Simulation of Half Wave and Full Wave Rectifier in Orcad Capture | applied electronics engineering

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Simulation of Half Wave and Full Wave Rectifier in Orcad Capture

By Applied Electronics - Friday, January 6, 2017 No Comments
Power Companies supplies an average rms AC voltage of 120V at a frequency of 60Hz to the civilian homes. This line voltage has to down converted to DC voltage required by electronics equipment. Transformers together with rectifiers is used in the electronics equipment for this purpose. Transformers only downconverts the magnitudes, that is, the signal at the output of the transformer or the secondary winding is still an AC signal with lower magnitude and the frequency is the same. This lowered AC signal is then converted to a subsequently converted to DC by full wave rectifier. Also after the full wave rectifier is a filter that smooths the noise DC signal from the full wave rectifier.

The voltage required at the rectifier side can be calculated using a formula that relates voltages, the inductance and turn ratio. This is given below-
\[\frac{V_2}{V_1}=\frac{N_2}{N_1}=\sqrt{\frac{L_2}{L_1}}\]
If we require a voltage of 50V at the secondary then V2=50 and V1=120V. If we take primary inductor value as L1=10mH then the formula above gives L2=1.73mH.

Using these values we can stimulate how the AC signal conversion happens with the transfomer design. Shown below is a schematic for simulation for the verification of operation of transformer action in Orcad capture.
In the above schematic, the AC source signal enters the transformer with the calculated inductor values above. The resistors of 0.2ohm are there to model the small resistance values of the inductor. The K linear part is required in Orcad capture, and with this part we can specify the strength of the coupling. Here the coupling strength is 0.9, so 90% is coupled from the primary to secondary and 10% is lost in the air core(say). Orcad capture also requires a dc path to ground. This is provided by a high resistance wire of 1Gohm path.

Now a simulation of this circuit gives us a waveform graph as shown above.


The graph shows the voltage at the primary side in red and the voltage at the secondary side in green. We observe that the circuit did reduce the voltage but not down to 50V. The observed voltage is 41 to 42V. This is because of the coupling strength provided as 0.9. If we put the strength as 1 then the voltage does convert to 50V exactly. Also the down converted voltage is still an AC signal.

Let's change the coupling strength to 0.999 so that we get a 50V at the secondary circuit.


Now re stimulate the circuit gives us the following waveform graph.


The signal at the secondary has become larger to 50V.

Next we will add a diode to the secondary circuit to see its consequence.


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