0
\$\begingroup\$

In the centre tapped rectifier:

enter image description here

It seems as if the highlighted ground line would provide a short circuit for any current developed between the top/bottom wires and the centre tap, eliminating any output voltage.

Additionally, in the bridge rectifier:

enter image description here

The current is supposed to flow through D1/D2 or D3/D4, but the presence of the ground makes it seem like the current would get through the resistor and then short to ground. This wouldn't affect the output voltage but it still seems strange to me.

I'm guessing I'm not grasping how grounding actually works, so a simple explanation for these circuits would be appreciated. Thanks.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ This is simply a notation for the 0V level for later discussions of the potential of individual nodes. \$\endgroup\$
    – Janka
    Oct 14, 2018 at 21:24
  • \$\begingroup\$ "Ground" on the right side is not the same as the AC ground on the left side. \$\endgroup\$
    – JRE
    Oct 14, 2018 at 21:30
  • \$\begingroup\$ @JRE - it could be the same 'ground', but it doesn't make any difference either way... \$\endgroup\$
    – brhans
    Oct 14, 2018 at 21:57

2 Answers 2

2
\$\begingroup\$

Perhaps you are not understanding that the transformer isolates the secondary winding and the connected circuit from the primary power source. The primary power source may be grounded, but there is no other connection between the primary and secondary. All of the secondary current must flow through the secondary winding even though a point on the secondary circuit is grounded.

A short circuit only exists if two points in the circuit are connected through an unintended path. The only path for current is from one end to the secondary winding through one diode or the other then through the load and through the diode on the other side of the load that matches the polarity of the first diode to the other end of the secondary winding. There is no path through the ground connection back to the other end of the secondary winding or to any other point in the circuit.

\$\endgroup\$
1
\$\begingroup\$

All your assumptions are wrong. You have a secondary winding, which is not referenced to any potential. Applying a ground only gives to the secondary circuit a reference potential, nothing else. This point could be placed anywhere on the secondary circuit without affecting the working of diodes and all remaining components.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.