1 Determine The Peak Output Voltage For The Bridge Rectifier In Figure ✓ Solved

1 Determine the peak output voltage for the bridge rectifier in Figure 1 shown. Assuming the practical model, what PIV rating is required for the diodes? The transformer is specified to have a 12 Vrms secondary voltage for the standard 120 V across the primary. Figure 1. Bridge Full Wave Rectifier 2 In the full-wave center-tapped rectifier circuit, the transformer has a turns ratio of 1:2.

The transformer primary winding is connected across an AC source of 230V (rms), 50 Hz. The load resistor is 50Ω. For this circuit, determine the DC output voltage, peak-to-peak ripple in the output voltage, and output ripple frequency. Show all work. 3 In an NPN transistor, the emitter and collector are both N-type materials.

A True B False 4 The dc beta is defined as the dc collector current divided by the dc emitter current. A True B False 5 When the ground side of each voltage source is connected to the emitter of a bipolar junction transistor, it is called a common collector. A True B False 6 The middle region of a transistor is the ________. 7 A transistor has a collector current of 12 mA and a base current of 40 µA. What is the current gain of the transistor?

Show all work. 8 A transistor has a current gain of 260. If the base current is 90 µA, what is the collector current?

Paper for above instructions

Assignment Solution: Electrical Circuits and Transistors

1. Peak Output Voltage for the Bridge Rectifier

The bridge rectifier is a circuit that transforms AC voltage into DC voltage. The output voltage (V_dc) for an ideal bridge rectifier is given by the equation:
\[
V_{dc} = V_{peak} - 2V_d
\]
Where:
- \( V_{peak} \) is the peak voltage from the transformer's secondary winding.
- \( V_d \) is the forward voltage drop across a diode. For this analysis, we will assume \( V_d \) is about 0.7V for silicon diodes.
Given that the transformer has a secondary voltage of 12 Vrms, the peak voltage (\( V_{peak} \)) can be calculated by converting the root mean square (RMS) voltage to peak voltage using the formula:
\[
V_{peak} = V_{rms} \times \sqrt{2}
\]
So, the calculation is:
\[
V_{peak} = 12 \, V_{rms} \times 1.414 \approx 16.97 \, V
\]
Using the formula for \( V_{dc} \):
\[
V_{dc} = 16.97 \, V - 2 \times 0.7 \, V
\]
\[
V_{dc} \approx 16.97 \, V - 1.4 \, V = 15.57 \, V
\]
PIV Rating for the Diodes:
The Peak Inverse Voltage (PIV) rating for the diodes in a bridge rectifier is twice the peak voltage from the transformer. Therefore:
\[
PIV = 2 \times V_{peak} = 2 \times 16.97 \, V \approx 33.94 \, V
\]

2. Full-Wave Center-Tapped Rectifier Circuit Analysis

In the center-tapped rectifier circuit, the transformer has a turns ratio of 1:2. This means that the secondary voltage will be half that of the primary voltage. Given the primary AC voltage is 230Vrms:
\[
V_{secondary} = \frac{230 \, V}{2} = 115 \, V_{rms}
\]
The peak voltage for half the secondary is:
\[
V_{peak} = 115 \, V_{rms} \times \sqrt{2} \approx 162.63 \, V
\]

DC Output Voltage

For a center-tapped full-wave rectifier, the DC output voltage can be calculated as:
\[
V_{dc} = \frac{V_{peak}}{2} - V_d
\]
Assuming \( V_d \) is approximately 0.7 V:
\[
V_{dc} = \frac{162.63 \, V}{2} - 0.7 \, V = 81.315 \, V - 0.7 \, V = 80.615 \, V
\]

Peak-to-Peak Ripple Calculation

The peak-to-peak ripple voltage (\( V_{ripple} \)) can be determined using the formula:
\[
V_{ripple} = \frac{I_{load}}{f \times C}
\]
Where:
- \( I_{load} = \frac{V_{dc}}{R} = \frac{80.615 V}{50 \, \Omega} = 1.6123 \, A \)
- \( f = 100 \, Hz \) (because the ripple frequency is double the AC supply frequency)
- \( C \) must be defined. Assuming \( C \) for smooth ripple is 1000 µF (0.001 F):
\[
V_{ripple} = \frac{1.6123 \, A}{100 \, Hz \times 0.001 \, F} = \frac{1.6123}{0.1} = 16.123 \, V_{pp}
\]

Output Ripple Frequency

For a full-wave rectifier, the output ripple frequency is double that of the AC source:
\[
f_{ripple} = 2 \times f_{AC} = 2 \times 50 \, Hz = 100 \, Hz
\]

3. True/False Questions

3. In an NPN transistor, the emitter and collector are both N-type materials.
- Answer: B (False). In an NPN transistor, the emitter is N-type, while the base is P-type and the collector can be either N-type or P-type depending on circuit configurations.
4. The DC beta is defined as the DC collector current divided by the DC emitter current.
- Answer: B (False). The DC beta is defined as the DC collector current divided by the DC base current, not emitter current.
5. When the ground side of each voltage source is connected to the emitter of a bipolar junction transistor, it is called a common collector.
- Answer: A (True). This configuration is referred to as a common collector or emitter follower.
6. The middle region of a transistor is the ________.
- Answer: Base.

4. Transistor Current Gain Calculations

7. A transistor has a collector current of 12 mA and a base current of 40 µA. What is the current gain of the transistor?
Current gain, \( \beta \), is calculated as:
\[
\beta = \frac{I_C}{I_B}
\]
Where:
- \( I_C = 12 \, mA = 0.012 \, A \)
- \( I_B = 40 \, µA = 0.00004 \, A \)
Now, substituting the values:
\[
\beta = \frac{0.012 \, A}{0.00004 \, A} = 300
\]
8. A transistor has a current gain of 260. If the base current is 90 µA, what is the collector current?
Collector current can be calculated using the formula:
\[
I_C = \beta \times I_B
\]
Substituting values gives:
\[
I_C = 260 \times 90 \, µA = 260 \times 0.00009 = 0.0234 \, A = 23.4 \, mA
\]

Conclusion

This assignment explored calculations involved with bridge and center-tapped rectifiers, provided fundamental electronics knowledge through true/false questions, and analyzed transistor behavior through current gain calculations. Electrical principles remain vital to engineers and technicians in effectively managing circuits and designing systems.

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