Zener Diode Q-Point Calculation

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The Q-point for the Zener diode in the figure given is ?
 
Attachments
  • Circuit schematic of a Zener diode regulator: 20 V source, 11 kΩ series resistor, Zener diode labeled Vz = 4 V and Rz = 0, and 3.6 kΩ load resistor in parallel with the diode.
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  • Handwritten solution to problem 21.39: a 20 V source feeds an 11 kΩ series resistor to a Zener diode shunting a 3.6 kΩ load, with IL = 4/3.6 kΩ = 1.11 mA, I = (20 - 4)/11 kΩ = 1.4545 mA, I2 = 0.34 mA and V2 = 4 V boxed.
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But sir my answer does no matches any option,and solution to this problem was given by making thevenin equivalent of the circuit across zener diode !
 
Attachments
  • Multiple-choice answer options from a printed textbook page under the heading 'de Circuits': (A) (0.34 mA, 4 V), (B) (0.34 mA, 4.93 V), (C) (0.94 mA, 4 V), (D) (0.94 mA, 4.93 V).
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lazyaditya said:
But sir my answer does no matches any option,and solution to this problem was given by making thevenin equivalent of the circuit across zener diode !

It doesn't match option A?
 
lazyaditya said:
Answer given behind book is option (B)

Book answer is wrong. If Vz = 4.9V the zener would be burned to a crisp! (internal R = 0).
 
I have attached the solution given behind the book, please tell me if the solution is valid and can i use it in other problems similar to this ?
 
Attachments
  • Photo of a textbook page showing a Thevenin equivalent calculation: R1 = 12 kΩ, R2 = 2 kΩ, RTH = 1.71 kΩ, VTH = -3.57 V, a simplified BJT bias circuit with +5 V, -5 V, 5 kΩ, 0.5 kΩ and 1.71 kΩ resistors, and the equation -3.57 = IBQ(1.71k) + VBE + (β+1)IBQ(0.5k) - 5.
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  • Printed solution page titled BJT Biasing: IBQ = 14 μA, IEQ = (100 + 1)IBQ = 1.412 mA, ICQ = 100IBQ = 1.4 mA, VCEQ = 5 − RCICQ − REIEQ + 5 = 5 − (5)(1.4) − (0.5)(1.412) + 5 = 2.3 V, Hence (B) is correct option.
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lazyaditya said:
I have attached the solution given behind the book, please tell me if the solution is valid and can i use it in other problems similar to this ?

These attachments are irrelevant to your zener problem.
 
I am sorry i posted the wrong the image.
 
Attachments
  • Printed solution text from a textbook: Thevenin voltage v_TH = 3.6(20)/(11+3.6) = 4.93 V > Vz, R_TH = 11 || 3.6 = 2.71 kΩ, i_z = (4.93 - 4)/2.71k = 0.34 mA, then 'Hence (B) is correct option.'
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Both methods of analysis are valid, providing you draw the logical conclusion. Determining the Thévenin equivalent is fine, but you don't have to.

Answer (a) is correct. The zener voltage here must be 4 volts if it is conducting.