Question

A full-wave bridge-rectifier circuit with a 1-k load operatesfroma120-V(rms)60-Hzhouseholdsupplythrough a 12-to-1 step-down transformer having a single secondary winding. It uses four diodes, each of which can be modeled to have a 0.7-V drop for any current. What is the peak value oftherectifiedvoltageacrosstheload?Forwhatfractionofa cycle does each diode conduct? What is the average voltage across the load? What is the average current through the load?

Answer 1

Vp=12.74V

both pairs of diodes will conduce only half of the cycle.

`Vo=8.11V`

`Io=8.11mA`

Step-by-step explanation:

The output of the transformer is:

`Vt=120Vrms*(1)/(12)=10Vrms`

the average voltage is given by:

`Vo=(1)/(T)*\int\limits^T_0 {10sin(wt)} \, dwt\\`

For a full wave rectifier
`Vo=(2*Vp)/(\pi)`

We model the diodes to have a drop of 0.7V, in ta full wave rectifier we have two diodes in series for each cycle (positive and negative) so:

`V=10*√(2) -1.4=12.74V`

we have to transform 10Vrms to its peak value in order to substract the diode voltage drop.

the peak voltage on the resistor is Vp=12.74V

So two diodes will conduce when the input is positive and the other two will conduce on the negative input. That is, both pairs of diodes will conduce only half of the cycle.

`Vo=(2*12.74)/(\pi)=8.11V`

`Io=(Vo)/(R)=8.11mA`