For a common base configuration BJT such as the one shown here:
How can I derive the voltage gain and current gain?
Electronic – Derivation of current gain and voltage gain for common base BJT
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Related Solutions
Understanding transistors is a bit like peeling an onion- there are many layers. At the simplest large-signal level you can consider the transistor as a current sink that's controlled by the current through the base-emitter junction. The latter behaves like a forward-biased diode. Not much current until you get to some hundreds of mV, and way too much current if you put volts across the junction. As you say, the transistor will conduct excessive current and will be destroyed if you simply connect (say) 5V to the base with emitter grounded. This is in stark contrast to the behavior of a MOSFET.
At a more sophisticated level of understanding (which is required if you want to predict how most amplifiers work) and for small signals the base-emitter junction behaves like a resistor of Vt/Ib where Vt is the thermal voltage, about 26mV at room temperature. So if your base current is 2.5uA (say the beta is 300 and the transistor is biased with 0.75mA collector current), the base-emitter junction looks like about a 10K resistor for small signals. You can consider the transistor as a (somewhat imperfect because of r0) voltage controlled current source with an input resistance of Vt/Ib. This is the hybrid-\$\pi\$ model. Note that the transconductance gm (and thus the voltage gain in a common emitter configuration) is a function of the collector bias current and temperature and beta does not enter into it at all.
I must emphasize that this model is a linearized model about a bias point and is quite invalid if the (change in) input voltage is large (more than some millivolts). In other words we're talking about relatively small changes on top of a fixed base-emitter voltage of perhaps 600 or 700mV.
One way to find the gains is to pick a operating point, analyze at that, then perturb it a little and analyze at that. It takes separate pertubations for a common mode change and a differential mode change.
For example, analyze everything at 0 V on both inputs. For common mode gain, raise each input 1 V and analyze what happens to the output. The change in output divided by the change in input (1 V in this example) is the common mode gain.
Similarly, starting with the previously analyzed case of both inputs at 0, raise the positive input 1 mV and see what you get. The differential mode gain is then the resulting output change divided by the differential input voltage change (1 mV in this example). If the output clips, then the input change was too large. In that case, make the input change smaller and try again.
Related Topic
- Electronic – Confusion in the difference between DC current gain and AC current gain in common emitter amplifier
- Electrical – voltage gain and current gain in a common emitter
- Electrical – Determining base-emitter voltage in a CE BJT with base supplied by constant current source
- Electronic – BJT Differential Amplifier Common Mode & Differential Mode Gain
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Best Answer
Forgive me for using another schematic for common base configuration, but I found it more understandable in education purpose.
First, voltage gain:\$A_{v}=\frac{u_{iz}}{u_{ul}}\$
Usual way to deal with this fraction is to derive the expression for both of the voltages separately and then evaluate.
Output voltage is equal to: \$u_{iz}= -h_{fe}\cdot i_b \cdot R_c ||R_T\$
Input voltage is equal to: \$u_{ul}=i_{ul}\cdot R_E = -i_b \cdot r_{be}\$. Notice the sign of the current and why we took \$-i_b\$(because the current flows in the negative terminal of the predefined voltage \$u_{ul}\$)
Therefore, voltage gain is equal to: \$A_{v}=\frac{u_{iz}}{u_{ul}}=\frac{-h_{fe}\cdot i_b \cdot R_C||R_T}{-i_b \cdot R_E}=h_{fe}\frac{R_C||R_T}{R_E}\$
Current gain is a bit difficult to derive, but don't give up :)
First, we need to see what is the expression for input resistance of the amplifier \$R_{ul}\$ (which is also an important parameter while designing an amplifier)
Input resistance is: \$R_{ul}=\frac{u_{ul}}{i_{ul}}\$
Currents are: \$i_{ul}=i_{Re}+i_e; i_{Re} = \frac{u_{ul}}{R_E}; i_b = - \frac{u_{ul}}{r_{be}}\$
If we apply KCL for node E we have: \$i_e+i_b+h_{fe}i_b=0, i_e=-i_b(1+h_{fe})\$
Expression for input current is: \$i_{ul}=i_{Re}+i_e=i_{Re}-i_b(1+h_{fe})=\frac{u_{ul}}{R_E}-(1+h_{fe})=u_{ul}\cdot (\frac{1}{R_E}+\frac{1}{\frac{r_{be}}{1+h_{fe}}})\$
Now when we take a step back and take a look at the input resistance expression: \$R_{ul}=\frac{u_{ul}}{i_{ul}}=R_E || (\frac{r_{be}}{1+h_{fe}})\$
From the last expression we can notice that the input resistance is equal to 2 resistor in parallel. First one is \$R_E\$ and the second one we can call \$R_{ul}'\$.
We are finally here, \$i_{iz}=\frac{u_{iz}}{R_T}; i_{ul}=\frac{u_{ul}}{R_E || R_{ul}'}\$
Current gain: \$A_{i}=\frac{i_{iz}}{i_{ul}}=\frac{\frac{u_{iz}}{R_T}}{\frac{u_{ul}}{R_E || R_{ul}'}}=\frac{u_{iz}}{u_{ul}} \cdot \frac{R_E || R_{ul}'}{R_T} = A_V \cdot \frac{R_E ||R_{ul}'}{R_T}\$
From here we can see that the current gain of these configuration can never be bigger than 1 and always positive.
I hope everything is clear now!