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A canter-tapped full-wave rectifier supplies 01A and 15 (peak) to the load. The ripple voltage at 0.4V peak to peak. The input signal from the transformer (primary) is 120V (rms) at 60Hz. Silicon diodes will be used

Determine the ratio of turns (2) needed.

User Burndog
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Final answer:

The question relates to calculating the turns ratio for a transformer used in a rectifier circuit. Understanding the transformer's principles and the full-wave rectifier configuration, the needed secondary to primary turns ratio is approximately 0.0967, when accounting for the diode voltage drop.

Step-by-step explanation:

The question involves determining the turns ratio of a transformer in a full-wave rectifier circuit. Given the primary input of 120V (rms) from the transformer and a peak output required for the load being 15V, alongside the use of silicon diodes which typically have a forward voltage drop of approximately 0.7 volts per diode, we can calculate the turns ratio. The peak secondary voltage needs to be higher than the load voltage by at least the forward voltage drop of two diodes (since it is a full-wave rectifier which uses two diodes in both halves of the cycle).

For a center-tapped full-wave rectifier:

  • The peak secondary voltage Vp(sec) is equal to the load voltage plus twice the diode drop: Vp(sec) = 15V + 2(0.7V) = 16.4V
  • To find the rms value of the secondary voltage Vrms(sec), the peak voltage Vp(sec) is divided by √2: Vrms(sec) = 16.4V / √2 ≈ 11.6V (rms)
  • The turns ratio Nsecondary/Nprimary (Ns/Np) can be calculated by dividing the rms secondary voltage by the rms primary voltage: (11.6V rms) / (120V rms)

Performing this calculation:

  • Ns/Np = 11.6 / 120 ≈ 0.0967.

Therefore, the turns ratio needed for the secondary to primary winding is approximately 0.0967 or 9.67% of the primary turns.

User Abass A
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