Electronics 2 study guide with Sample Problems and Solutions to apply the basic principle and concept of Frequency Response of BJT and FET

Bipolar Junction Transistor (BJT)




- Bipolar transistors are so named because the controlled current must go through
two types of semiconductor material: P and N. The current consists of both
electron and hole flow, in different parts of the transistor.
- Bipolar transistors consist of either a P-N-P or an N-P-N semiconductor
“sandwich” structure.
- The three leads of a bipolar transistor are called the Emitter, Base, and
Collector.
- Transistors function as current regulators by allowing a small current to control a
larger current. The amount of current allowed between collector and emitter is
primarily determined by the amount of current moving between base and emitter.
- In order for a transistor to properly function as a current regulator, the controlling
(base) current and the controlled (collector) currents must be going in the proper
directions: meshing additively at the emitter and going in the direction of the
emitter arrow symbol.

BJT Small Signal Analysis Model
The re and hybrid model will be used to analyse small signal AC analysis of
standard transistor in the region of interest.

,Common Emitter Fixes Bias Configuration




AC Equivalent Ckt
Input Impedance:
𝑍𝑖 = 𝑅𝑠 //𝛽𝑟𝑒

Output Impedance:
𝑍𝑜 = 𝑟𝑜 //𝑅𝑐
If 𝑟𝑜 = ∞,
𝑍𝑜 = 𝑅𝑐

Voltage Gain:
Looking at the output side of AC equivalent circuit,
𝑉𝑜 = −(𝛽𝐼𝑏 )(𝑟𝑜 //𝑅𝑐 )
But
𝑉𝑖
𝐼𝑏 =
𝛽𝑟𝑒
So that
𝑉𝑖
𝑉𝑜 = − (𝛽 ) (𝑟 //𝑅𝑐 )
𝛽𝑟𝑒 𝑜
Therefore
𝑉𝑜 𝑟𝑜 //𝑅𝑐
=−
𝑉𝑖 𝑟𝑒
If 𝑟𝑜 = ∞,
𝑉𝑜 𝑅𝑐
=−
𝑉𝑖 𝑟𝑒
Current Gain:
Applying KCL on the input and output side.
(𝑟𝑜 )(𝛽𝐼𝑏 ) 𝐼𝑜 (𝑟𝑜 )(𝛽)
𝐼𝑜 = 𝑎𝑛𝑑 =
𝑟𝑜 + 𝑅𝑐 𝐼𝑏 𝑟𝑜 + 𝑅𝑐

, (𝑅𝐵 )(𝐼𝑖 ) 𝐼𝑏 (𝑅𝐵 )
𝐼𝑏 = 𝑎𝑛𝑑 =
𝑅𝐵 + 𝛽𝑟𝑒 𝐼𝑖 𝑅𝐵 + 𝛽𝑟𝑒
The result is
𝐼𝑜 𝐼𝑜 𝐼𝑏 (𝑟𝑜 )(𝛽) (𝑅𝐵 )
𝐴𝑖 = = ( )( ) = ( )( )
𝐼𝑖 𝐼𝑏 𝐼𝑖 𝑟𝑜 + 𝑅𝑐 𝑅𝑏 + 𝛽𝑟𝑒
Another formula
𝑉𝑜 𝑉𝑖
𝐼𝑜 = − 𝑎𝑛𝑑 𝐼𝑖 =
𝑍𝑜 𝑍𝑖
The result is
𝐼𝑜 𝑉𝑜 𝑍𝑖 𝑍𝑖
𝐴𝑖 = = (− ) ( ) = −𝐴𝑣
𝐼𝑖 𝑍𝑜 𝑉𝑖 𝑍𝑜

Voltage Divider Bias




AC Equivalent Circuit
Input Impedance
𝑍𝑖 = 𝑅′//𝛽𝑟𝑒
Where
𝑅 ′ = 𝑅1 //𝑅2
Output Impedance
𝑍𝒐 = 𝑟𝑜 //𝑅𝑐
If 𝑟𝑜 = ∞,
𝑍𝒐 = 𝑅𝑐
Voltage Gain
Looking at the output side of AC equivalent circuit,
𝑉𝑜 = −(𝛽𝐼𝑏 )(𝑟𝑜 //𝑅𝑐 )
But
𝑉𝑖
𝐼𝑏 =
𝛽𝑟𝑒
So that

, 𝑉𝑖
𝑉𝑜 = − (𝛽 ) (𝑟 //𝑅𝑐 )
𝛽𝑟𝑒 𝑜
Therefore
𝑉𝑜 𝑟𝑜 //𝑅𝑐
=−
𝑉𝑖 𝑟𝑒
If 𝑟𝑜 = ∞,
𝑉𝑜 𝑅𝑐
=−
𝑉𝑖 𝑟𝑒
Current Gain
Applying KCL on the input and output side.
(𝑟𝑜 )(𝛽𝐼𝑏 ) 𝐼𝑜 (𝑟𝑜 )(𝛽)
𝐼𝑜 = 𝑎𝑛𝑑 =
𝑟𝑜 + 𝑅𝑐 𝐼𝑏 𝑟𝑜 + 𝑅𝑐

(𝑅′)(𝐼𝑖 ) 𝐼𝑏 (𝑅′)
𝐼𝑏 = 𝑎𝑛𝑑 =
𝑅′ + 𝛽𝑟𝑒 𝐼𝑖 𝑅′ + 𝛽𝑟𝑒
The result is
𝐼𝑜 𝐼𝑜 𝐼𝑏 (𝑟𝑜 )(𝛽) 𝑅′
𝐴𝑖 = = ( )( ) = ( )( )
𝐼𝑖 𝐼𝑏 𝐼𝑖 𝑟𝑜 + 𝑅𝑐 𝑅′ + 𝛽𝑟𝑒
Another formula
𝑉𝑜 𝑉𝑖
𝐼𝑜 = − 𝑎𝑛𝑑 𝐼𝑖 =
𝑍𝑜 𝑍𝑖
The result is
𝐼𝑜 𝑉𝑜 𝑍𝑖 𝑍𝑖
𝐴𝑖 = = (− ) ( ) = −𝐴𝑣
𝐼𝑖 𝑍𝑜 𝑉𝑖 𝑍𝑜