Cmos 4017. The 4017 is a decade counter. The count starts at zero. And it advances by one - each time pin 14 is taken high. When the count reaches nine - it goes back to zero - and starts all over again. = As the count progresses - each of the output pins goes high in turn. The first is pin 3 - it represents zero. The second is
How To Use A Cmos 4017 To Build A Sequential Timer - schematic

pin 2 - it represents one. Pin 4 represents two - and so on. The number represented by each output pin is shown in red. = Although the 4017 will count up to nine - we are only using it to count up to four. Consequently - some of the output pins are unused. Unused Cmos output pins should always be left unconnected. = The circuit uses the first five outputs. These are - pins 3, 2, 4, 7 & 10 The first four outputs pins are each connected to a 2k7 resistor. With a 12v supply - the resistor limits the current available to 12v G· 2k7 = just under 5mA. = This is more than enough to operate a transistor switch. And the transistor can be used to sound a buzzer - or energize a relay. If you increase the current to about 10mA - it will drive an optical isolator. = To increase the current - you should reduce the value of the resistor. As a rough guide - it`s safe to draw up to about 1mA per supply volt. This means that the value of the resistor - should never be below 1k. = Every step in the sequence is the same. Each output charges C5 through a timing resistor. And - when the voltage on C5 takes pin 14 high - the sequence moves on to the next output. For clarity - I`ve only highlighted the first step in the sequence. Once you understand the first step - you`ll understand the remaining three steps as well. = Consider what happens when we turn on the power. The count begins at zero. In other words - pin 3 will be high. And the rest of...

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