BJT output voltage calculation not matching textbook answer

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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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  • Textbook problem 2.28: NPN transistor circuit with 5 V source through 1 MΩ base resistor, +12 V supply through 2 kΩ collector resistor, hFE = 100, and 4 kΩ load to ground at output Vo. Asks for output voltage Vo.
    photo 1.JPG
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  • Handwritten working on lined paper, rotated sideways, showing a BJT circuit sketch with 12V supply, 2kΩ and 1kΩ resistors, 10kΩ and 100 resistances, and an output node Vo, followed by equations for IB, IC = βIB = 100(43μA) = 4.3mA, and nodal equations for the voltage Vo.
    photo 2.JPG
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  • Handwritten calculation on lined paper: 24 - 2v0 - v0 = 4.30 mA, 4kΩ; 24 - 3v0 = 43.0 x 10^-6 x 4 x 10^3; 24 - 3v0 = 172.0 x 10^-3; 24 - 1.72 = 3v0; 22.28 = 3v0; v0 = 7.4266 volts, boxed.
    photo 3.JPG
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  • Handwritten solution to a BJT problem: I_CS = 9.7 mA; V_C = V_CC = +5 V, V_E = 0 V, V_CE = V_C - V_E = 5 V; I_b = (5 - 0.7)/10^6 = 4.3 µA; I_c = βI_B = 100 × 4.3 µA = 0.43 mA; V_o = 12 - 2 × 0.43 = 12 - 0.86 = 11.14 V.
    photo(7).JPG
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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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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: