Full-Wave Rectifier w/ 2 P-N Junction Diodes & Transformer

In summary: Yes, you can use V/R = I to find the current in the load. Just make sure you use the correct values for voltage and resistance. And yes, you can use the same equation for the other parts as well.
  • #1
Solidsam
23
0
Show how two P-N junction diodes can be used in conjunction with a
transformer that has two secondary windings, to form a full-wave rectifier.
(Draw a circuit diagram including connections to the mains transformer).
Explain the operation of your circuit using sketches of waveforms. (3 marks)

(ii) If the output from your rectifier is 25V(PEAK) at 50 Hz and is applied to a 1000W
resistive load, find:

a) The peak current in the load

b) The mean current (3 marks)

c) The RMS current



how do i solve a,b and c? i know how to sketch the p-n junction.
 
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  • #2
Solidsam said:
Show how two P-N junction diodes can be used in conjunction with a
transformer that has two secondary windings, to form a full-wave rectifier.
(Draw a circuit diagram including connections to the mains transformer).
Explain the operation of your circuit using sketches of waveforms. (3 marks)

(ii) If the output from your rectifier is 25V(PEAK) at 50 Hz and is applied to a 1000W
resistive load, find:

a) The peak current in the load

b) The mean current (3 marks)

c) The RMS current


how do i solve a,b and c? i know how to sketch the p-n junction.

Why don't you post us a sketch of your schematic? I realize you've said you understand that part, but a sketch of the circuit is always a good place to start.

You need to show us your work and a reasonable attempt at the solution before we can help!
 
  • #3
1000w should be 1000ohms


Vpk=25/1000=0.25A

Vrms=25/sqroot2=17.68v Irms=17.68/1000=0.0177Arms

Vmean=Vrms*0.9=15.9V Imean=15.9/1000=0.016Amean


Are these result and equations correct? sry don't have a scanner so can't post sketches
 
  • #4
Solidsam said:
1000w should be 1000ohms


Vpk=25/1000=0.25A

Vrms=25/sqroot2=17.68v Irms=17.68/1000=0.0177Arms

Vmean=Vrms*0.9=15.9V Imean=15.9/1000=0.016Amean


Are these result and equations correct? sry don't have a scanner so can't post sketches

You have Vpk = 0.25A. Vpk represents a voltage not a current.

[tex]P = I_{pk}V_{pk}[/tex]

[tex]I_{pk} = \frac{P}{V_{pk}}[/tex]
 
  • #5
ok well since i have the mean voltage, rms voltage and peak voltage can i no just use V/R=I


so Vpk/1000ohms=Ipk


then do the same for the rest of them?
 
  • #6
Solidsam said:
ok well since i have the mean voltage, rms voltage and peak voltage can i no just use V/R=I


so Vpk/1000ohms=Ipk


then do the same for the rest of them?


Oh you where pointing out a typo, not your actual reasoning? (This with regards to the 1000W or 1000 ohms)
 

1. How does a full-wave rectifier work?

A full-wave rectifier uses two P-N junction diodes and a transformer to convert alternating current (AC) to direct current (DC). The transformer steps down the voltage from the AC power source, and then the two diodes allow the positive portion of the AC signal to pass through one diode and the negative portion to pass through the other diode, resulting in a pulsating DC output.

2. What are the advantages of using a full-wave rectifier?

A full-wave rectifier has a higher efficiency compared to a half-wave rectifier, as it utilizes both halves of the AC input signal. This results in a smoother DC output with less ripple. It also allows for a higher output voltage, making it useful for applications that require a higher voltage.

3. How do the P-N junction diodes in a full-wave rectifier function?

The P-N junction diodes act as one-way valves for electric current. When the AC input signal is positive, the diode connected to the positive terminal of the transformer allows current to flow through it, while the other diode blocks the negative current. When the AC input signal is negative, the roles of the diodes are reversed, allowing current to flow through the other diode and blocking the negative current.

4. How does a transformer contribute to the operation of a full-wave rectifier?

The transformer in a full-wave rectifier steps down the voltage from the AC power source, making it more manageable for the diodes to convert to DC. It also isolates the circuit from the high voltage of the power source, providing safety for the user and any connected devices.

5. What are some common applications of a full-wave rectifier with 2 P-N junction diodes and a transformer?

A full-wave rectifier is commonly used in power supplies for electronic devices, such as laptops, TVs, and radios. It is also used in battery charging circuits and in DC motor control circuits. Additionally, it is used in welding machines and some types of voltage regulators.

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