Calculate Thevenin voltage in a Common Emitter Circuit

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.

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

Both approaches look perfectly fine. Which one is correct and why?

• 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? Commented Feb 15, 2018 at 21:30
• Use Vbe for one part and beta for the other part. Commented Feb 15, 2018 at 22:01
• @JohnD I agree with you. Since the one affects the other this is rather an odd assignment. Commented Feb 15, 2018 at 22:36
• Plz friends take a look at my edited problem. Commented Feb 16, 2018 at 0:58
• 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... Commented Feb 16, 2018 at 1:42

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.