BJT Early Model Analysis: Proving r0 = (Va + VCE)/IC for Resistance Calculation

[Trentonx]

Homework Statement

Working with BJT analysis, and varying models for the analysis.
Prove $r_{0}=\frac{V_{a} + V_{CE}}{I_{C}}$, where $r_{0}$ is the resistance for the Early model. I haven't been able to find a circuit or much of anything about this, let alone proving it.

Homework Equations

$i_{C} = I_{s}e^{\frac{V_{BE}}{V_{A}}} (1+\frac{V_{CE}}{V_{A}})$
$r_{0} = (\frac{\delta i_{C}}{\delta v_{CE}})$

The Attempt at a Solution

I did the partial derivative (that is what Eq2 mean right?) of $i_C$, and that gives me
$i^{'}_{C}=\frac{I_{s}e^{\frac{V_{BE}}{V_{A}}}}{V_{A}}$, but this doesn't simplify in anyway, and it's $i^{'}_{C}$, not just $i_{C}$. I'm guessing this is the wrong equation, but it's what we've been given.

 

[KataKoniK]

Dear poster,

Thank you for your question. I understand your struggle in finding information on this topic, as it is not a commonly discussed concept. However, I am happy to assist you in proving this relationship between r_0, V_a, V_CE, and I_C for the Early model.

To begin, let's first define the Early model for a BJT. The Early model is a simplified model used to approximate the output characteristics of a BJT in the active region. It assumes that the collector current is directly proportional to the collector-emitter voltage, and that the base-emitter voltage remains constant. This model is useful for small-signal analysis and can help us understand the behavior of a BJT in different circuit configurations.

Now, let's look at the equations you have provided. The first equation, i_C = I_s e^(V_BE/V_A) (1 + V_CE/V_A), is known as the Ebers-Moll equation. This is the fundamental equation for the BJT, describing the relationship between the collector current and the base-emitter voltage. The second equation, r_0 = (delta i_C/delta v_CE), is the definition of the Early voltage, which is the slope of the output characteristics curve in the Early model.

To prove the relationship r_0 = (V_a + V_CE)/I_C, we will use the fact that the Early voltage, V_a, is defined as the voltage at which the collector current is equal to 63% of its maximum value. In other words, when V_CE = V_a, the collector current is equal to 0.63 times its maximum value. This can be expressed mathematically as i_C = 0.63I_C, where I_C is the maximum collector current.

Now, let's take the derivative of the Ebers-Moll equation with respect to V_CE:

delta i_C/delta v_CE = (delta/delta v_CE)(I_s e^(V_BE/V_A) (1 + V_CE/V_A))

Using the chain rule, we can rewrite this as:

delta i_C/delta v_CE = (I_s e^(V_BE/V_A))(1/V_A)

Finally, substituting in the expression for i_C at V_CE = V_a, we get:

delta i_C/delta v_CE = (0.63I_C)(1/V_A)

Recall that r_0 is defined as the slope of the