![]() The transistor is now working within (mostly) its linear range, and for the full duration of the input signal cycle, its current gain is nearly constant hfe=100. Base bias currents from R1, R2 (about 25uA) now dominate over transistor base current. If you decrease input voltage from 0.1V peak to 0.01V peak, base current becomes more sinusoidal shape, with average value near 2.5uA. ![]() Logic circuits and switching circuits use BJT. When over-driven by the signal generator, current into the base peaks at 44uA - this current dominates the static bias current of 25uA, and causes base voltage to change from its static value ( where you have no AC input). BJT finds application in clipping circuits so that the waves can be shaped. As these circuits are used only for clipping input waveform as per the requirement and for transmitting the waveform, they do not contain any energy storing element like a capacitor. Base bias current of R1, R2 is only about 25uA with no input signal. The clipper circuit can be designed by utilizing both the linear and nonlinear elements such as resistors, diodes, or transistors. The difference between I(R1) and I(R2) is base current: about 10uA. When the transistor is over-driven, the base-emitter junction pulls a lot more current on positive peaks, because during these positive peaks the transistor has no current gain. The base biasing resistors R1, R2 weakly establish base voltage because their values are large. You are driving the base with an input signal too large for linear operation where the transistor has current gain. But perhaps you're investigating behaviour of an overdriven amplifier. ![]() I'm assuming that you're trying to create a linear amplifier. ![]()
0 Comments
Leave a Reply. |