Calculating IE in Amplifier Circuit with Thevenin's Theorem

[Lunat1c]

Given the circuit of the amplifier shown in the uploaded figure, I'm trying to find I_E (the dc emitter current). For some reason, the book I'm using, when calculating V_b which is equivalent to V_TH took in consideration the resistances in the emitter which are in parallel with R₂ (the given solution is also uploaded, check figure 2 please).

I know this makes sense because the current passing through R₁ and R₂ is not equal since some current goes through the base. However for the thevenin equivalent circuit shouldn't the part where there's R₁ and R₂ be open circuit?

If i were to calculate I_E myself I'd do it as follows:

V_TH = (15/25.1k)*5.1k = 3.0478V
R_TH = R1||R2 = 4064ohms

Considering the input loop of the first stage, V_B = V_TH= I_BR_TH + V_BE + I_E(R_E1 + R_E2)

Using this i got I_E = 5.85mA which is quite close to what the book gave as an answer however I'd like to confirm my doubts.

Thanks in advance

 

[The Electrician]

Given the circuit of the amplifier shown in the uploaded figure, I'm trying to find I_E (the dc emitter current). For some reason, the book I'm using, when calculating V_b which is equivalent to V_TH took in consideration the resistances in the emitter which are in parallel with R₂ (the given solution is also uploaded, check figure 2 please).

I know this makes sense because the current passing through R₁ and R₂ is not equal since some current goes through the base. However for the thevenin equivalent circuit shouldn't the part where there's R₁ and R₂ be open circuit?

If i were to calculate I_E myself I'd do it as follows:

V_TH = (15/25.1k)*5.1k = 3.0478V
R_TH = R1||R2 = 4064ohms

Considering the input loop of the first stage, V_B = V_TH= I_BR_TH + V_BE + I_E(R_E1 + R_E2)

Using this i got I_E = 5.85mA which is quite close to what the book gave as an answer however I'd like to confirm my doubts.

Thanks in advance

First off, V_B = V_TH is not true; V_B is going to be less than V_TH.

Proceeding with your equation:

V_TH= I_BR_TH + V_BE + I_E(R_E1 + R_E2)

And substituting I_B=I_E/(beta)

We get:

V_TH= I_ER_TH/(beta) + V_BE + I_E(R_E1 + R_E2)

Plugging in the numbers and solving, I get I_E=5.909 mA. I don't know why you got 5.85 mA.

This is a somewhat more accurate method of solution than that used by the book solution.

They just used a voltage divider method, but didn't include the effect of the .7 volt V_be when they calculated the voltage out of the voltage divider. They included it in the last calculation, but a small error results from not including it in the earlier calculation. Your method includes it properly, hence gets a slightly more accurate result.

In both cases, where "beta" is used, more exactly it should be "beta+1", but that's a small difference.

 

[Lunat1c]

With V_B I meant the voltage at the point between R1 and R2 (i.e. Vcc minus the voltage drop on R1), why is that not equal to V_TH?

 

[The Electrician]

With V_B I meant the voltage at the point between R1 and R2 (i.e. Vcc minus the voltage drop on R1), why is that not equal to V_TH?

I assumed that's what you meant; that's the voltage at the base, hence the designation V_B. V_TH, on the other hand, is the open-circuit voltage out of the R1-R2 divider. Once the base is connected to the R1-R2 junction, and draws some current, the voltage out of the divider will decrease.

V_TH is the voltage at the R1-R2 junction when the base of the transistor is not connected there.

V_B is the voltage at the R1-R2 junction when the base is connected.

Does this make sense?

 

[Lunat1c]

Yes, I understood what you just said. So my method and the one used by the book would have produced almost identical results if the book had taken into consideration the 0.7V. At the end of the day i know that it doesn't make that much of a difference but it is good to know.

Thanks a lot for your help!

 

[The Electrician]

Yes, I understood what you just said. So my method and the one used by the book would have produced almost identical results if the book had taken into consideration the 0.7V.

Yes, and you can see that this is true in the following way:

Leave out the .7V from your equation and solve; you will get 7.67 mA.

Leave out the .7V from their last calculation and solve, giving 2.89/377 = 7.67 mA.