How to explain FWR circuit if diodes are reversed?
The circuit of FWR is not difficult to understand. But when the diodes in that circuit are reversed, the thing become somewhat tricky.
So we shall discuss the working of the circuit given below. In this circuit, both diodes D1 and D2 are connected in reverse direction.
When a student starts explaining the circuit, he gets confused, as he CANNOT START the flow of current either from A or B, when anyone terminal is positive (+).
So here comes the basic concept of DUAL NATURE OF POWER SUPPLY. So we can explain the working as follows…FWR circuit with reverse connections of diodes D1 and D2
The waveforms in the circuit are shown OUT OF PHASE. This is because, the input and output waveforms of a transformer are always, out-of-phase.
During positive (+) half cycle of AC voltage AT SECONDARY, let A is positive (+) and B is negative (-) with respect to center tap C. Now we say that, since C is at zero (0) potential, IT IS MORE POSITIVE than point B. So we start the flow of CONVENTIONAL CURRENT from C…!
Thus the path of current can be traced as: C → E → RL → D → D2 → B and back to C. Thus the current will flow through the load resistor from E → D. So point E is positive (+) and point D is negative (-).
During negative (-) half cycle of AC voltage AT SECONDARY, let B is positive (+) and A is negative (-) with respect to center tap C. Since C is again at zero (0) potential, IT IS MORE POSITIVE than point A. So we again start from C, as follows…
The traced path will be: C → E → RL → D → D1 → A and back to C. Again the direction of current will be same through the load resistor i.e. from E → D. So again E is positive (+) and point D is negative (-).
Thus we will get TWO CONSECUTIVE NEGATIVE HALF CYCLES during one complete cycle of the AC voltage. In this way, we will get NEGATIVE VOLTAGE across the output terminals of the circuit, such that E will be positive (+) and D will be negative (-).