Question
The following data was collected from a common emitter circuit diagram: B=240, Vcc= 2,4V, Ic=Icq=4mA. Calculate Rb and Rc
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GPT-4o mini
To calculate the bias resistor \(R_B\) and the collector resistor \(R_C\) for a common emitter amplifier with the given parameters, we can use the following relationships.
1. **Determine \(V_E\)**: Since \(V_{CE}\) needs to be at least \(0.2V\) for the transistor to be in active mode, let's first determine \(V_E\) (the emitter voltage). Assume a reasonable \(V_{CE}\). For this example, let's take \(V_{CE} = V_{CC} - V_E - I_C R_C\).
2. **Calculate \(R_C\)**: We will use the given collector current \(I_C\):
\[
I_C = 4 \, \text{mA} = 0.004 \, \text{A}
\]
The value of \(R_C\) can be found using:
\[
V_C = V_{CC} - I_C R_C
\]
The output voltage \(V_C\) should be around \(V_{CC}/2\) for maximum swing in some cases. However, it's often calculated or set by specific design parameters. Assuming we want the collector voltage \(V_C\) to be about 1V (you can adjust as needed), we can substitute:
\[
1.0V = V_{CC} - I_C R_C
\]
Now substituting the known values:
\[
1.0V = 2.4V - 0.004A \cdot R_C
\]
Therefore:
\[
0.004A \cdot R_C = 2.4V - 1.0V
\]
\[
0.004A \cdot R_C = 1.4V
\]
\[
R_C = \frac{1.4V}{0.004A} = 350 \, \Omega
\]
3. **Calculate \(R_B\)**: This resistor is dependent on the base current \(I_B\). First, we must calculate \(I_B\) using the relation considering the current gain \( \beta \):
\[
I_B = \frac{I_C}{\beta} = \frac{4 \, \text{mA}}{240} \approx 0.01667 \, \text{mA}
\]
The voltage across \(R_B\) can be determined from the base-emitter voltage (\(V_{BE}\)), which is typically around \(0.7V\):
\[
V_B = V_E + V_{BE}
\]
Assuming we want the emitter voltage to be close to \(0.6V\) if the collector current is 4mA, we calculate \(R_B\) under the assumption \(V_E \approx 0.6\,V\):
\[
V_B \approx 0.6V + 0.7V = 1.3V
\]
Thus, the voltage across \(R_B\) is:
\[
V_{RB} = V_{CC} - V_B = 2.4V - 1.3V = 1.1V
\]
Now we can calculate \(R_B\):
\[
R_B = \frac{V_{RB}}{I_B} = \frac{1.1V}{0.01667 \, \text{mA}} = \frac{1.1V}{0.00001667A} \approx 66,000 \, \Omega \approx 66 \, k\Omega
\]
So finally:
- \(R_C \approx 350 \, \Omega\)
- \(R_B \approx 66 \, k\Omega\)
This calculation provides a basic design for the resistors \(R_B\) and \(R_C\) in a common emitter transistor amplifier given your specifications. Adjustments may be needed based on further design details and requirements.
1. **Determine \(V_E\)**: Since \(V_{CE}\) needs to be at least \(0.2V\) for the transistor to be in active mode, let's first determine \(V_E\) (the emitter voltage). Assume a reasonable \(V_{CE}\). For this example, let's take \(V_{CE} = V_{CC} - V_E - I_C R_C\).
2. **Calculate \(R_C\)**: We will use the given collector current \(I_C\):
\[
I_C = 4 \, \text{mA} = 0.004 \, \text{A}
\]
The value of \(R_C\) can be found using:
\[
V_C = V_{CC} - I_C R_C
\]
The output voltage \(V_C\) should be around \(V_{CC}/2\) for maximum swing in some cases. However, it's often calculated or set by specific design parameters. Assuming we want the collector voltage \(V_C\) to be about 1V (you can adjust as needed), we can substitute:
\[
1.0V = V_{CC} - I_C R_C
\]
Now substituting the known values:
\[
1.0V = 2.4V - 0.004A \cdot R_C
\]
Therefore:
\[
0.004A \cdot R_C = 2.4V - 1.0V
\]
\[
0.004A \cdot R_C = 1.4V
\]
\[
R_C = \frac{1.4V}{0.004A} = 350 \, \Omega
\]
3. **Calculate \(R_B\)**: This resistor is dependent on the base current \(I_B\). First, we must calculate \(I_B\) using the relation considering the current gain \( \beta \):
\[
I_B = \frac{I_C}{\beta} = \frac{4 \, \text{mA}}{240} \approx 0.01667 \, \text{mA}
\]
The voltage across \(R_B\) can be determined from the base-emitter voltage (\(V_{BE}\)), which is typically around \(0.7V\):
\[
V_B = V_E + V_{BE}
\]
Assuming we want the emitter voltage to be close to \(0.6V\) if the collector current is 4mA, we calculate \(R_B\) under the assumption \(V_E \approx 0.6\,V\):
\[
V_B \approx 0.6V + 0.7V = 1.3V
\]
Thus, the voltage across \(R_B\) is:
\[
V_{RB} = V_{CC} - V_B = 2.4V - 1.3V = 1.1V
\]
Now we can calculate \(R_B\):
\[
R_B = \frac{V_{RB}}{I_B} = \frac{1.1V}{0.01667 \, \text{mA}} = \frac{1.1V}{0.00001667A} \approx 66,000 \, \Omega \approx 66 \, k\Omega
\]
So finally:
- \(R_C \approx 350 \, \Omega\)
- \(R_B \approx 66 \, k\Omega\)
This calculation provides a basic design for the resistors \(R_B\) and \(R_C\) in a common emitter transistor amplifier given your specifications. Adjustments may be needed based on further design details and requirements.
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