Designing Full-Wave Rectifier: 110V AC to 11V DC

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Discussion Overview

The discussion focuses on designing a full-wave rectifier to convert 110V AC from a wall socket to a lower DC voltage, specifically 11V. It involves concepts related to transformers, voltage ratios, and time measurement in the context of AC waveforms.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes using a transformer with a turn ratio of 10:1 to lower the voltage from 110V to 11V.
  • Another participant clarifies that the turns ratio is equivalent to the ratio of the two voltages, emphasizing the importance of understanding RMS and peak values.
  • There is a discussion about calculating the peak voltage from the RMS value, with a formula provided: Peak voltage = 1.414 * RMS voltage.
  • A participant questions how to determine the unit of time for the horizontal axis in the context of a 100 Hz sine wave.
  • One participant suggests using the period T, calculated as 1/100, indicating that the time scale could be in centiseconds.
  • Another participant suggests that milliseconds would be a more elegant choice for the time scale, proposing to test this idea.

Areas of Agreement / Disagreement

Participants generally agree on the use of a transformer and the relationship between RMS and peak voltages. However, there is a lack of consensus on the most appropriate unit of time for the horizontal axis, with differing opinions on using centiseconds versus milliseconds.

Contextual Notes

There are unresolved aspects regarding the exact calculations for peak voltage and the implications of diode drops in the rectification process. Additionally, the choice of time unit remains open to interpretation.

Who May Find This Useful

This discussion may be useful for students or individuals interested in electrical engineering, specifically in the design and analysis of rectifier circuits.

calvert11
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Homework Statement



Design a full-wave rectifier that converts 110 [V], 100[Hz] ac electricity from a regular wall socket to a low DC voltage.

a) (Fill in the blanks) To lower the voltage from 110 [V] to 11[V], we use a _________, with a turn ration of ___:___. The units of the horizontal axis (time) should be in ______ .

Homework Equations



I assume V1/V2 = N1/N2

The Attempt at a Solution



To lower the voltage from 110 [V] to 11[V], we use a transformer with a turn ratio of 10:1. The units of the horizontal axis (time) should be in ______ .

1. Is the turn ratio calculated from the transformer equation V1/V2 = N1/N2?
2. How does one determine the unit of time?
 
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You just about have it.
N1 / N2 is the turns ratio, so it is just the same as the ratio of the two voltages.

Now, 110 volts is the RMS value of the supply voltage, so what is the peak value?
Peak voltage = 1.414 * RMS voltage.


If you wanted to produce 12 volts DC after filtering, what RMS voltage from the transformer would produce this? Peak voltage = 1.414 * RMS voltage.

(The 12 volts has to be the peak value of the transformer output minus two diode drops from the bridge rectifier. This is the value the filter capacitor will charge to. So the peak value of the output from the transformer is 12 + 0.6 + 0.6.)

If you have a 100 Hz sinewave, how long does one cycle take to complete? This should tell you the answer to the time scale question.
 
vk6kro said:
If you have a 100 Hz sinewave, how long does one cycle take to complete? This should tell you the answer to the time scale question.

Thanks for the help.

I would just need to use the period T, right? Which would be 1/100. So the time scale would be in centiseconds?
 
You could, but milliseconds would look more elegant. You would get 5 steps per half cycle.

Try it and see if you agree.
 

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