Question

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

Answer:

To analyze this transistor circuit (likely a BJT voltage - divider bias circuit), we follow these steps:

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

The resistors $R_1$ and $R_2$ form a voltage divider across $V_{CC}$. The formula for the base voltage is $V_B=\frac{R_2}{R_1 + R_2}\times V_{CC}$.
Given that $R_1 = 7\space k\Omega$, $R_2=7\space k\Omega$ 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 the base - emitter voltage ($V_{BE}$)

For a silicon BJT, we assume $V_{BE}\approx0.7\space V$. Then we can calculate the emitter voltage ($V_E$) using the formula $V_E=V_B - V_{BE}$.

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

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

Using Ohm's law, $I_E=\frac{V_E}{R_E}$. 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 collector current ($I_C$)

For a BJT in the active region, $I_C\approx I_E$ (since $\alpha\approx1$ for most BJTs, and $I_C=\alpha I_E$). But we can also use the relationship $I_C=\beta I_B$, and $I_E=I_C + I_B=(1 + \beta)I_B$. However, when $\beta$ is not too small, $I_C\approx I_E$.
If we use the $\beta$ value, we know that $I_B=\frac{I_C}{\beta}$. And from the voltage - divider, we can also analyze the base current. But as a first approximation, since $I_E\approx0.841\space mA$, $I_C\approx0.841\space mA$ (if we consider the active region operation).

Step 5: Calculate the collector voltage ($V_C$)

Using the formula $V_C=V_{CC}-I_C\times R_C$. Given $R_C = 4.7\space k\Omega$ and $V_{CC}=12\space V$, and $I_C\approx0.841\space mA$.

$$V_C=12\space V-0.841\space mA\times4.7\space k\Omega=12\space V - 3.95\space V=8.05\space V$$

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

The formula for $V_{CE}$ is $V_{CE}=V_C - V_E$.

$$V_{CE}=8.05\space V - 5.3\space V = 2.75\space V$$

If we want to check the active region condition: For a BJT to be in the active region, $V_{CE}>V_{BE}$ (or $V_{CE}>0.7\space V$ for silicon). Here $V_{CE} = 2.75\space V>0.7\space V$, so the transistor is in the active region.

If we want to calculate the base current ($I_B$), we can use $I_B=\frac{I_C}{\beta}$. Since $I_C\approx0.841\space mA$ and $\beta = 15$.

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

We can also calculate $I_B$ from the base voltage: $I_B=\frac{V_B - V_{BE}}{R_{th}}$, where $R_{th}=\frac{R_1\times R_2}{R_1 + R_2}=\frac{7\times7}{7 + 7}\space k\Omega = 3.5\space k\Omega$.

$$I_B=\frac{V_B - V_{BE}}{R_{th}}=\frac{6 - 0.7}{3.5\space k\Omega}=\frac{5.3}{3.5\space k\Omega}\approx1.514\space mA$$

There is a discrepancy here. The reason is that when $\beta$ is small ( $\beta = 15$ is relatively small), the approximation $I_C\approx I_E$ and the voltage - divider approximation (assuming that the base current is small compared to the current through $R_1$ and $R_2$) may not hold well.

Let's do a more accurate analysis:

The current through $R_1$ is $I_{R1}=\frac{V_{CC}-V_B}{R_1}=\frac{12 - 6}{7\space k\Omega}=\frac{6}{7}\space mA\approx0.857\space mA$

The current through $R_2$ is $I_{R2}=\frac{V_B}{R_2}=\frac{6}{7}\space mA\approx0.857\space mA$

The base current $I_B$ should satisfy $I_{R2}=I_B + I_{R_{th}\text{-to - emitter}}$ (but in our case, the emitter current is $I_E=(1 + \beta)I_B$)

We know that $V_E = I_E\times R_E=(1 + \beta)I_B\times R_E$

And $V_B=V_{BE}+V_E=V_{BE}+(1 + \beta)I_B\times R_E$

So, $I_B=\frac{V_B - V_{BE}}{(1 + \beta)R_E}$

Substitute $V_B = 6\space V$, $V_{BE}=0.7\space V$, $\beta = 15$ and $R_E = 6.3\space k\Omega$

$$I_B=\frac{6 - 0.7}{(1 + 15)\times6.3\space k\Omega}=\frac{5.3}{16\times6.3\space k\Omega}=\frac{5.3}{100.8\space k\Omega}\approx0.0526\space mA = 52.6\space\mu A$$

Then $I_C=\beta I_B=15\times0.0526\space mA = 0.789\space mA$

$I_E=I_C + I_B=0.789\space mA+ 0.0526\space mA=0.8416\space mA$ (which matches our earlier approximation of $I_E$)

$V_E=I_E\times R_E=0.8416\space mA\times6.3\space k\Omega\approx5.302\space V$

$V_C=V_{CC}-I_C\times R_C=12\space V-0.789\space mA\times4.7\space k\Omega=12 - 3.708\space V = 8.292\space V$

$V_{CE}=V_C - V_E=8.292 - 5.302=2.99\space V\approx3\space V$

This circuit analysis falls under the subfield of Engineering (Electrical Engineering, specifically Electronics Engineering) within the Natural Science discipline. The calculations involve using Ohm's law, transistor biasing principles (voltage - divider bias for BJT), and understanding the operation of a bipolar junction transistor (BJT) in the active region.