Test: Collector To Base Bias - UPSC MCQ

The collector current is analysed by the DC analysis of a transistor. It involves the DC equivalent circuit of a transistor. The base current is first found and the collector current is obtained from the relation, I_C=I_Bβ.

The collector to emitter voltage is obtained in order to find the operating point of a transistor. It is taken when there is no signal applied to the transistor. The point thus obtained lies in the cut off region when the transistor is used as a switch.

When in a transistor is driven into saturation, we use V_CE(SAT) as another linear parameter. In, addition when a transistor is biased in saturation mode, we have I_C < βI_B. This characteristic used to prove that the transistor is indeed biased in saturation mode.

The self destruction of a transistor due to increase temperature is called thermal run away. It is avoided by the negative feedback produced by the base resistor in a collector to base bias. The I_C which is responsible for the damage is reduced by decreased output signal.

Before the feedback is applied, when the temperature is increased, the reverse saturation increases. The collector current also increases. When the feedback is applied, the base current increases with decreasing collector current and the thermal runway too.

When the stability factor S=1, the collector resistor value should be very large when compared to the base resistor. So, when RC is large we need to provide large power supply which increases the cost. At the same time, as the base resistor is small we need to provide small power supply.

The resistor R_B can provide negative feedback for both AC and DC signals. The negative feedback for DC signal is done good as it can provide stable operating point. On the other side, the negative feedback is badly done for AC signal by decreasing the voltage gain.

We know, I_C=(V_CC-V_BE)/R_B
By putting the values, we have I_C=5.9mA. I_E=I_C/α. So, I_E=5.99mA.
V_CE= V_CC-RC(I_C+I_B). We have V_CE=6V.

 We know, by KVL -12+(I_C+I_B)1K+6+V_BE=0
We have I_E=5.3. I_C= αIE=5.24mA. From another loop, -12+I_EI_K+V_BE=0
We have, V_CE=12-5.3m*1000=6.7V. Hence the Q point is (6.7V, 5.3mA).

The base current I_B is zero when β value is large. So, the V_CE changes to V_CC-R_CI_C. The collector current I_C is changed to (V_CC-V_BE)/R_B from β(V_CC-V_BE)/(1+ β)R_E+ R_B.