review

Please review questions 1(a-b),2(all parts),3(ripple voltage), & 4(a-b, d-e).

1. Assume the input signal to a rectifier circuit has a peak value of Vm = 12 V and is at a frequency of 60 Hz. Assume the output load resistance is R = 2kΩ and the ripple voltage is to be limited to Vr= 0.4 V. Determine the capacitance required to yield this specification for a (a) full-wave rectifier and (b) half-wave rectifier. Show all work.

Given Vm=12V

f = 60Hz

R = 2kΩ

Vr = 0.4V

For a full wave rectifier the ripple voltage is given by

For half wave rectifier the ripple voltage is given by

The capacitance C =

2. A full-wave rectifier is to be designed to produce a peak output voltage of 12 V, deliver 120 mA to the load, and produce an output with a ripple of not more than 5 percent. An input line voltage of 120 V (rms), 60 Hz is available. Consider a bridge type rectifier. Specify the transformer ratio and the size of the required filter capacitor. Show all work.

The peak secondary voltage considering a diode drop of 0.7V is equal to:

2*(12+0.7) =25.4V

The peak primary voltage is equal to

The transformer ratio is equal to

The transformer ratio is 7:1

The DC voltage output is 12*0.637=7.644V

The required secondary current is 120mA. The value of the resistor is equal to

The ripple voltage is required to be 5% of the DC voltage 7.644V.

Required ripple voltage is equal to 0.05*7.644V= 0.3822V

The formula for calculating the ripple voltage in a full wave rectifier is

3. Silicon diodes are used in a two-diode full-wave rectifier circuit to supply a load with 12 volts D.C. Assuming ideal diodes and that the load resistance is 12 ohms, compute the secondary transformer voltage, the load ripple voltage, and the efficiency of the rectifier. Show all work.

The peak output voltage is equal to

The secondary transformer voltage is 2* 18.838= 37.677Vpeak

The secondary rms voltage= rms

Load ripple voltage=Vmax-Vdc=

The formula for calculating the efficiency of a full wave transformer is

RF=diode forward resistance. Since the diodes are ideal diodes the forward resistance is equal to zero

The efficiency is thus equal to 0.812*100%

= 81.2%

4. A half-wave rectifier using silicon diode has a secondary emf of 14.14 V (rms) with a resistance of 0.2 Ω. The diode has a forward resistance of 0.05 Ω and a threshold voltage of 0.7 V. If load resistance is 10 Ω, determine the following:

· dc load current

The DC voltage of the half wave rectifier is equal to

The DC load current is equal to

· dc load voltage

VRL=Idc*RL= 0.87875*10= 8.7875V

· voltage regulation

The voltage regulation is calculated

· circuit efficiency

RF=diode resistance

The efficiency in percentage is 40.398%

· diode PIV and current rating

The peak output voltage is

PIV≥ 19.997V

Current rating≥19.997/0.05=400A