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So I have been trying to solve the problem below for quite some time but I can't seem to do it right.

I have mainly tried to do KVL in order to get iE.

For example: 

15 = 3iE + vEC 3iC = vBC

I have used the fact that C is grounded so vEC is just vE and vBC is just vB.

I also used iE = ( Beta+1 ) iB and iC = Beta iB

However, I keep getting wrong answers.

Is there something I am missing to see?

Thank you!

enter image description here

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    \$\begingroup\$ Welcome to the site :-) On Stack Exchange sites, the standard is not to mark a problem as solved by editing the title. Instead, please "accept" (add the "green tick") to the answer which best helped you to solve the problem, to show that the issue is resolved for you (see here). You can do this even when you solved the problem yourself. Therefore I have reversed (rolled-back) your change to the question's title. Please mark your answer with that green tick, to indicate that your answer is the solution you needed. Thanks :-) \$\endgroup\$ Commented Dec 14, 2019 at 9:53
  • \$\begingroup\$ @SamGibson Hi! Thanks for the heads up :) However, when I try to click on the check mark it tells me I have to wait till tomorrow to accept my answer. I'l try again tomorrow :) \$\endgroup\$ Commented Dec 14, 2019 at 22:11
  • \$\begingroup\$ @RayanAlHobayb Do you need something different? Or have you simply given up? \$\endgroup\$ Commented Dec 31, 2019 at 5:25

2 Answers 2

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Since you haven't selected your own answer yet, here's mine that may reflect your understanding:

schematic

simulate this circuit – Schematic created using CircuitLab

All I've done in this transition is to set up the Thevenin \$V_\text{TH}\$ and \$R_\text{TH}\$ from the original biasing pair, \$R_1\$ and \$R_2\$. This makes it very easy to perform the following KVL:

$$V_\text{CC}-I_\text{E}\cdot R_E-V_\text{BE}-I_\text{B}\cdot R_\text{TH}=V_\text{TH}$$

Some very simple algebraic manipulation results in:

$$I_\text{E}=\frac{V_\text{CC}-V_\text{TH}-V_\text{BE}}{\frac{R_\text{TH}}{\beta+1}+R_E}$$

Plugging in the values of \$V_\text{TH}=3\:\text{V}\$, \$R_\text{TH}=2.4\:\text{k}\Omega\$, \$V_\text{CC}=15\:\text{V}\$, \$\beta=49\$, \$V_\text{BE}=700\:\text{mV}\$, and \$R_\text{E}=3\:\text{k}\:\Omega\$, we find that \$I_\text{E}\approx 3.71\:\text{mA}\$.

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  • \$\begingroup\$ Maby I'm cherry-picking but why Vbe is positive? \$\endgroup\$ Commented Dec 21, 2019 at 15:10
  • \$\begingroup\$ @G36 I looked at the answers available to the OP and chose the direction of current from that (which forced my hand regarding the sign of Vbe.) If you look at the datasheets for the 2N3904 and the 2N3906, for example, they both provide positive values for the maximum Ic. There's some inconsistencies in the world and I just roll with the punches. \$\endgroup\$ Commented Dec 21, 2019 at 18:00
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The answer was quite simple!

KVL : 15 = 12I + 3(I + IB ) where I is the current passing through the 12 K resistor

you get I = 1-0.2IB

KVL : 3 IE + VEB = 12 I

with VEB = 0.7, replace I and you will have the answer.

Also use (Beta + 1) IB = IE

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  • \$\begingroup\$ Thank you for posting the answer. On SE you indicate "solved" by accepting an answer rather than putting it in the title. You can accept your own answer (but you may have to wait a day). It will then turn the Answers button green on the home page to indicate that it has been solved. \$\endgroup\$ Commented Dec 14, 2019 at 9:50
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    \$\begingroup\$ Using KVL I get the following equation: \$I_\text{E}=\left(\beta+1\right)\frac{V_\text{TH}-V_\text{BE}}{R_\text{TH}+\left(\beta+1\right)R_\text{E}}\approx 3.71\:\text{mA}\$. \$\endgroup\$ Commented Dec 14, 2019 at 18:42

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