Full-Wave Bridge Rectifiers Electronics Help

Figure 17-6 shows another circuit used to perform full-wave rectification. One advantage of this circuit is that it does not require a transformer (although a transformer is often used to isolate the ac power line from the rest of the power supply). The diodes are arranged in the form of a bridge, and the circuit is called a full-wave bridge rectifier.

Demonstrates that current flows through RL in the same direction when Vin is positive as it does when Vin is negative. In Figure 17-7(a), v'” is positive, so O2 and OJ are forward biased, while 01 and 04 are reverse biased. Therefore current flows through R,. from right to left, as shown. In (b), ViII is negative, so 01 and D4 arc forward biased, and O2 and D.I are reverse biased. Notice that current stillllows through RI. from right to left. Therefore, the polarity of the voltage across R,. is always the same, confirming that full-wave rectification occurs. Note that the common side of the ac source must be isolated from the common side of the load

voltage. Thus, the negative terminal of RL in Figure .7-7(a) cannot be the same as the negative terminal of Vi”. As previously mentioned, a transformer is often used to provide this isolation.
Since load current flows through two forward-biased diodes during each halfcycle of Vi”, the peak rectified voltage across RL is the peak input voltage reduced Another advantage of the bridge rectifier is that the reverse bias across each diode never exceeds VPH• Thus, the PlV rating of the diodes is half that required in the rectifier using a center-tapped transformer.

Diode bridges are now commonly available in single-package units. The packages have a pair of terminals to which the ac input is connected and another pair at which the Iull-wuve -rectifled output is taken. Figure 17-l:l shows a typical manufacturer’s specification sheet for a line of single-package bridges with current ratings from 1 to 100 A. The VRRM specifications refer to the maximum repetitive reverse voltage ratings of each, i.e., the peak inverse voltage ratings when operated with repetitive inputs, such as sinusoidal voltages. These ratings range from 50 to 1200 V. Note the Im( specifications, which refer to maximum forward surge current. As we will learn in the next section, a filter is often connected to the output of a bridge, and the nature of the filter determines the amount of current that surges through the rectifier when ac power is first applied. For example, if a large capacitor is connected directly across the bridge output, then the rectifier is essentially shorted out when power is first applied. The filter must be designed so that this initial surge current doe s not exceed the bridge rating. The specifications show that bridges are available with IFsM ratings from 30 to 1000 A.by 2(0.7 V) = 1.4 V:

A bridge rectifier isolated from the 120-V-rms power-line by a transformer.
1. What turns ratio should the transformer have to produce an average current of 1 A in RL? .
2. What is the average current in each diode under the conditions of (1)?
3. What minimum PIV rating should each diode have?
4. How much power is dissipated by each diode?

1. I.l’g = 2IpR/TT, where In is the peak value of the rectified current in RI.o’ Thus,From equation 17-5, the peale value of the voltage applied across the bridge is Vp = VPR + 1.4 “” 15.7 + 1.4 = 17.1 V. Since the peak value of the 120-V-rms primary is (Yz)(120 V rms) =. 169.7 V-pk,

2. Each diode conducts during one half-cycle only, so the average current in each is the same as that of a half-wave-rectified sine wave

Posted on November 19, 2015 in Power Supplies and Voltage Regulators

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