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schematic

simulate this circuit – Schematic created using CircuitLab

In my book, this diagram is used to explain how a bridge rectifier works, in positive and negative half cycles. But it does not explain the significance and necessity of the ground in the circuit. What is the significance of the ground here?

If we consider ground's voltage as zero voltage (generally what I do in circuit analysis) and for any half cycle (just say positive half cycle), there will be 0.7V voltage drop across D1. Doesn't it make positive half cycle start from -0.7 volt? It seems odd to me.

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The ground is not necessary and has no significance in reality. If the ground is not present, the lower voltage end of the resistor will float with respect to the AC supply, which is fine for most cases.

The wave form for each half cycle will be offset by the voltage across the diodes, as current will only start to flow then the voltage at the low-voltage side of the transformer exceeds two diode drops. There will be a small gap of zero potential ( V=IR across the resistor ) across the resistor when the AC crosses between positive and negative. At this point, the voltage of the more positive wire of the low-voltage side of the transformer will be increasing from zero to say 1.4V wrt the other wire of that side of the transformer. The voltage dropped across the diodes increases until they start to conduct then remains constant ( to a first approximation ), they don't drop 0.7V if you only apply 0.1V to them. So the half cycle starts at 0V then stays there until 1.4V is applied to the bridge, then tracks the sine-wave less the 1.4V diode drops until it hits zero again, then stays at zero until the next AC crossing.

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Without a ground (or some path such as a 1G resistor to ground) a SPICE simulation will fail, and it also provides a reference voltage for discussion.

The node that is shown as grounded is the one that would normally be grounded in a circuit with a positive supply voltage. Say a \$\mu\$A7805 is used as a regulator, that's where the 'common' terminal is connected. The 'ground', outside of simulation world, is just a shorthand for a common connection, but of all the possibilities, this one is the most sensible in terms of something that is likely to be common with a lot of other points.

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I think in that specyfic case ground is present on that drawing to show where is point considered at 0V on other related drawings (like voltage vs time graph).

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Here, I built the circuit which carries resistive load.

  1. If we put GND to the bridge side, we get full-wave rectified output signal.
  2. If we put GND to the source voltage side, we will see divided alternances. Positives are on the top, negatives are below located.

To me, I generally use GND to bridge on my univ. projects. :)

GND to VS- GND to Bridge

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  • \$\begingroup\$ What is the name of this simulation software? I kind of like it :p @Bay \$\endgroup\$ – Anklon Apr 22 '14 at 6:47
  • \$\begingroup\$ Check EveryCircuit website: everycircuit.com \$\endgroup\$ – Alexandre Teles Aug 22 '14 at 22:07

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