Question

r1 = 7 kω, vcc = 12 v; r2 = 7 kω, rc = 4.7 kω; β = 15, re = 6.3 kω (circuit diagram with r1, r2, rc, re, vcc, and transistor u1)

Explanation:

To solve for the operating point (Q - point) parameters (like \( V_{BE} \), \( I_B \), \( I_C \), \( I_E \), \( V_{CE} \)) of this BJT (Bipolar Junction Transistor) circuit, we follow these steps:

Step 1: Calculate the Base Voltage (\( V_B \))

The base voltage is determined by the voltage - divider formed by \( R_1 \) and \( R_2 \). The formula for the voltage - divider is \( V_B=\frac{R_2}{R_1 + R_2}\times V_{CC} \).

Given \( R_1 = 7\space k\Omega \), \( R_2 = 7\space k\Omega \), and \( V_{CC}=12\space V \).

Substitute the values into the formula:
\( V_B=\frac{7\space k\Omega}{7\space k\Omega + 7\space k\Omega}\times12\space V=\frac{7}{14}\times12\space V = 6\space V \)

Step 2: Assume \( V_{BE}\approx0.7\space V \) (for a silicon transistor) and calculate the Emitter Voltage (\( V_E \))

We know that \( V_E=V_B - V_{BE} \).

Substitute \( V_B = 6\space V \) and \( V_{BE}=0.7\space V \):
\( V_E=6\space V - 0.7\space V = 5.3\space V \)

Step 3: Calculate the Emitter Current (\( I_E \))

Using Ohm's Law (\( I=\frac{V}{R} \)) for the emitter resistor \( R_E \), we have \( I_E=\frac{V_E}{R_E} \).

Given \( R_E = 6.3\space k\Omega \) and \( V_E = 5.3\space V \):
\( I_E=\frac{5.3\space V}{6.3\space k\Omega}\approx0.841\space mA \)

Step 4: Calculate the Base Current (\( I_B \))

For a BJT, \( I_E=I_B + I_C \) and \( I_C=\beta I_B \) (where \( \beta \) is the current gain). So, \( I_E=I_B+\beta I_B=I_B(1 + \beta) \).

We can re - arrange this formula to solve for \( I_B \): \( I_B=\frac{I_E}{1 + \beta} \).

Given \( \beta = 15 \) and \( I_E\approx0.841\space mA \):
\( I_B=\frac{0.841\space mA}{1 + 15}=\frac{0.841\space mA}{16}\approx0.0526\space mA = 52.6\space\mu A \)

Step 5: Calculate the Collector Current (\( I_C \))

Using the relationship \( I_C=\beta I_B \):

Substitute \( \beta = 15 \) and \( I_B\approx0.0526\space mA \):
\( I_C=15\times0.0526\space mA\approx0.789\space mA \)

Step 6: Calculate the Collector - Emitter Voltage (\( V_{CE} \))

First, find the voltage drop across \( R_C \) (\( V_{RC}=I_CR_C \)) and the voltage drop across \( R_E \) (\( V_{RE}=I_ER_E \)). Then, use \( V_{CE}=V_{CC}-V_{RC}-V_{RE} \).

• Calculate \( V_{RC} \): \( V_{RC}=I_CR_C=0.789\space mA\times4.7\space k\Omega\approx3.71\space V \)
• Calculate \( V_{RE} \): \( V_{RE}=I_ER_E = 0.841\space mA\times6.3\space k\Omega\approx5.3\space V \)

Now, calculate \( V_{CE} \):
\( V_{CE}=12\space V-3.71\space V - 5.3\space V=3\space V \) (approximate, considering rounding errors in intermediate steps)

Final Answers (Approximate Values)

• \( V_B = 6\space V \)
• \( V_E = 5.3\space V \)
• \( I_B\approx52.6\space\mu A \)
• \( I_C\approx0.79\space mA \)
• \( I_E\approx0.84\space mA \)
• \( V_{CE}\approx3\space V \)

If you want to calculate a specific parameter (e.g., \( I_C \), \( V_{CE} \)), you can use the relevant step from above. For example, if you want \( I_C \), the final answer (approximate) is \( \boldsymbol{0.79\space mA} \) (or more precisely \( \approx0.789\space mA \)).

