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I am confused on how to proceed with the above problem. I have solved many problems on thevenin voltage but never faced one involving BJT. Please help me with this problem.

EDIT: I have 2 approaches in my mind.

1st Approach is short ckt a and b and get Isc. Calculate Voc for open circuit case and divide Voc by Isc to get Rth.

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2nd Approach is to calculate re using DC Analysis. Then draw small signal model short circuiting DC and use this circuit to get Rth.

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enter image description here Both approaches look perfectly fine. Which one is correct and why?

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    \$\begingroup\$ It's not entirely fair to throw a non-linear device in a Thevenin equivalent problem, IMO. However, you can find a Thevenin equivalent by evaluating the open-circuit voltage and the short circuit current. Can you calculate the current when you short the a-b terminals? And the voltage with a-b open? \$\endgroup\$
    – John D
    Commented Feb 15, 2018 at 21:30
  • \$\begingroup\$ Use Vbe for one part and beta for the other part. \$\endgroup\$
    – Daniel P
    Commented Feb 15, 2018 at 22:01
  • \$\begingroup\$ @JohnD I agree with you. Since the one affects the other this is rather an odd assignment. \$\endgroup\$
    – Trevor_G
    Commented Feb 15, 2018 at 22:36
  • \$\begingroup\$ Plz friends take a look at my edited problem. \$\endgroup\$
    – Nikhil Kashyap
    Commented Feb 16, 2018 at 0:58
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    \$\begingroup\$ I'd say they're both right... and both wrong. Because you are using approximations that make the result with two decimal places wrong, but they are reasonable approximations nonetheless. As a matter of fact, one might just 'look' into the emitter and find the resistance in the base circuit 'amplified' by beta or so (even more crude approximation). - In one case you are approximating Vbe with 0.7V; in the other you are using an approximate formula to compute re... \$\endgroup\$
    – Sredni Vashtar
    Commented Feb 16, 2018 at 1:42

1 Answer 1

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It's been +25 years but I gonna give it a try ...

\$R_{th}=\dfrac{V_{oc}}{I_{sc}}\$

Calculating \$I_{sc}\$ :

\$I_{b}=\dfrac{10.7V - 0.7V}{10k\Omega}=1mA\$

\$I_{c}=I_{b}\cdot\beta =1mA\cdot100=100mA\$

\$I_{sc}=I_{b}+I_{c}=1mA+100mA=101mA\$

Calculating \$V_{oc}\$ :

\$-10.7+10000\cdot I_{b}+0.7+1000\cdot(\beta+1)\cdot I_{b}=0\$

\$10000\cdot I_{b}+1000\cdot101\cdot I_{b}=10\$

\$I_{b}=\dfrac{1}{11100} A\$

\$I_{e}=(\beta+1)\cdot I_{b}=\dfrac{101}{11100}A\$

\$V_{oc}=I_{e}\cdot 1k\Omega=\dfrac{101\cdot1000}{11100}V=\dfrac{1010}{111}V\$

\$R_{th}=\dfrac{V_{oc}}{I_{sc}}=\dfrac{\dfrac{1010}{111}V}{101mA}=90.09\Omega\$

PS : It's 5:49AM here, time to get some sleep.

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