Question

r1 =7 kω, vcc = 12 v; r2 =7 kω, rc = 4.7 kω; β = 15, re = 6.3 kω

Explanation:

To solve for the operating point (Q - point) of this BJT (Bipolar Junction Transistor) circuit, we typically calculate the base voltage ($V_B$), emitter voltage ($V_E$), emitter current ($I_E$), collector current ($I_C$), base current ($I_B$), and collector - emitter voltage ($V_{CE}$). Here are the key steps:

Step 1: Calculate the base voltage ($V_B$)

The resistors $R_1$ and $R_2$ form a voltage divider across the supply voltage $V_{CC}$. The formula for the base voltage is given by the voltage - divider rule:

$$V_B=\frac{R_2}{R_1 + R_2}\times V_{CC}$$

We know that $R_1 = 7\space k\Omega$, $R_2=7\space k\Omega$, and $V_{CC} = 12\space V$.
Substituting 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: Calculate the emitter voltage ($V_E$)

For a BJT in active mode, the base - emitter voltage $V_{BE}\approx0.7\space V$ (for a silicon transistor). Using Kirchhoff's voltage law in the base - emitter loop: $V_B=V_{BE}+V_E$.
So, we can solve for $V_E$:

$$V_E=V_B - V_{BE}$$

Assuming $V_{BE} = 0.7\space V$, we have:

$$V_E=6\space V- 0.7\space V = 5.3\space V$$

Step 3: Calculate the emitter current ($I_E$)

The emitter resistor is $R_E = 6.3\space k\Omega$. Using Ohm's law ($I=\frac{V}{R}$) in the emitter circuit:

$$I_E=\frac{V_E}{R_E}$$

Substituting $V_E = 5.3\space V$ and $R_E=6.3\space k\Omega$:

$$I_E=\frac{5.3\space V}{6.3\space k\Omega}\approx0.841\space mA$$

Step 4: Calculate the collector current ($I_C$)

In the active mode of a BJT, the collector current $I_C\approx\beta I_B$ and also $I_E=I_C + I_B$. But for a transistor with a relatively large $\beta$ (here $\beta = 15$), we can approximate $I_C\approx I_E$ (since $I_B=\frac{I_E}{1 + \beta}$, and when $\beta$ is not extremely small, $I_C\approx I_E$). So, $I_C\approx I_E\approx0.841\space mA$

Step 5: Calculate the base current ($I_B$)

We know that $I_C=\beta I_B$, so we can solve for $I_B$:

$$I_B=\frac{I_C}{\beta}$$

Substituting $I_C\approx0.841\space mA$ and $\beta = 15$:

$$I_B=\frac{0.841\space mA}{15}\approx0.056\space mA = 56\space\mu A$$

Step 6: Calculate the collector - emitter voltage ($V_{CE}$)

Using Kirchhoff's voltage law in the collector - emitter loop: $V_{CC}=I_CR_C+V_{CE}+I_ER_E$. But since $I_C\approx I_E$, we can rewrite it as:

$$V_{CE}=V_{CC}-I_C(R_C + R_E)$$

Substituting $V_{CC}=12\space V$, $I_C\approx0.841\space mA$, $R_C = 4.7\space k\Omega$ and $R_E=6.3\space k\Omega$:
First, calculate $R_C + R_E=4.7\space k\Omega+6.3\space k\Omega = 11\space k\Omega$
Then, $I_C(R_C + R_E)=0.841\space mA\times11\space k\Omega = 9.251\space V$

$$V_{CE}=12\space V-9.251\space V = 2.749\space V$$

If we want to present the Q - point values:

• $I_C\approx0.84\space mA$
• $V_{CE}\approx2.75\space V$
• $I_B\approx56\space\mu A$

(Note: The approximations and exact values may vary slightly depending on the precision of $V_{BE}$ and the level of approximation used for $I_C\approx I_E$)

Answer:

To solve for the operating point (Q - point) of this BJT (Bipolar Junction Transistor) circuit, we typically calculate the base voltage ($V_B$), emitter voltage ($V_E$), emitter current ($I_E$), collector current ($I_C$), base current ($I_B$), and collector - emitter voltage ($V_{CE}$). Here are the key steps:

Step 1: Calculate the base voltage ($V_B$)

The resistors $R_1$ and $R_2$ form a voltage divider across the supply voltage $V_{CC}$. The formula for the base voltage is given by the voltage - divider rule:

$$V_B=\frac{R_2}{R_1 + R_2}\times V_{CC}$$

We know that $R_1 = 7\space k\Omega$, $R_2=7\space k\Omega$, and $V_{CC} = 12\space V$.
Substituting 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: Calculate the emitter voltage ($V_E$)

For a BJT in active mode, the base - emitter voltage $V_{BE}\approx0.7\space V$ (for a silicon transistor). Using Kirchhoff's voltage law in the base - emitter loop: $V_B=V_{BE}+V_E$.
So, we can solve for $V_E$:

$$V_E=V_B - V_{BE}$$

Assuming $V_{BE} = 0.7\space V$, we have:

$$V_E=6\space V- 0.7\space V = 5.3\space V$$

Step 3: Calculate the emitter current ($I_E$)

The emitter resistor is $R_E = 6.3\space k\Omega$. Using Ohm's law ($I=\frac{V}{R}$) in the emitter circuit:

$$I_E=\frac{V_E}{R_E}$$

Substituting $V_E = 5.3\space V$ and $R_E=6.3\space k\Omega$:

$$I_E=\frac{5.3\space V}{6.3\space k\Omega}\approx0.841\space mA$$

Step 4: Calculate the collector current ($I_C$)

In the active mode of a BJT, the collector current $I_C\approx\beta I_B$ and also $I_E=I_C + I_B$. But for a transistor with a relatively large $\beta$ (here $\beta = 15$), we can approximate $I_C\approx I_E$ (since $I_B=\frac{I_E}{1 + \beta}$, and when $\beta$ is not extremely small, $I_C\approx I_E$). So, $I_C\approx I_E\approx0.841\space mA$

Step 5: Calculate the base current ($I_B$)

We know that $I_C=\beta I_B$, so we can solve for $I_B$:

$$I_B=\frac{I_C}{\beta}$$

Substituting $I_C\approx0.841\space mA$ and $\beta = 15$:

$$I_B=\frac{0.841\space mA}{15}\approx0.056\space mA = 56\space\mu A$$

Step 6: Calculate the collector - emitter voltage ($V_{CE}$)

Using Kirchhoff's voltage law in the collector - emitter loop: $V_{CC}=I_CR_C+V_{CE}+I_ER_E$. But since $I_C\approx I_E$, we can rewrite it as:

$$V_{CE}=V_{CC}-I_C(R_C + R_E)$$

Substituting $V_{CC}=12\space V$, $I_C\approx0.841\space mA$, $R_C = 4.7\space k\Omega$ and $R_E=6.3\space k\Omega$:
First, calculate $R_C + R_E=4.7\space k\Omega+6.3\space k\Omega = 11\space k\Omega$
Then, $I_C(R_C + R_E)=0.841\space mA\times11\space k\Omega = 9.251\space V$

$$V_{CE}=12\space V-9.251\space V = 2.749\space V$$

If we want to present the Q - point values:

• $I_C\approx0.84\space mA$
• $V_{CE}\approx2.75\space V$
• $I_B\approx56\space\mu A$

(Note: The approximations and exact values may vary slightly depending on the precision of $V_{BE}$ and the level of approximation used for $I_C\approx I_E$)