CALCULATING CURRENTS Once the peak and average load voltage values are known, it is easy to determine the values of IL(pk) and Iave with the help of Ohm’s law.

EXAMPLE Determine the values IL(pk) and Iave for the circuit in last lecture.

SOLUTION In the previous lecture, we calculated peak and average output voltages for the circuit. Using these value we can determine current as follows

SOLUTION IL(pk) = VL(pk)/RL = 20.5/5.1k =4.02 mApk And Iave = Vave/RL = 2.56mA dc

EXAMPLE Determine the values of VL(pk), Vave, IL(pk)and Iave for the circuit shown in the fig.

SOLUTION Rated value of the transformer is given so the peak value can be calculated as V2(pk) =24Vac/0.707 =33.94Vpk

SOLUTION Now VL(pk) = V2/2 – 0.7 =16.27Vpk And Vave = 2VL(pk)/ Π =10.34Vdc

SOLUTION Now for currents IL(pk) = VL(pk)/RL =16.27/8k =2.03mApk And Iave = Vave/RL =10.34/8k =1.29mA

NEGATIVE FULL WAVE RECTIFIER The main differences between the positive and negative full wave rectifier are the direction that the diodes are pointing and the polarity of the output voltage. The analysis of the negative full wave rectifier circuit is same as negative half wave rectifier.

NEGATIVE FULL WAVE RECTIFIER If we reverse the direction of the diodes in the positive full wave rectifier, we will have a negative full wave rectifier as shown in the fig.

PEAK INVERSE VOLTAGE When one of the diodes in a full wave rectifier is reverse biased, the voltage across it is approximately equal to V2. The peak load voltage supplied by the full wave rectifier is equal to the one half of the secondary voltage V2. Therefore, the reverse voltage will be

PEAK INVERSE VOLTAGE twice the peak load voltage. By formula PIV = 2VL(pk) Since the peak load voltage is half the secondary voltage, we can also find value of PIV as PIV = V2(pk)

PEAK INVERSE VOLTAGE When calculating the PIV in practical diode circuit, the 0.7 drop will be also be taken into consideration. In the fig. when D1 is on and will have voltage drop of 0.7

PEAK INVERSE VOLTAGE Since D1 is in series with D2 the PIV across D2 will be reduced by the voltage across D1. The equation PIV = V2(pk) – 0.7 Gives a more accurate PIV voltage.

FULL WAVE BRIDGE RECTIFIER The bridge rectifier is the most commonly used full wave rectifier circuit for several reasons (1) It does not require the use of center-tapped transformer, and therefore can be coupled directly to the ac power line, if desired.

FULL WAVE BRIDGE RECTIFIER (2) Using a transformer with the same secondary voltage produces a peak output voltage that is nearly double the voltage of the full wave center-tapped rectifier. This results in the higher dc voltage from the supply.

BASIC CIRCUIT OPERATION The bridge rectifier shown in the fig. consists of four diodes and a resistor. The full wave rectifier produces its output by alternating the circuit conduction

BASIC CIRCUIT OPERATION Between the two diodes . When one is on( conducting), the other if off( non conducting), and vice versa. The bridge rectifier works basically in the same way. The main difference is that the bridge rectifier alternates conduction between two diodes pairs.

BASIC CIRCUIT OPERATION When D1 and D3 are on, D2 and D4 are off and vive versa. This circuit operation is illustrated in the fig. The current direction will not change in the either condition.

CALCULATING the LOAD VOLTAGE and CURRENT VALUES The full wave rectifier has an output voltage equal to the one half of the secondary voltage. The center-tapped transformer is essential for the full wave rectifier to work, but it cuts the output voltage to half.

CALCULATING the LOAD VOLTAGE and CURRENT VALUES The bridge rectifier does not require the use of center-tapped transformer. Assuming the diodes to be ideal, the rectifier will have a peak output voltage of VL(pk) = V2(pk) (ideal)

CALCULATING the LOAD VOLTAGE and CURRENT VALUES When calculating the circuit output values, we will get more accurate results if we take the voltage drops across the two conducting diodes into account. To include these values,we will use of following equation VL(pk) =V2(pk) – 1.4 The 1.4 volts represents the sum of diode voltage dropes.

EXAMPLE Determine the dc load voltage and current values for the circuit shown in the fig.

SOLUTION With the 12Vac rated transformer, the peak secondary voltage is found as V2(pk) = 12/0.707 =16.97Vpk

SOLUTION The peak load voltage is now found as VL(pk) = V2 – 1.4 =15.57Vpk The dc load voltage is found as

SOLUTION Vave = 2VL(pk)/Π =31.14/Π =9.91 Vdc Finally the dc load current is found as Iave = Vave/RL = 9.91/12k = 825.8microA