Answer
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Hint: By using current formula $V = IR$, the current flowing through a voltage V and resistance R. show the figure in common emitter transistor circuit, and given values are base current is \[40\mu {\text{A}}\], voltage current is \[{V_{CC}} = 10V\]and the base resistance is $245K\Omega $. Find the base emitter voltage\[{V_{BE}}\].
Complete step by step answer:
The given values are,
Base current is\[40\mu {\text{A}}\]
Voltage current\[{I_V} = {V_{CC}} = 10V\]
Base resistance ${R_B} = 245K\Omega $
We know,
Input voltage ${I_V} = {V_{BE}} + ({I_B}) \times ({R_B})$
Where,
${I_B}$ Is the base current, and
${R_B}$ Is the resistance offered in the base region.
${V_{BE}} = {I_V} - {I_B} \times {R_B}$
\[{I_V} = {V_{CC}} = 10V\] And
${I_B} = 40\mu A$
$ \Rightarrow 40 \times {10^{ - 6}}A$
Also,${R_B} = 245K\Omega $
$ \Rightarrow 245 \times {10^3}\Omega $ [From the figure]
Putting their respective values, we get:
${V_{BE}} = 10 - (40 \times {10^{ - 6}}A)(245 \times {10^3}\Omega )$
On simplification the bracket term and we get
$ \Rightarrow (10 - 9.8)V$
On subtracting we get,
$ \Rightarrow 0.2V$
Therefore, option (B) is the correct answer.
Additional information:
Common emitter transistor:
As such the common emitter configuration may be a good all round circuit to be used in many applications. It's also worth noting at this stage that the common emitter transistor amplifier inverts the signal at the input. Therefore if a waveform that's rising enters the input of the common emitter amplifier, it'll cause the output voltage to fall.
Note: Advantages of Common Emitter amplifier:
1. A common emitter amplifier is inverting and has low I/p impedance
2. High output impedance
3. High voltage gain.
4. High current gain.
Disadvantages of Common Emitter amplifier:
1. It’s a high output resistance.
2. It responds poorly to high frequencies.
3. Its high thermal instabilities.
4. Its voltage gain is extremely unstable
Complete step by step answer:
The given values are,
Base current is\[40\mu {\text{A}}\]
Voltage current\[{I_V} = {V_{CC}} = 10V\]
Base resistance ${R_B} = 245K\Omega $
We know,
Input voltage ${I_V} = {V_{BE}} + ({I_B}) \times ({R_B})$
Where,
${I_B}$ Is the base current, and
${R_B}$ Is the resistance offered in the base region.
${V_{BE}} = {I_V} - {I_B} \times {R_B}$
\[{I_V} = {V_{CC}} = 10V\] And
${I_B} = 40\mu A$
$ \Rightarrow 40 \times {10^{ - 6}}A$
Also,${R_B} = 245K\Omega $
$ \Rightarrow 245 \times {10^3}\Omega $ [From the figure]
Putting their respective values, we get:
${V_{BE}} = 10 - (40 \times {10^{ - 6}}A)(245 \times {10^3}\Omega )$
On simplification the bracket term and we get
$ \Rightarrow (10 - 9.8)V$
On subtracting we get,
$ \Rightarrow 0.2V$
Therefore, option (B) is the correct answer.
Additional information:
Common emitter transistor:
As such the common emitter configuration may be a good all round circuit to be used in many applications. It's also worth noting at this stage that the common emitter transistor amplifier inverts the signal at the input. Therefore if a waveform that's rising enters the input of the common emitter amplifier, it'll cause the output voltage to fall.
Note: Advantages of Common Emitter amplifier:
1. A common emitter amplifier is inverting and has low I/p impedance
2. High output impedance
3. High voltage gain.
4. High current gain.
Disadvantages of Common Emitter amplifier:
1. It’s a high output resistance.
2. It responds poorly to high frequencies.
3. Its high thermal instabilities.
4. Its voltage gain is extremely unstable
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