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Thevenin's Theorem

  • Thread starter Lunat1c
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Given the circuit of the amplifier shown in the uploaded figure, I'm trying to find IE (the dc emitter current). For some reason, the book I'm using, when calculating Vb which is equivalent to VTH 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 IE myself I'd do it as follows:

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

Considering the input loop of the first stage, VB = VTH= IBRTH + VBE + IE(RE1 + RE2)

Using this i got IE = 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
 

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  • #2
The Electrician
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Given the circuit of the amplifier shown in the uploaded figure, I'm trying to find IE (the dc emitter current). For some reason, the book I'm using, when calculating Vb which is equivalent to VTH 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 IE myself I'd do it as follows:

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

Considering the input loop of the first stage, VB = VTH= IBRTH + VBE + IE(RE1 + RE2)

Using this i got IE = 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, VB = VTH is not true; VB is going to be less than VTH.

Proceeding with your equation:

VTH= IBRTH + VBE + IE(RE1 + RE2)

And substituting IB=IE/(beta)

We get:

VTH= IERTH/(beta) + VBE + IE(RE1 + RE2)

Plugging in the numbers and solving, I get IE=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 Vbe 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.
 
  • #3
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With VB 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 VTH?
 
  • #4
The Electrician
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With VB 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 VTH?
I assumed that's what you meant; that's the voltage at the base, hence the designation VB. VTH, 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.

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

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

Does this make sense?
 
  • #5
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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!
 
  • #6
The Electrician
Gold Member
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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.
 

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