Electronic – getting a half-wave rectified output when i am using a full-wave rectifier

The current should follow sign of the voltage. If you measure the current after the rectifier, it has already been rectified and so the current will only flow in one direction as the diodes will prevent it from flowing the 'wrong way'. However, if you measure the current before the rectifier, then it will follow the sign of the input signal. Since the input signal in this case is a sinewave that has both positive and negative peaks, you will see both positive and negative current flow.

It looks like you are measuring the current at the input of the bridge rectifier and the voltage at the output of the rectifier.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. (a) A single winding with full-wave bridge rectifier. (b) A centre-tapped transformer with rectifier diodes.

• I am assuming that the centre-tapped transformer has twice the number of turns as the single-winding transformer. e.g., If (a) is a 0 - 12 V transformer then (b) will be a 12 - 0 - 12 V transformer.
• If you trace the current flow in (a) when the dot end is positive you will see that the current will flow through D2, out to the load via V+, back on V- and through D3 back to the secondary. Each diode will create a voltage drop of 0.7 V so the voltage out will be down about 1.4 V.
• You can see in (b) that when the dot end is positive that D5 will conduct and the return will be straight to the centre-tap. The voltage drop will be 0.7 V approx.

The centre-tapped version is more efficient but probably more expensive to make.

Note that the XFMR2 wire gauge can be lighter than XFMR1 as it only carries current on alternate half-cycles. Total weight of copper will be about the same.