TRANSISTOR AMPLIFICATION IN AC DOMAIN:

• The transistor can be employed as an amplifying device, that is, the output ac power is greater than the input ac power.
• The factor that permits an ac power output greater than the input ac power is the applied DC power.
• The amplifier is initially biased for the required DC voltages and currents. Then the ac to be amplified is given as input to the amplifier.
• If the applied ac exceeds the limit set by dc level, clipping of the peak region will result in the output.
• Thus, proper (faithful) amplification design requires that the dc and ac components be sensitive to each other’s requirements and limitations.
• The superposition theorem is applicable for the analysis and design of the dc and ac components of a BJT network, permitting the separation of the analysis of the dc and ac responses of the system.

BJT TRANSISTOR MODELING:

• The key to transistor small-signal analysis is the use of the equivalent circuits (models).
• A Model is a combination of ircuit elements like voltage or current sources,resistors,capacitors etc that best approximates the behavior of a device under specific operating conditions.
• Once the model (ac equivalent circuit) is determined, the schematic symbol for the device can be replaced by the equivalent circuit and the basic methods of circuit analysis applied to determine the desired quantities of the network. Hybrid equivalent network – employed initially.
• The only drawbak is that It is defined for a set of operating conditions that might not match the actual operating conditions.

COMMON EMITTER-UNBYPASSED EMITTER BIAS CONFIGURATION:

Applying KVL to the input side:

   $V_{i} = I_{b}\beta_{re} + I_{e}R_{E}$ ..

  $V_{i} = I_{b} \beta_{ re} +( \beta+1) I_{b}R_{E}$

Input impedance looking into the network to the right of RB is

 $Z_{b} = V_{i} \dfrac I \beta_{ re}+ ( \beta +1)R_{E}$

Since b>>1, $( \beta +1) = \beta$...

Thus

 $Z_{b} = V_{i} I_{b} (\beta_{re}+R_{E})$

 Since RE is often much greater than re,

   $Z_{b} = \beta R_{E}, Z_{i} = R_{B}||Z_{b}$

 Zo is determined by setting Vi to zero, Ib = 0 and b Ib can be replaced by open circuit equivalent.

The result is,

   $Z_{o} = R_{C}$

AV : We know that,

    $V_{o} = - I_{o}R_{C} = - \beta I_{b}R_{C} = - \beta(\dfrac{V_{i}}{Z_{b}})R_{C}$

    $A_{V} =\dfrac{ V_{o}} { V_{i}} = - \beta(\dfrac{R_{C}}{Z_{b}})$

Substituting,

  $Z_{b} = \beta(re + R_{E}) A_{V} =\dfrac{ V_{o}} { V_{i}} = - \beta[\dfrac{R_{C}} {re + R_{E}}] $

RE >>re,

  $A_{V} =\dfrac {V_{o}} { V_{i}} = \beta[\dfrac{R_{C}}{ R_{E}}]$

Phase relation: The negative sign in the gain equation reveals a 180 degree phase shift between input and output.