Computing input resistance of two common-base configurations

When computing the input resistance of this common base configuration, the collector current $$\I_C\$$ is approximated as equal to the emitter current $$\I_E\$$:

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It is $$\I_C = g_m V_{BE} = g_m(-V_{test}) = -g_m V_{test} = -I_{test}\$$ and therefore

$$R_{in} = \frac{V_{test}}{I_{test}} = \frac{V_{test}}{g_m V_{test}} = \frac{1}{g_m}$$

But when a base resistor is added, the approximation $$\I_C \simeq I_E\$$ is no more used:

simulate this circuit

Now instead the computation is more complex:

$$I_C = g_m V_{BE} = g_m(V_B - V_E) = g_m \left( -R_B I_B - V_{test} \right)$$

$$I_{test} = - I_E = - (I_C + I_B) = - \beta I_B - I_B = - \left(\beta + 1 \right)I_B$$

$$I_C = g_m \frac{R_B I_{test}}{\beta + 1} - g_m V_{test}$$

$$I_{test} = -I_C - \frac{I_C}{\beta} = -\left( I_C + \frac{I_C}{\beta} \right) = -I_C \left( 1 + \frac{1}{\beta} \right)$$

$$I_C = -I_{test} \frac{\beta}{\beta + 1}$$

$$-I_{test} \frac{\beta}{\beta + 1} = g_m \frac{R_B I_{test}}{\beta + 1} - g_m V_{test}$$

$$I_{test} \left( g_m \frac{R_B}{\beta + 1} + \frac{\beta}{\beta + 1} \right) = g_m V_{test}$$

$$R_{in} = \frac{V_{test}}{I_{test}} = \frac{\frac{g_m R_B + \beta}{\beta + 1}}{g_m} = \frac{R_B}{\beta + 1} + \frac{\beta}{g_m(\beta + 1)}$$

Why here (and only here) the results are radically changed if $$\I_C \simeq I_E\$$?

• In a circuit without RB resistor Rin is equal to $R_{IN} = r_e = \frac{V_T}{I_E} = r_{\pi}||g_m = \frac{\beta}{g_m(\beta + 1)} \approx \frac{1}{g_m}$ But if we include RB resitor the Rin will increse to $R_{IN} = \frac{R_B}{\beta +1} + r_e \approx \frac{R_B}{\beta +1} + \frac{1}{g_m}$ So, where is the problem?
– G36
Jul 16, 2023 at 6:27
• Where $R_B$ was zero you assumed $\beta+1 = \beta$, so why not here? Jan 23 at 8:57