Let's solve for $V_O$ and $V_{CE}$ in this transistor circuit.

Given Data:

1. $V_{BE} = 0.7 \, \text{V}$
2. $\beta = 100$
3. $R_B = 10 \, \text{k}\Omega$
4. $R_C = 500 \, \Omega$
5. $R_E = 200 \, \Omega$
6. Supply voltages:
   • $V_{CC} = 12 \, \text{V}$ (for the collector side)
   • $V_{BB} = 5 \, \text{V}$ (for the base side)

Step-by-Step Solution:

1. Calculate Base Current $I_B$:

Using Kirchhoff's Voltage Law (KVL) for the base circuit: $V_{BB} = I_B R_B + V_{BE} + I_E R_E$ Since $I_E = (1 + \beta) I_B$, we substitute $I_E$ and rearrange to find $I_B$: $5 = I_B (10 \times 10^3) + 0.7 + I_B (1 + 100)(200)$ $5 = I_B (10,000 + 20,200)$ $I_B = \frac{5 - 0.7}{30,200} = \frac{4.3}{30,200} \approx 1.424 \, \mu A$

2. Calculate Collector Current $I_C$:

Using $I_C = \beta I_B$: $I_C = 100 \times 1.424 \, \mu A \approx 0.1424 \, \text{mA}$

3. Calculate $V_O$:

Using Ohm's law on the collector resistor $R_C$: $V_O = V_{CC} - I_C R_C$ $V_O = 12 - (0.1424 \times 500)$ $V_O = 12 - 0.0712 = 2.876 \, \text{V}$

4. Calculate $V_{CE}$:

The voltage $V_{CE}$ is given by: $V_{CE} = V_O - V_E$ Since $V_E = I_E R_E$ and $I_E \approx I_C$ (assuming $\beta$ is large): $V_E = 0.1424 \times 200 = 2.848 \, \text{V}$ $V_{CE} = 2.876 - 0.7 = 2.005 \, \text{V}$

Answers:

1. $V_O = 2.876 \, \text{V}$
2. $V_{CE} = 2.005 \, \text{V}$

Would you like further details, or do you have questions?

Related Questions:

1. How would changing $\beta$ affect $V_O$ and $V_{CE}$?
2. What happens to $V_O$ if $R_E$ is increased?
3. How would the circuit behave if $V_{BE}$ was different?
4. How can we determine if the transistor is in saturation?
5. What is the role of $R_C$ in setting the output voltage?

Tip:

To verify if a transistor is in the active region, check if $V_{CE} > V_{BE}$.