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Above is two identical bridge rectifier circuits. Input is a 12V sinusoidal voltage source. Red is the source. Blue is the output from the circuit on the right side. Green is the output from the circuit on the left side. It seems like the left circuit is wrong and the one on the right side has true output. But I don't understand why it is not the other way around.

The only difference is that the circuit on the right side has a ground between the load and two diodes.

On the other hand, the circuit on the left side has a ground connected to the negative terminal of the supply.

Normally I connect the negative terminal of the power supply to the common ground.

An example to this is the circuit on the left side.

But when I plot the output for both circuits, the circuit on the left side acts weird for negative alternating input.

But my logic says:

When V2 goes positive the current must flow through D7 R2 D6 and finally to the supply terminal.

When V2 goes negative the current must flows throug D8 R2 D5 and to the supply terminal.

So I would expect the same output as in the circuit on the right side. Hence I would expect a rectification for negative halves as well.

What is wrong in my logic here?

Regarding the working circuit in the right side:

I dont understand the circuit on the right side by the way. Because:

When V1 goes positive the current must flow through D3 R1 and finally to the ground or to D2 ???.

When V1 goes negative the current must flow through D4 R1 and finally to the ground or to D1 ???.

Why do we need ground here? Current still should flow to the terminal of the supply without a ground there.

My question is:

Why is the common ground connected like in the circuit on the right?

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    \$\begingroup\$ tl;dr It's a simulation. You need a ground somewhere for it to have a reference point. Where you put it is irrelevant as long as you know the results are relative to that point. \$\endgroup\$ – Matt Young Feb 2 '16 at 15:07
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    \$\begingroup\$ If you connect both sides to ground, you have shorted out one of the diodes and you no longer have a bridge rectifier. Therefore one side MUST float. If you need to ground the DC supply, the AC input must float (as a transformer secondary does). \$\endgroup\$ – Brian Drummond Feb 2 '16 at 15:07
  • \$\begingroup\$ @BrianDrummond Which diode is shorted? \$\endgroup\$ – user16307 Feb 2 '16 at 15:09
  • \$\begingroup\$ @MattYoung Do you mean that common is just needed for simulation? \$\endgroup\$ – user16307 Feb 2 '16 at 15:10
  • \$\begingroup\$ In reality the AC and DC sides will have their own commons, like Brian said. For simulation, you can only have one, and where it is doesn't matter. \$\endgroup\$ – Matt Young Feb 2 '16 at 15:15
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Both circuits output rectified unsmoothed DC, and if you plot the voltage across R1 and R2 you will find that both waveforms are absolutely identical, looking like this:

rectified AC

However you are not plotting the voltage across R1 and R2. You are plotting the positive output of the rectifier against ground. In the first case ground is tied to the incoming AC, causing the weird output you see. In fact, grounding the other side of the AC supply would result in the exact same effect:

other side

If you are ever building a real world power supply, ground the negative DC output of the diode bridge (like on the right of your schematic), as grounding the input of the bridge is not only useless but will short out a diode if you also ground the negative output of the bridge.

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  • \$\begingroup\$ amazing answer thx \$\endgroup\$ – user16307 Feb 2 '16 at 15:19
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Both circuits are correct as they stand.

As there are diode components between the source and the load, the two sides may not share a ground, and still work properly.

The circuit on the left has a grounded source, and a floating load. This is a fairly unusual configuration, there are not many cases where you would have a floating DC load. One good example would be charging an auto battery from an old fashioned transformer and rectifier. The absence of a ground at the car would allow the source to be grounded without problems, though it would not need to be.

The circuit on the right is the more common configuration, where a floating transformer winding powers a DC load that is grounded.

Don't get confused between what simulators demand, and real life. Both of the circuits are OK in real life. Simulators OTOH usually demand that at least one ground is present, and may automatically ground other nodes if you ask for certain measurements. This unintentional ground may be causing your problem. Simulators are often a bit tricky to use with floating measurements. Make sure that you understand exactly what you are asking your simulator to do in both cases.

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