This problem involves the Pass-Transistor circuit you will test in the Bipolar T

Question

This problem involves the Pass-Transistor circuit you will test in the Bipolar Transistors II Lab. This circuit is essentially an emitter follower and it is useful for the construction of stable constant voltage sources. For the following questions, use the resistor values shown in the figure below and assume the transistor has a very high input impedance (i.e., assume beta is very large). For Vin= 15 V DC, calculate Vx (the voltage at point X) and Vout. For Vin = 10 V DC, calculate Vx and Vout. Now to make the output more stable to input fluctuations, suppose you replace the 6.8k resistor with a reversed-biased 5.6 V Zener diode. For Vin= 15 V DC, calculate Vx and Vout. For Vin = 10 V DC, calculate Vx and Vout.

Explanation / Answer

Since the input impedance of the transistor is very high and also IBis negligible, all the current from source Vin flows through the 10k and 6.8k resistors.
For Vin = 15V, Using voltage divider rule VX = (15)(6800) / (6800 + 10000) = 6.07V Vout = VX + VBE For VBE = 0.7V Vout = 6.07 + 0.7 = 6.14V
For Vin = 10V, Using voltage divider rule VX = (10)(6800) / (6800 + 10000) = 4.04V Vout = VX + VBE For VBE = 0.7V Vout = 4.04 + 0.7 = 4.12V

Part B Replacing 6.8k resistor by a 5.6V zener diode, the voltage at X gets regulated
For Vin = 15V and Vin = 10V, VX = 5.6V Vout = 5.6 + 0.7 = 6.3V




For VBE = 0.7V Vout = 6.07 + 0.7 = 6.14V
For Vin = 10V, Using voltage divider rule VX = (10)(6800) / (6800 + 10000) = 4.04V Vout = VX + VBE For VBE = 0.7V Vout = 4.04 + 0.7 = 4.12V

Part B Replacing 6.8k resistor by a 5.6V zener diode, the voltage at X gets regulated
For Vin = 15V and Vin = 10V, VX = 5.6V Vout = 5.6 + 0.7 = 6.3V




For Vin = 10V, Using voltage divider rule VX = (10)(6800) / (6800 + 10000) = 4.04V Vout = VX + VBE For VBE = 0.7V Vout = 4.04 + 0.7 = 4.12V For VBE = 0.7V Vout = 4.04 + 0.7 = 4.12V

Part B Replacing 6.8k resistor by a 5.6V zener diode, the voltage at X gets regulated
For Vin = 15V and Vin = 10V, VX = 5.6V Vout = 5.6 + 0.7 = 6.3V

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