Answer:

To solve for the operating point (Q - point) parameters (like \( V_{BE} \), \( I_B \), \( I_C \), \( I_E \), \( V_{CE} \)) of this BJT (Bipolar Junction Transistor) circuit, we follow these steps:

Step 1: Calculate the Base Voltage (\( V_B \))

The base voltage is determined by the voltage - divider formed by \( R_1 \) and \( R_2 \). The formula for the voltage - divider is \( V_B=\frac{R_2}{R_1 + R_2}\times V_{CC} \).

Given \( R_1 = 7\space k\Omega \), \( R_2 = 7\space k\Omega \), and \( V_{CC}=12\space V \).

Substitute the values into the formula:
\( V_B=\frac{7\space k\Omega}{7\space k\Omega + 7\space k\Omega}\times12\space V=\frac{7}{14}\times12\space V = 6\space V \)

Step 2: Assume \( V_{BE}\approx0.7\space V \) (for a silicon transistor) and calculate the Emitter Voltage (\( V_E \))

We know that \( V_E=V_B - V_{BE} \).

Substitute \( V_B = 6\space V \) and \( V_{BE}=0.7\space V \):
\( V_E=6\space V - 0.7\space V = 5.3\space V \)

Step 3: Calculate the Emitter Current (\( I_E \))

Using Ohm's Law (\( I=\frac{V}{R} \)) for the emitter resistor \( R_E \), we have \( I_E=\frac{V_E}{R_E} \).

Given \( R_E = 6.3\space k\Omega \) and \( V_E = 5.3\space V \):
\( I_E=\frac{5.3\space V}{6.3\space k\Omega}\approx0.841\space mA \)

Step 4: Calculate the Base Current (\( I_B \))

For a BJT, \( I_E=I_B + I_C \) and \( I_C=\beta I_B \) (where \( \beta \) is the current gain). So, \( I_E=I_B+\beta I_B=I_B(1 + \beta) \).

We can re - arrange this formula to solve for \( I_B \): \( I_B=\frac{I_E}{1 + \beta} \).

Given \( \beta = 15 \) and \( I_E\approx0.841\space mA \):
\( I_B=\frac{0.841\space mA}{1 + 15}=\frac{0.841\space mA}{16}\approx0.0526\space mA = 52.6\space\mu A \)

Step 5: Calculate the Collector Current (\( I_C \))

Using the relationship \( I_C=\beta I_B \):

Substitute \( \beta = 15 \) and \( I_B\approx0.0526\space mA \):
\( I_C=15\times0.0526\space mA\approx0.789\space mA \)

Step 6: Calculate the Collector - Emitter Voltage (\( V_{CE} \))

First, find the voltage drop across \( R_C \) (\( V_{RC}=I_CR_C \)) and the voltage drop across \( R_E \) (\( V_{RE}=I_ER_E \)). Then, use \( V_{CE}=V_{CC}-V_{RC}-V_{RE} \).

• Calculate \( V_{RC} \): \( V_{RC}=I_CR_C=0.789\space mA\times4.7\space k\Omega\approx3.71\space V \)
• Calculate \( V_{RE} \): \( V_{RE}=I_ER_E = 0.841\space mA\times6.3\space k\Omega\approx5.3\space V \)

Now, calculate \( V_{CE} \):
\( V_{CE}=12\space V-3.71\space V - 5.3\space V=3\space V \) (approximate, considering rounding errors in intermediate steps)

Final Answers (Approximate Values)

• \( V_B = 6\space V \)
• \( V_E = 5.3\space V \)
• \( I_B\approx52.6\space\mu A \)
• \( I_C\approx0.79\space mA \)
• \( I_E\approx0.84\space mA \)
• \( V_{CE}\approx3\space V \)

If you want to calculate a specific parameter (e.g., \( I_C \), \( V_{CE} \)), you can use the relevant step from above. For example, if you want \( I_C \), the final answer (approximate) is \( \boldsymbol{0.79\space mA} \) (or more precisely \( \approx0.789\space mA \)).