Question

A circuit is arranged as shown in figure. The output voltage \(V_o\) is equal to ________ V. 

 

Hint:

In transistor circuits, the first step is always to find the base voltage. Look for voltage dividers or diode logic that sets the base potential. Once you have \(V_B\), calculate \(V_{BE}\) to determine if the transistor is ON (active/saturation) or OFF (cut-off). This will usually determine the output state.

Answer 1

Correct Answer: 5

Step 1: Understanding the Question:
We are given a circuit containing diodes and a transistor and asked to find the output voltage \(V_o\). We need to determine the operating state of the transistor (cut-off, active, or saturation).
Step 2: Key Formula or Approach:
1. Analyze the input diode section to find the voltage at the node connected to the transistor's base. 2. Use this base voltage to determine if the transistor's base-emitter junction is forward-biased. 3. The transistor is in cut-off if the base-emitter voltage \(V_{BE}\) is less than the cut-in voltage (typically \(\sim\)0.7 V for silicon). 4. If the transistor is in cut-off, it acts as an open circuit, and no collector current flows. The output voltage will be the collector supply voltage.
Step 3: Detailed Explanation:
Let's analyze the circuit. The input to diode D1 is 0 V, and the input to diode D2 is +5 V. Let's assume the diodes are ideal for simplicity.
- Diode D2 has +5 V at its anode. - Diode D1 has 0 V at its anode. - Since the anode of D2 is at a higher potential, D2 will be forward biased (ON), and D1 will be reverse biased (OFF). - Therefore, the junction point P (where the diodes meet the first resistor) will be at +5 V.
Now, consider the base of the NPN transistor. The base is connected to a voltage divider formed by two equal resistors R. One resistor is connected to point P (+5 V), and the other is connected to -5 V.
The voltage at the base, \(V_B\), can be calculated using the voltage divider rule: \[ V_B = \frac{(-5 \text{ V}) \cdot R + (+5 \text{ V}) \cdot R}{R + R} = \frac{-5R + 5R}{2R} = \frac{0}{2R} = 0 \text{ V} \] So, the potential at the base of the transistor is 0 V.
The emitter of the transistor is connected directly to the ground, so its potential is \(V_E = 0\) V.
The base-emitter voltage is \(V_{BE} = V_B - V_E = 0 \text{ V} - 0 \text{ V} = 0\) V.
For an NPN transistor to turn on and conduct, the base-emitter voltage \(V_{BE}\) must be positive and greater than its cut-in voltage (around 0.6 V to 0.7 V). Since \(V_{BE} = 0\) V, the transistor is in the cut-off region.
When the transistor is in cut-off, it acts like an open switch between its collector and emitter. No current flows through the collector (\(I_C = 0\)).
The output voltage \(V_o\) is the voltage at the collector. Since no current flows through the collector resistor R, there is no voltage drop across it. Therefore, the output voltage \(V_o\) is equal to the collector supply voltage.
\[ V_o = 5 \text{ V} \] Step 4: Final Answer:
The output voltage \(V_o\) is equal to 5 V.