0
\$\begingroup\$

In a full wave rectifier (consisting of 2 diodes), ac is converted into DC. Confusing part is I know how the rectification is done, but still how the input ac current (consisting of both negative and positive currents) is converted into positive DC current????? enter image description here

How this negative current is converted into positive DC current?

\$\endgroup\$
  • \$\begingroup\$ Can you add a schematic of the circuit you have in mind. It sounds like it's a centre-tapped transformer with two diodes rather than a single winding with a full bridge rectifier (4 diodes). There's a CircuitLab button on the editor toolbar. 45° diodes are down at the bottom if you need them. Double-click a component to edit its properties. \$\endgroup\$ – Transistor Aug 9 '18 at 19:04
  • \$\begingroup\$ The minor current loops alternate as polarity inverts about the centre tap=0V whichever winding output has more voltage than the resistor relative to centre conducts. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 9 '18 at 23:23
1
\$\begingroup\$

That schematic is confusing as it appears that everything is happening simultaneously. Let's look at each half-cycle of the AC supply.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) A centre-tapped full-wave rectifier. (b) Effective circuit when dot-end is positive. (c) Effective circuit when dot-end is negative.

  • The diodes only conduct in the direction of the arrow in the diode symbol.
  • When the dot end of the transform is positive the lower diode is reverse biased, does not conduct and so is effectively out of circuit as shown in Figure 1b.
  • When the dot end of the transform is negative the upper diode is reverse biased, does not conduct and so is effectively out of circuit as shown in Figure 1c.

The end result is that a pulse of current is alternately given by the upper and lower diode on each half-cycle of the AC supply voltage.

schematic

simulate this circuit

Figure 2. Schematic redrawn in same format as that in the question.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

During one half-cycle of the AC, the top terminal of the transformer seconodary will be positive relative to the centre tap, so the top diode will conduct. During the other half of the AC cycle, the bottom terminal of the transformer secondary will be positive, so the bottom diode conducts.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

How this negative current is converted into positive DC current?

The words are confusing, because current is induced (not converted) by the mutual inductor, the transformer. And, 'positive' current just means (by convention) current going clockwise in a circuit diagram. The diagram can be drawn as a mirror-image, without changing the circuit to which it refers, so positive or negative current is just... a conventional direction-naming scheme.

The most important thing to notice, is that current in the load resistor is going right-to-left when it is CW in one (the upper) branch of the transformer output, or when it is CCW in the other (lower) branch of the transformer output windings.

The load resistor, therefore, can be part of an active CW and also an active CCW current-carrying secondary circuit. We use a completely different convention when calling its current positive or negative, and that is the right-to-left current direction being positive.

Then you look at the rectifiers: those diodes allow only positive current in the resistor. This, we call DC (because the resistor current does not change its direction).

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.