BJT Problem: Calculating Output Voltage (Q.28) - Solution and Comparison

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Discussion Overview

The discussion revolves around calculating the output voltage in a BJT circuit, specifically addressing a problem labeled Q.28. Participants share their approaches to solving the problem and compare their results with a solution provided in a textbook.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant shares their calculations and asks for feedback on their approach, noting a discrepancy with the textbook solution.
  • Another participant agrees with the initial calculations but points out an oversight regarding the current through a resistor.
  • A different participant argues that the output voltage will decrease when the transistor is connected to the circuit, suggesting a different expected behavior.
  • Multiple participants highlight that the initial calculations were correct until a specific substitution was made, indicating a misunderstanding of the current values.
  • One participant presents a detailed calculation of the output voltage, providing specific values for currents and junction drops, arriving at a result of 7.4133 volts.
  • Another participant comments on the significant digits used in the calculations, suggesting that the precision may be excessive given the known variability of certain parameters.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the calculations and the interpretation of the circuit behavior. There is no consensus on the final output voltage, as multiple approaches and interpretations are presented.

Contextual Notes

There are unresolved issues regarding the assumptions made about current values and the impact of the transistor's connection on the output voltage. The discussion reflects varying degrees of precision in calculations and the significance of junction voltage drops.

Who May Find This Useful

Readers interested in BJT circuit analysis, output voltage calculations, and those seeking to understand common pitfalls in circuit interpretation may find this discussion relevant.

lazyaditya
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I have attached the images for the question and the way i attempted it please refer to it and tell me have i made any mistake.I have also attached the image for the solution provided by book, and my answer is not matching to the answer given in the book.

Question has asked to calculate the output voltage and it is (Q.28) for solution given in my book.
 

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For the circuit in your first pic, your working & consequent answer are correct.
 
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Correct. They forgot the current through the 4kΩ resistor (I2).
 
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Ok. Thank you.
 
Lazyaditya,

Your work was correct until the last line of the second picture you attached.

You substituted 430 uA for Ic which is not correct.

430uA is the total current = Ic + I(4 Kohm resistor).

The solution in the book is correct, you don't need to substitute each current value, you just calculate the voltage on 2K resistor, & substitute from 12 V -> you will get Vo.
 
^^^^ That does not seem right. If the transistor was not connected then 2.0 milli-amps would be flowing through the 4k Ω resistor making 8.0 volts at Vo. Once the transistor is connected in parallel to the the 4k Ω resistor voltage at Vo will only get lower than 8.0 volts.
 
hisham.i said:
Your work was correct until the last line of the second picture you attached.

You substituted 430 uA for Ic which is not correct.

430uA is the total current = Ic + I(4 Kohm resistor).

That is wrong. Posts #2 and #3 are right.
 
I computed Vo = 7.4133 volts. The Vbe forward junction drop is 0.60V at such low currents. Ib is 5.0V - 0.60V, divided by 1.0 megohm, which is 4.4 uA. Multiplying by beta value of 100 gives 0.440 mA for Ic, the collector current. This Ic equals the current in the 2.0 kohm resistor minus that in the 4.0 kohm resistor.

Using algebra and solving for Vo gives 7.4133V. The 2.0 kohm resistor current is 2.2933 mA, and the 4.0 kohm current is 1.8533 mA. The difference is 0.440 mA, exactly equal to collector current Ic.

Claude
 
cabraham said:
I computed Vo = 7.4133 volts. The Vbe forward junction drop is 0.60V at such low currents. Ib is 5.0V - 0.60V, divided by 1.0 megohm, which is 4.4 uA. Multiplying by beta value of 100 gives 0.440 mA for Ic, the collector current. This Ic equals the current in the 2.0 kohm resistor minus that in the 4.0 kohm resistor.

Using algebra and solving for Vo gives 7.4133V. The 2.0 kohm resistor current is 2.2933 mA, and the 4.0 kohm current is 1.8533 mA. The difference is 0.440 mA, exactly equal to collector current Ic.

Claude
At two Significant Digits, you both got 7.4 volts. :biggrin:
 
  • #10
Posts 1-8 have too many significant digits. Vbe is known to about 0.65 +/- 0.10 V. So post 9 is the best.
 

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