# Fullwave Rectifier: Calculate Peak Diode Current

• exis
In summary, a 50V peak AC signal at 50Hz is fed to a 1:1 turns ratio transformer full-wave rectifier circuit. The full-wave rectifier waveform is smoothed by the use of a capacitor C=47uF, R=4.3kohms. The Attempt at a Solution calculates the peak diode current (however before this i was asked to calculate the average voltage across the load resistor R and the peak-peak rippe voltage across R). The peak diode current occurs when Vin=Vpeak. Therefore I_max=Vpeak/R=50/4300=11.6mA.
exis

## Homework Statement

A 50V peak AC signal at 50Hz is fed to a 1:1 turns ratio transformer full-wave rectifier circuit. The full-wave rectifier waveform is smoothed by the use of a capacitor C=47uF, R=4.3kohms.

I need to calculate the peak diode current (however before this i was asked to calculate the average voltage across the load resistor R and the peak-peak rippe voltage across R)

## The Attempt at a Solution

Vrippe(peak-peak) = Vpeak/(2fRC) = 50/(2*50*4300*47*10^-6) = 2.47V
Vdc = Vpeak - Vripple(peak-peak)/2 = 50 - 2.47/2 = 48.76V

I reasoned that the maximum current through the diode occurs when Vin=Vpeak. Therefore I_max=Vpeak/R=50/4300=11.6mA.

I tried considering the current through the capacitor as well but didnt know how to find the max current through it.

#### Attachments

• circuit.jpg
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Note: are you to assume the diodes have zero voltage across them, or would they have 0.6 to 0.7 V across them?

Once you answer that, the key here is: You know the peak voltage across the resistor. You know R for the resistor. Find I for the resistor.

I assumed the diodes are ideal, therefore zero voltage drop across them. I found the peak diode current exactly the way you just said and the answer I got isn't correct. Peak voltage = 50V, R=4300ohms, therefore I_max = 11.6mA. The answer i should be getting is 127mA

Weird. Perhaps you are also supposed to calculate a current going into the capacitor when the voltage increases from it's minimum to maximum values (i.e., the rising part of the ripple in the voltage).

I usually work these things by first guestimating then refining. The rough, peak-to-peak ripple is about 0.6 volts across the capacitor--very small.

The discharge curve is nearly a straight line and will intercept the rectified voltage curve very close to 90 degrees. About 81 degrees. From dv/dt, at this point on the curve, you can obtain the peak capacitor current.

I(t) = C dv/dt

btw, I get a combined peak current of about 126 mA

Phrak, how did you work it out exactly please?

## 1. What is a fullwave rectifier?

A fullwave rectifier is an electronic circuit that converts alternating current (AC) into direct current (DC). It consists of diodes that allow current to flow in only one direction, effectively converting the negative portions of the AC signal into positive DC output.

## 2. How does a fullwave rectifier work?

A fullwave rectifier works by using two diodes in a bridge configuration. The AC signal is applied across the two diodes, and the output is taken across the other two diodes. As the AC signal changes polarity, the diodes allow current to flow in opposite directions, effectively converting the signal into DC.

## 3. What is peak diode current?

Peak diode current refers to the maximum amount of current that flows through a diode during a cycle of the AC input signal. It is important to calculate this value to ensure that the diodes can handle the current without overheating or failing.

## 4. How do you calculate peak diode current in a fullwave rectifier?

To calculate peak diode current, you need to know the peak voltage of the AC signal, the forward voltage drop of the diode, and the resistance in the circuit. The formula for peak diode current is Ipeak = (Vpeak - Vf) / R, where Vpeak is the peak voltage, Vf is the forward voltage drop of the diode, and R is the resistance in the circuit.

## 5. What are some common uses of fullwave rectifiers?

Fullwave rectifiers are commonly used in power supplies for electronic devices, battery chargers, and inverters. They are also used in audio amplifiers to convert AC signals into DC before amplification. Additionally, fullwave rectifiers are used in many household appliances and electronic equipment that require a steady DC power source.

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