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Why don't I use this circuit for full wave rectifier instead of Bridge rectifiers or Center tapped transformer type rectifiers?

enter image description here

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    \$\begingroup\$ That is a crazy circuit. The 40 Hz is floating, for one thing. \$\endgroup\$ – jonk May 11 '19 at 7:16
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    \$\begingroup\$ It might help if you explained why you think that circuit is a full wave rectifier. \$\endgroup\$ – Blair Fonville May 11 '19 at 7:22
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    \$\begingroup\$ I think you need to review what a diode does, what voltage is, what current is, what AC is, and what a short-circuit is. \$\endgroup\$ – DKNguyen May 11 '19 at 7:32
  • \$\begingroup\$ Maybe because it is not very good for your diodes. \$\endgroup\$ – Oldfart May 11 '19 at 7:32
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Because you will short circuit your power supply and burn out your diodes.

Both diodes will conduct at the same time. So, for half the time, your diodes form a short circuit across your generator.

In real life, either the diodes would burn out or the transformer would overheat and burn out.

That looks like you are using the Falstad circuit simulator.

Put a current meter in series with your diodes (between the generator and the upper diode.) Have Falstad plot the current over time.

You will see periods of zero current (the diodes are not conducting) and periods of high current.

Or, get a small AC transformer (like a wall wart with 12VAC output) and a couple of small signal diodes (like the ubiquitous 1N4148) and build your circuit and try it out. No need to measure current - something will get noticeably hot or the diodes will explode.

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  • \$\begingroup\$ Now I understood clearly. Thanks. \$\endgroup\$ – Lakshmanan R May 11 '19 at 9:00
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enter image description here

Figure 1. The original circuit.

Why don't I use this circuit for full wave rectifier instead of Bridge rectifiers or Center tapped transformer type rectifiers?

When the top of the AC source goes positive a very large current will flow through Dtop and Dbottom which will create an almost dead short across the supply. If the supply is not current limited the diodes will be burnt out. If the supply is current limited there will be 2 x Vf across the diodes. (Vf, the forward voltage, is usually about 0.7 V for silicon diodes.

No current will flow through the 1k as there is no way for it to return to the 40 Hz supply. Remember that current must flow in a circuit and although you have an earth connection there is no earth connection on the 40 Hz supply.

So, to answer your question, we don't use that circuit because it doesn't work.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2. A circuit that uses the same number of components but does work. The return current path is indicated in orange.

  • This circuit uses a centre-tapped transformer.
  • When the centre-tap is grounded the two outputs are 180° out of phase. When the top terminal is positive the bottom is negative, etc.
  • When the top terminal is positive D1 conducts current flows through R1 and back to the transformer via GND. No current flows in the lower half of XFMR1.
  • When the top terminal goes negative the bottom terminal will go positive and D2 will conduct. Current again flows through R1 and back to the transformer via GND. No current flows in the upper half of XFMR1.

Unfortunately it is too late to patent this idea. It has been standard practice for well over a century.

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