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

[calvert11]

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?

 

[vk6kro]

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.

 

[calvert11]

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?

 

[vk6kro]

You could, but milliseconds would look more elegant. You would get 5 steps per half cycle.

Try it and see if you agree.

 

[DeeNos]

I would like to clarify a few things before providing a response. Firstly, the statement mentions using a transformer to lower the voltage from 110V to 11V, but it does not specify the type of transformer (step-up or step-down). Assuming it is a step-down transformer, the turn ratio would indeed be 10:1, as you have correctly calculated using the transformer equation V1/V2 = N1/N2.

Secondly, the statement does not mention the type of full-wave rectifier being designed (half-wave or full-wave). A full-wave rectifier typically uses four diodes in a bridge configuration to convert AC to DC, whereas a half-wave rectifier uses only two diodes. In this case, I will assume we are designing a full-wave rectifier.

Now, to answer your questions:

1. Yes, the turn ratio is calculated using the transformer equation V1/V2 = N1/N2. This equation relates the voltage on the primary (input) side of the transformer (V1) to the voltage on the secondary (output) side (V2) in terms of the number of turns on each side (N1 and N2).

2. The unit of time would depend on the type of rectifier being designed. If we are designing a full-wave rectifier, the unit of time on the horizontal axis would be in cycles per second (Hz) since the AC input is given in terms of frequency. However, if we are designing a half-wave rectifier, the unit of time would be in seconds, as only half of the AC cycle is being rectified.

In summary, to design a full-wave rectifier that converts 110V AC to 11V DC, we would use a step-down transformer with a turn ratio of 10:1 and the unit of time on the horizontal axis would be in cycles per second (Hz).