Common Base Structure

#Electronics In this note, we are going to learn about the common base structure. The AC equivalent circuit of this structure looks like this: The input goes into the emitter and the output is on the collector side. The intrinsic gain is the gain when $R_s = 0$. We can easily realize that the intrinsic gain is $g_m R_c$,, because the structure would be no different from the structure of a common emitter (imagine their hybrid pi models). The only difference is that this time, the gain is positive, meaning the circuit is non-inverting. Now how do we bias this structure? The base voltage is fixed at a value $V_b$. The resistor $R_E$ is calculated such that it allows for the emitter (collector) current we desire. Note: I think I might have missed something here. Because $R_E$ does appear in the bias circuit but not in the AC equivalent circuit, I guess there might have been a capacitor in parallel with $R_E$! Let’s calculate the gain of this structure. We use the hybrid pi model. A simple KCL gives us \(g_m v_{\pi} + \frac{v_{\pi}}{r_{\pi}} + \frac{v_i - (- v_{\pi})}{R_s} = 0 \Rightarrow v_{\pi} = - \frac{v_i \frac{1}{R_s}}{g_m + \frac{1}{r_{\pi}} + \farc{1}{R_s}} = - \frac{v_i}{1 + g_m R_s + \frac{R_s}{r_{\pi}}} ,\) and we have \(v_o = - g_m v_{\pi} R_c = \frac{v_i g_m R_c}{1 + g_m R_s + \frac{R_s}{r_{\pi}}} \Rightarrow A_v = \frac{v_o}{v_I} = \frac{R_c}{\frac{1}{g_m} + R_s + \frac{R_s}{\beta}} = \frac{R_c}{r_m + R_s \underbrace{(1 + \frac{1}{\beta})}_{\approx 1}} = \frac{R_c}{r_m + R_s} .\) We can see that $R_s$ has a very significant role in the gain. It actually drops the gain quite much! Although we managed to solve for the gain using the hybrid pi model we always used, it is actually better if we derive a second equivalent hybrid pi model for the common base structure. We start with our standard model. Next, we remove the current source from its place. Not to disturb the KVL and KCL equations, we replace it with two current sources as shown in the figure below. Now we note that the left current source is actually a current source whose current depends on the potential difference of its own two terminals. This is therefore simply a resistor. Now we just factorize the two resistors on the left to one resistor $r_e$: \(r_e = r_{\pi} \parallel \frac{1}{g_m} = r_{\pi} \parallel r_m = \beta r_m \parallel r_m = \alpha r_m \approx r_m .\) Therefore, the new hybrid pi model looks like this: We can see that \(v_o = - g_m v R_c , \quad v = - \frac{r_e}{r_e + R_s} v_i \Rightarrow v_o = \frac{\overbrace{g_m r_e}^{\approx 1} R_c}{r_e + R_s} v_i = \frac{R_c}{r_e + R_s} v_i .\) Now let’s calculate the output resistance of the common base structure, We use the standard hybrid pi model for this. We have \(v_o = - i \frac{R_s r_{\pi}}{R_s + r_{\pi}} .\) Now we write KVL and use the relation above: \(- v + r_o (i - g_m v_{\pi}) - VV_{\pi} = 0 \Rightarrow v = r_o i - v_{\pi} (g_m r_o + 1) \Rightarrow v = r_o i + i \frac{R_s r_{\pi}}{R_s + r_{\pi}} (\underbrace{g_m r_o}_{>> 1} + 1) .\) Note that \(g_m r_o = \frac{I_c}{V_T} \frac{V_A}{I_c} = \frac{V_A}{V_T} ,\) and because $V_A$ is usually in the order of $10 - 120 \text{V}$ and $V_T = 25 \text{mV}$, this fraction is definitely always much larger than $1$. Therefore, we get \(R_o = \frac{v}{i} = r_o [ \frac{R_s + r_{\pi} + g_m R_s r_{\pi}}{R_s + r_{\pi}} ] = r_o [ \frac{\frac{R_S}{r_{\pi}} + 1 + g_m R_s}{\frac{R_s}{r_{\pi}} + 1} ] \approx r_o [ \frac{1 + g_m R_s}{1 + g_m \frac{R_s}{\beta}} ] .\) Finally, we know that the output resistance of the common base structure is \(R_{out} = R_o \parallel R_c \approx R_c .\)

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