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Operation - unijunction transistor:
Operation : Imagine that the emitter supply voltage is turned biased and a small emitter reverse current flows. Then the intrinsic stand off voltage reverse biases the emitter diode, as mentioned above. If VB is the barrier voltage of the emitter diode, then the total reverse bias voltage is VA + VB = VBB + VB. For silicon VB = 0.7 V. Now let the emitter supply voltage VE be slowly increased. When VE becomes equal to n VBB, n will be reduced to zero. With equal voltage levels on each side of the diode, neither reverse nor forward current will flow. When emitter supply voltage is further increased, the diode becomes forward biased as soon as it exceeds the total reverse bias voltage. This value of emitter voltage VE is called the peak point voltage and is denoted by Vp. When VE = VP, emitter current IE starts to flow through RB to ground. This is the minimum current that is required to trigger the UJT. This is called the peak point emitter current and is denoted by Ip. Ip is inversely proportional to the inter base voltage, VBB. Now when the emitter diode starts conducting, charge carriers are injected into the RB region of the bar. Since the resistance of a semiconductor material depends upon doping, the resistance of region RB decreases rapidly due to additional charge carriers. With this decrease in resistance, the voltage drop across RB also decreases causing the emitter diode to be more heavily forward biased. This, in turn, results in larger forward current, and consequently more charge carriers are injected causing still further reduction in the resistance of the RB region. Thus the emitter current goes on increasing until it is limited by the emitter power supply. Since VA decreases with the increase in emitter current, the UJT is said to have negative resistance characteristic. It is seen that the base 2 (B2) is used only for applying external voltage VBB across it. Terminals E and B1 are the active terminals. UJT is usually triggered into conduction by applying a suitable positive pulse to the emitter. It can be turned off by applying a negative trigger pulse.
Use Norton Theorem, find the current flow through resistor R=10Ω.
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