Does the following tutorial neglect the dynamic resistance across the base and the emitter in derivation of the output resistance:?
It assumes Vbe constant and ignores the dynamic resistance? But dynamic resistance still exists even there is no change in Vbe(?) Because it says the voltage drop Vin - Vout = (I/beta)* Rsource. Dynamic resistance seems to be ignored(?)
Answer
Yes you are right the dynamic resistance of \$V_{be}\$ is neglected because it is usually very small in comparison with rest of the resistances in the circuit. If you write the exact equation (neglecting the output resistance of BJT), you will get: $$V_{in} - \frac{I}{\beta}r_{source} - \frac{\beta + 1}{\beta}I(\frac{\alpha}{g_m}) = V_{out}$$ where \$\frac{\alpha}{g_m}\$ is the dynamic resistance of \$V_{be}\$ Thus, $$V_{out} - V_{in} = -\frac{I(\beta + 1)}{\beta}(\frac{r_{source}}{\beta + 1} + \frac{\alpha}{g_m})$$ Since, \$Z_{out} = (V_{out} - V_{in})/(\frac{-I(\beta + 1)}{\beta})\$. Finally,
$$Z_{out} = \frac{r_{source}}{\beta + 1} + \frac{\alpha}{g_m}$$ To get the sense of numbers, usually \$\beta\$ ~ 100 so, \$\alpha = \frac{\beta}{\beta + 1}\approx 1\$.
\$\frac{1}{g_m} = \frac{V_T}{I_c}\$, at room temperature \$V_T \approx 25mV\$ and assuming \$I_c\$ ~ \$1mA\$, \$\frac{1}{g_m}\$ ~ \$25\Omega\$.
Thus last term in the expression for output impedance is of the order of few ohms.
The source resistance, on the other hand, is the coming from the output impedance of a previous amplifier stage and is usually high ~ \$10^5\Omega -10^6\Omega\$. So usually the first term dominates. So we can approximate the output impedance as:
\$Z_{out} \approx \frac{r_{source}}{\beta + 1} \approx \frac{r_{source}}{\beta}\$.
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