Position: Index > Power Supply Circuit >

Rectification using a Gyrator Circuit

2014-11-28 18:31  
Declaration:We aim to transmit more information by carrying articles . We will delete it soon, if we are involved in the problems of article content ,copyright or other problems.

To avoid excess ripple output on a power supply feeding a heavy load, usually a large value capacitor ischosen following the rectifier. In this circuit, C1 is only a 470uF capacitor. The gyrator principle uses theeffect that the value of input capacitance at the base of a transitor is effectively multiplied by the currentgain of the transistor. Here C2 which is 100u appears at the ouput ( Vreg ) to be 100 x current gain ofthe 2N3055 power transistor. If you assume a dc current gain of 50, then the smoothing across thesupply, would be as though you had chosen a 5000uF capacitor. visit page.Rectification using a Gyrator Circuit - schematic
The load draws nearly 400mA. With the output directly from the rectifier there is about 5v pk-pk ripplein the output. Using the output at the emitter of the transistor things are much better. The circuit willtake a few hundred milliseconds for the visit page.