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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\$:

schematic

simulate this circuit – Schematic created using CircuitLab

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:

schematic

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\$?

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  • \$\begingroup\$ 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? \$\endgroup\$
    – G36
    Jul 16, 2023 at 6:27
  • \$\begingroup\$ Where \$R_B\$ was zero you assumed \$\beta+1 = \beta\$, so why not here? \$\endgroup\$
    – HarryH
    Jan 23 at 8:57

1 Answer 1

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As for the base impedance, the Ic = Ie approximation is equivalent to setting beta very high. The approximate value of 1/(beta +1) is then... zero, and the approximate value of beta/(beta + 1) = 1

That means that Rin is very similar in the two calculations.

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  • \$\begingroup\$ What means "very similar"? The difference between both cases is RB/beta+1). \$\endgroup\$
    – LvW
    Jul 16, 2023 at 8:26
  • \$\begingroup\$ @LvW Division by a large number (beta plus one) is... nearly multiplication by zero; that difference is small. It is the nature of the 'large beta' approximation to omit such terms. \$\endgroup\$
    – Whit3rd
    Jul 17, 2023 at 22:26
  • \$\begingroup\$ No, for a ratio of two numbers, you must not consider the denominator only. A commin collector stage is used very often to transform a large output resistance into a smaller one. For example, when RB=1E4 Ohms and beta=100, the additional term is 1E4/1E2 and, thus, even larger than 1/gm. \$\endgroup\$
    – LvW
    Jul 18, 2023 at 6:17
  • \$\begingroup\$ @LvW I'm not considering the ratio; I'm only pointing out that this is a consequence of the approximation. By setting beta=100, you're choosing not to approximate. Beta might be loosely known in any case (typically 200 to 600), so the detailed calculation will rarely be a precise value. \$\endgroup\$
    – Whit3rd
    Jul 18, 2023 at 18:47

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