# Morse Code astable Oscillator

Posted on Jul 18, 2012

Astable or free-running multivibrators have been used in home-built amateur radio equipment for many years. The basic circuit is a two stage amplifier with AC-coupled feedback from output to input. One transistor stage is on ( conducting current ) while the other is off ( not conducting current ) until the stages switch conducting states repeatedly at a specific frequency. The oscillation frequency is set by the resistor and capacitor values connected to the base terminal of each stage. This RC network determines how long the transistor stays in the off position. Presented are two projects which utilize astable multivibrators built using the ubiquitous 2N3904 BJT. The first project is a code practice audio-frequency oscillator while the second is a simple , no-frills electronic keyer for keying a transmitter. Either circuit would be a great first project to learn how to build circuits using Ugly Construction.

Above is the schematic for a simple Morse code practice oscillator. This circuit was originally built with 2N3904 transistors, however many different NPN transistors could be substituted as required. Tracing the circuit from left to right first brings us to the multivibrator circuit which is composed of Q1 and Q2. The oscillation frequency of the multivibrator is ~ 700 hertz and is set by the RC network formed by the 100K resistor and the 0.01 uF capacitor connected to each transistor base terminal. The approximate time off for each transistor is given by the following formula: [ Time 0ff = 0.7 * R * C ] with R in ohms and C in farads. It maybe more practical to leave the resistance value fixed and vary the capacitor value to obtain a desired oscillation frequency. Rearranging the above formula allows this : [ Total Time Off = 1 / Frequency ] with Total Time Off being the total number of seconds that both transistors are off and Frequency is in hertz. Once the total time off is known, you must divide that answer by 2 as each transistor is off half of the total time off in this symmetrical circuit. Then you simply solve for the capacitor value: [ Time Half = Total Time Off / 2 ] [ Capacitor = Time Half / ( 0.7 * R )] with Capacitor answer in farads. Lets run the numbers to solve for the capacitor values in the schematic; R = 100K, desired frequency = 700 hertz. Total Time Off = 1 / F ----> 1 / 700 =...

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