Question

THEORY: Part 2-Relaxation Oscillator (See q() for an RC circuit in textbook.) Gas Tube An interesting circuit that produces a pulsating direct voltage car be constructed as shown in Figure 5. It is used to operate flashing neon bulbs, such as those found in blinking highway signs stroboscope, etc. It is also the basis of a type of oscillating circuit called a multivibrator. The circuit is often called a "relaxation oscillator, and consists of a resistor and capacitor in series with a DC emf A gas tube (such as a neon bulb) is in parallel with the capacitor. ε-loo 4μF can capacitor HD box on high scale しへ Before explaining the action of the circuit, we need to consider the behavior of a gas-filled tube such as the neon bulb. Such tubes do not conduct any current (i.e., there is no appreciable ionization in the tube) until the voltage across the tube is raised to some critical value called the "firing potential" Vf (usually around 75 volts). After the tube starts conducting it will continue conducting, even though the voltage across it is lowered, until it is lowered to some critical value called the "extinction potential", Ve, (usually around 55 volts). The values of Ve and Vr depend upon the gas in the tubes and the geometry of its electrodes, but they are about the same for similar tubes. When the tube conducts (or, as we often say, when a discharge is set up in the tube), light is produced because of the excited energy levels to which the gas atoms are raised. Relaxation Oscillator Consider now how the circuit of Figure 5 will cause the gas tube to start conducting at regular intervals, thus producing light flashes of a fixed frequency. This will be apparent from studying Figure 6, which shows how the potential Ve across the capacitor (and also across the gas tube) varies with time When the key is closed, the charge on the capacitor starts accumulating according the the well-known exponential law for RC series circuits, and the voltage across C (V. q/C) follows a similar curve, shown as OHAK in figure 6. This same voltage acts across the tube at all instants, but the tube does nothing until this voltage has increased to Vr, indicated by point A. The tube then starts conducting current and thus very quickly discharges the capacitor, as shown by AB ure This discharge time (during which the light is produced) is so small, compared to the charge tim that it can be taken as zero, as we show it on the graph, where ta and ta are at the same abscissa

Please get Flashking Frequency formula.

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Answer 1