Can Any one help what is wrong with this full wave rectifier circuit?

[rajbir_123]

 

[KlausST]

Hi,

The way it is connected .. it is no full wave rectifier.

A full wave rectifier has 4 connections:
* AC 1: your upper connection
* AC 2: your lower connection = GND
* "+" : your right connection to the capacitor
* "-" : your left connection ... wrongly not used in your circuit

Klaus

 

[Easy peasy]

it's just plain wrong due to gnd connections re-draw and start again

 

[crutschow]

You cannot have both the input source and the output load grounded with a full-wave rectifier.
Look up full-wave rectifier.

And 1GHz is a rather high frequency for a bridge.

 

[barry]

Pretty much EVERYTHING is wrong with that circuit.
3 out of 4 of those diodes do nothing.
You're uselessly showing undefined parameters.
1 GHz is an awfully high frequency

 

[danadakk]

Here notice ground :

 

[rajbir_123]

Hi,

The way it is connected .. it is no full wave rectifier.

A full wave rectifier has 4 connections:
* AC 1: your upper connection
* AC 2: your lower connection = GND
* "+" : your right connection to the capacitor
* "-" : your left connection ... wrongly not used in your circuit

Klaus

Sir As per the figure stated below I tried to modify the diagram in ADS but Stil now I am not receiving any output. Can anyone help me with the full wave rectifier design file of ADS circuit simulator

 

[KlausST]

Hi,

Please tell us about your knowledge level, so we can adjust our answers.

Now your lower diode has GND on both sides. What do you expect it to do?
Your left diode now shorts the negative part of your input....

A full wave rectifier is one of the most basic circuits. ... you will find many millions of correct schematics in the internet.
You need to be able to reproduce this ...

Both post#4 and post#6 mentioned the GND problem. Did you read the posts? Post#4 shows the schematic.


Klaus

 

[rajbir_123]

Hi,

Please tell us about your knowledge level, so we can adjust our answers.

Now your lower diode has GND on both sides. What do you expect it to do?
Your left diode now shorts the negative part of your input....

A full wave rectifier is one of the most basic circuits. ... you will find many millions of correct schematics in the internet.
You need to be able to reproduce this ...

Both post#4 and post#6 mentioned the GND problem. Did you read the posts? Post#4 shows the schematic.


Klaus

I am beginner school student so I ma very noob at this stage

 

[Easy peasy]

your grounds are now shorting out a diode ....

 

[BigBoss]

AC simulation is a Small Signal analysis and cannot be used for Large Signal simulations such as your circuit. It linearizes the circuit components and does not take any harmonics into account including DC. Harmonic Balance is appropriate choice.

 

[KerimF]

The full bridge rectifier has two sides AC and DC. The node 'GND' can exist on one of these two sides only.

For example, on your schematic post (#7), you need to connect the negative node of the bridge (between DIODE1 and DIODE3) with the lower node of the load (R1) after removing the two GNDs at these two nodes. By doing this, there is a GND on the AC side only and no GND on the DC side.

Or, you just need to remove the GND on the AC side which is at the bottom of your schematic (between SRC1 and DIODE3). By doing this, there is a GND on the DC side only and no GND on the AC side.

Kerim

 

[danadakk]

Basic approach, not at 1 Ghz, just where grounds make sense :

Regards, Dana.

 

[crutschow]

I am beginner school student so I ma very noob at this stage

Then you need to learn how to follow a simple schematic,
It should be apparent your circuit is not the same as the posted bridge schematics.

 

[danadakk]

@rajbir_123

As an aside when you look at a circuits, as you journey to being more comfortable
interpreting whats going on in the circuit, when you see differences in what you
are looking at and another web based solution, let the alarm bells go off in your
head and seek the answer on web, and other ref material. Its an accumulated art
to a significant extent, do not despair. Just keep asking questions and seeking on
your own.

Another tool, given you are using a simulator, using simulator probes look at the V, I, P
at various components, that will reveal clues, like your circuit driving a diode directly
with a V source. We know when we drive a diode with a V in forward direction small
changes in V create large changes in current thru it. Same in reverse direction but different
mechanism causing it, junction over voltage breakdown. So in your schematic you have an
AC source strapped right across one of the diodes, on one cycle the V source will cause
high currents thru the diode, in the other if V source generates high enough V that also will
occur. So by probing circuit you will see reasonable currents in some diodes and huge in
others, especially any directly connected to a V source. Keep in mind ideal V source can
deliver infinite current if not limited somehow. Just as a ideal current source can develop
infinite V if one places an infinite R in its path.

So in closing, and device placed in parallel with a ideal V source, does not affect the
V in the circuit because it always puts put, no matter what the load is, a constant V. Its'
like whatever you place in parallel with a V source it does not affect its V, its like the
component is not even there. It does, and can, affect the energy that is required out
of the V source. The inverse is true for ideal current source, no matter what you place
in series it always generates the same current in that branch. Its like the component
does not exist in the branch. Exception again is energy the current source has to deliver.


Regards, Dana.