# Single phase rectifier

I can't recognize where the voltage can turn because all the diodes are blocked.

• Assuming the arrows indicate instantaneous voltage and current, several arrows are pointing in wrong direction. Therefore you are excused for being confused as it is a poor diagram. Jun 14, 2015 at 11:51
• Remember that the output of a transformer is always AC. All polarities therefore flip twice per power cycle. Jun 14, 2015 at 12:04
• the problem is this is a question in our exam, so you say that this schema is wrong ? and the diodes should be reversed ? Jun 14, 2015 at 12:14
• Diodes are OK, it's some of the arrows that are wrong. Get the schematic corrected. Jun 14, 2015 at 12:50
• Also, drawing an arrow to indicate voltage, and then writing -V there is the perfect way to confuse the observer. Jun 14, 2015 at 15:09

By simply redrawing the circuit in a more conventional way it is easy to see that it is a full wave rectifier circuit. If transformer centre tap is taken as 0V then the joined anodes of the diodes form the negative voltage output terminal.

I've never heard the term "charge fortement inductive' but it sounds like some form of on-line translation of an inductive/capacitive smoothing filter.

Consider each half cycle of the input. The dot notation gives the phase relationship between the input and outputs. The output at cathode D1 will be more positive than the centre tap which will be more positive than the cathode of D2. Conventional current flows from positive to negative and the diode arrow symbol shows that direction. On the positive (input) half cycle the cathode of D1 is always more positive than its anode so no current can flow through it. D2 cathode is always more negative than its anode so current can flow through it and complete the circuit path.

On the negative half cycle input the output voltages are reversed and current can only flow through D1.

The LOAD will experience current flowing in the same direction on each half cycle.

• its in french its mean inductive load Jun 14, 2015 at 13:56
• thanx a lot but the current came from where and go to where ? i mean that the current came from the center and passed trough the load then the D1 ? Jun 14, 2015 at 13:57
• thank you so much MR.JIM Dearden, the extra diagram help me to understand the circuit. Jun 14, 2015 at 15:33
• "charge fortement inductive" means "highly inductive load", and it is certainly put here to mean that the current is close to constant (or at least it never goes to zero, so you are in Continuous-conduction-mode) Jun 9, 2016 at 12:51

The secondary voltage arrows are confusing this. As mentioned in the comments the input and output will be AC. Every 1/2 cycle one of the two diodes will be forward biased and thus will permit conduction

When Vin is positive D1 will be reverse bias (note the dot of the XFMR), but D2 will be forward bias. This facilitates current flowing around the bottom 1/2 of the output stage

When Vin is negative D2 will be reverse bias but D1 will be forward bias. This facilitates current flowing around the top 1/2 of the output stage.

The result is the load always see's positive current flow

simulate this circuit – Schematic created using CircuitLab

• Thanx a lot MR.jonRB this help me so much, can you please tel me when D1 is conductive and how the current well circulate in the schema ? excuse my bad English Jun 14, 2015 at 13:41