Bridge Rectifier Bipolar Junction Transistors: Determine The

Bridge Rectifier Bipolar Junction Transistors1 Determine The Peak Ou

Determine the peak output voltage for the bridge rectifier shown, assuming the practical model, and specify the required PIV rating for the diodes. The transformer has a secondary voltage of 12 Vrms, with a 120 V primary voltage. Additionally, analyze the full-wave center-tapped rectifier circuit with a turns ratio of 1:2, connected across a 230 V (rms), 50 Hz AC source, with a load resistor of 50 Ω, calculating the DC output voltage, peak-to-peak ripple, and ripple frequency. Address questions regarding NPN transistors, including the correctness of specific statements about their structure and parameters, and calculate the current gain given collector and base currents. Furthermore, determine the collector current for a given base current and current gain, and specify the capacitor needed to limit ripple voltage to 0.4 V for both full-wave and half-wave rectifiers with specified input parameters. Lastly, design considerations for a full-wave rectifier to deliver a 12 V peak output at 120 mA load with minimal ripple, including transformer ratio and filter capacitor size; compute secondary voltage, ripple voltage, efficiency, and diode PIV and current ratings for a circuit using silicon diodes and specific load conditions. Include all calculations, explanations, and reasoning for each part, ensuring comprehensive understanding of each circuit and transistor parameter involved.

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Rectification circuits, particularly bridge and center-tapped full-wave rectifiers, are fundamental in converting alternating current (AC) into direct current (DC). The primary goal of these circuits is to produce a stable DC voltage with minimal ripple, suitable for powering electronic devices. Their design involves careful selection of components such as diodes, transformers, and filtering elements to meet voltage and current requirements efficiently.

Peak Output Voltage of the Bridge Rectifier

In the given bridge rectifier circuit, the peak output voltage (V_peak) can be estimated considering the transformer's secondary voltage and the voltage drops across diodes. Since the transformer secondary voltage is specified as 12 Vrms, the peak voltage V_m (or V_peak) is given by:

V_m = Vrms × √2 ≈ 12 V × 1.414 ≈ 16.97 V

Practically, diode voltage drops must be accounted for, typically around 0.7 V for silicon diodes. Since a bridge rectifier involves two diodes in series during conduction, the total voltage drop is approximately 1.4 V. Therefore, the peak load voltage (V_load) is:

V_{load_peak} ≈ V_m - 1.4 V ≈ 16.97 V - 1.4 V ≈ 15.57 V

The PIV (Peak Inverse Voltage) rating for the diodes must exceed the peak voltage across them during the reverse bias phase. In a full-bridge rectifier, each diode experiences a voltage approximately equal to the applied peak voltage (excluding diode drops). Therefore, PIV should be at least:

PIV ≥ V_m ≈ 16.97 V, but to ensure safety margin, diodes with PIV ratings of at least 20 V are recommended.

Full-Wave Center-Tapped Rectifier Parameters

The transforms with a turns ratio of 1:2 and secondary voltage of 230 V rms produce a secondary voltage V_s of 230 V. The peak secondary voltage is:

V_{m_secondary} = 230 V × √2 ≈ 325.3 V

Considering the turns ratio, the secondary winding supplies a voltage suitable for stepping down to the desired load. The output DC voltage (V_DC) of a full-wave rectifier is approximately:

V_DC ≈ V_{m_secondary} - 2 × V_diode ≈ 325.3 V - 1.4 V ≈ 323.9 V

The ripple voltage (V_ripple) depends on the load current and the filter capacitance. Its peak-to-peak value can be calculated using:

V_ripple ≈ I_load / (f_ripple × C)

where f_ripple is twice the line frequency (since full-wave), i.e., 100 Hz, and C is the capacitance. The ripple frequency is therefore 100 Hz.

Transistor Parameters and Calculations

Regarding NPN transistors, the statement that both emitter and collector are N-type materials is false because in a typical NPN transistor, the emitter and collector are N-type, and the base is P-type.

The dc current gain (β_dc) is defined as the collector current divided by the base current:

β_dc = I_c / I_b

Given I_c = 12 mA, I_b = 40 µA, the gain is:

β_dc = 12 mA / 40 µA = 12×10^-3 A / 40×10^-6 A = 300

Another question relates to the collector current (I_c) given a base current and current gain:

I_c = β × I_b = 260 × 90 µA = 260 × 0.09 mA = 23.4 mA

Capacitor Calculation for Ripple Voltage

The required capacitance (C) to limit ripple to Vr across load R with peak voltage Vm involves the equation:

C = I_load / (f × Vr)

where I_load = (V_DC / R), assuming V_DC ≈ V_m. For the full-wave rectifier with Vm = 12 V, R = 2000 Ω, Vr = 0.4 V, and f = 120 Hz (twice 60 Hz line frequency):

I_load ≈ 12 V / 2000 Ω = 6 mA

C ≈ 6 mA / (120 Hz × 0.4 V) ≈ 6×10^-3 A / (120 × 0.4) ≈ 6×10^-3 / 48 ≈ 125 µF

Similarly, for a half-wave rectifier, the frequency is 60 Hz, and the capacitor can be slightly larger to compensate for different ripple characteristics.

Designing a Rectifier for a Specific Output

Design considerations include ensuring the transformer ratio can drop the line voltage to the desired peak load voltage, calculating the filter capacitor size to maintain ripple below specified limits, and verifying diode ratings.

For example, achieving a 12 V peak output at 120 mA from a 120 V, 60 Hz input involves a transformer with a turns ratio of approximately 10:1, and a capacitor of at least 220 µF to ensure ripple is controlled. The diode PIV must be higher than the peak secondary voltage, approximately 16 V, so selecting diodes rated at least 20 V is prudent.

Efficiency and Diode Ratings

The efficiency of the rectifier circuit depends on the power delivered to the load relative to input power, factoring in diode losses. With silicon diodes, forward resistance and voltage drops reduce efficiency slightly but are acceptable within the design parameters.

The secondary transformer voltage in a full-wave rectifier supplying a 12 V load at 12 Ω would be approximately 15 V peak, with diode drops considered. For a load of 12 Ω at this voltage, the load current and ripple are consistent with the calculations above.

In conclusion, these calculations and considerations emphasize the importance of component ratings, precise calculations for ripple, and efficient design choices to ensure stable and reliable rectifier operation. Proper understanding of transistor parameters further enhances the design of control circuits involved in power regulation and conversion tasks.

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