Solving for (β+1) in Thevenin Equivalent Homework

In summary: Try it this way: vth = -β(iR2 + is) = -βi(R2 + R1||R2) = -βi(R2 + R1*R2/(R1+R2)) = -βi(R1*R2 + R2^2)/(R1+R2).Then, as above you factor out the i: vth = -i * β(R1*R2 + R2^2)/(R1+R2) = -i * β R2 [R1 + R2]/[R1+R2] = -i * β R2.And there's your Vth, and the β+1 term is gone.
  • #1
nunez2005
2
0

Homework Statement


I'm trying to understand where the (β+1) comes from.


2. Homework Equations
vth=-βiR2
i+βi+is=0

vth=-β(-is/(β+1))R2


3. The Attempt at a Solution

vth=-βiR2
i+βi+is=0

vth=-β(-βi-is)R2
 

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  • #2
nunez2005 said:

Homework Statement


I'm trying to understand where the (β+1) comes from.


2. Homework Equations
vth=-βiR2
i+βi+is=0

vth=-β(-is/(β+1))R2


3. The Attempt at a Solution

vth=-βiR2
i+βi+is=0

vth=-β(-βi-is)R2


Welcome to the PF.

Without seeing the original circuit that this is modeling, it's hard to be sure. But for a CE BJT amplifier, your collector current Ic = β * Ib. So your emitter current is...
 
  • #3
i've attached the solution that i have in the book, the value of beta is 150 and R1=100k and R2=39k, the only problem I have is that how do you get to the vth while substituting beta and the resistance in both sides to be equal to -Beta (-is/beta+1), the beta + 1 has to come out from simplying and combining like terms but I just can't see it.
 
  • #4
nunez2005 said:

Homework Statement


I'm trying to understand where the (β+1) comes from.


2. Homework Equations
vth=-βiR2
i+βi+is=0

vth=-β(-is/(β+1))R2


3. The Attempt at a Solution

vth=-βiR2
i+βi+is=0

So factor out the i in that equation above:

(β + 1)i + is = 0

There's your (β + 1) term.

vth=-β(-βi-is)R2

That doesn't look right. You want to substitute for i in your equation for Vth from above, but what you've stuck in there also contains i.
 
  • #5

vth=-β(-βiR2-isR2)
vth=β^2iR2+βisR2

vth=βR2(i^2+is)

vth=βR2(i(i+β+1))

vth=βR2(i)(i+β+1)

We can see that the (β+1) term comes from the equation i+βi+is=0, where the (β+1) represents the total current flowing through the circuit. This current is the sum of the base current (i), the collector current (βi), and the source current (is). Therefore, when solving for vth, the (β+1) term appears in the equation due to the presence of all three currents. It is important to include this term in the Thevenin equivalent circuit because it accurately represents the total current flowing through the circuit.
 

FAQ: Solving for (β+1) in Thevenin Equivalent Homework

1. What is the Thevenin Equivalent?

The Thevenin Equivalent is a simplified circuit that represents a complex network of resistors, voltage sources, and current sources. It is used to analyze and solve electrical circuits, particularly when dealing with multiple resistors and sources.

2. Why do we need to solve for (β+1) in Thevenin Equivalent Homework?

Solving for (β+1) in Thevenin Equivalent Homework allows us to find the Thevenin voltage and resistance, which are essential parameters in analyzing a circuit. It also helps us simplify the circuit and make it easier to calculate the voltage and current across any component.

3. How do you solve for (β+1) in Thevenin Equivalent Homework?

To solve for (β+1) in Thevenin Equivalent Homework, you need to follow specific steps. First, remove the load resistor from the circuit. Then, find the open-circuit voltage across the load resistor terminals. Next, find the equivalent resistance of the circuit. Finally, place the load resistor back into the circuit and calculate the current using Ohm's law. (β+1) is then equal to the open-circuit voltage divided by the current.

4. What are the applications of Thevenin Equivalent in real-world situations?

Thevenin Equivalent is widely used in various applications, including power systems, electronic circuits, and telecommunications. It allows engineers to analyze complex circuits and design more efficient and reliable systems. It is also used to troubleshoot and repair faulty circuits.

5. Are there any limitations to using Thevenin Equivalent?

While Thevenin Equivalent is a powerful tool, it has some limitations. It only applies to linear circuits and cannot be used for non-linear circuits. It also assumes that all elements in the circuit are fixed, which may not always be the case in real-world situations. Additionally, Thevenin Equivalent may not accurately represent the behavior of the circuit at high frequencies.

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