AC source signal getting clipped

I'm new to the bridge rectifier circuit and just experimenting with various forms. This version of the circuit has the expected Vdc output signal, but the AC source voltage seems like it getting clipped when cycling negative. I'm not sure if this is expected or a byproduct of how I have this setup in Circuit Lab.

Circuit I'm using:

Observed voltages: [img]

by klh1228
March 22, 2023

The model does not match the intention.

Look at the same, with extra labels:

Run the simulation and note that the negativeSide of the source never goes under ... 0.7 volt. A diode voltage, it is surely a clue to what is going on!

If you consider the part of the cycle where the source try to be more negative than the ground, or, if you prefer, where the "GROUND is MORE POSITIVE than the foot or the AC source".

If you THEN follow the path of the (classical or conventional) current from the ground toward the AC source, 0 at the ground, lose 0.7 volt through D2, and that's all folks.

Now that you know the problem, how to fix the simulation? I am sure that the real pleasure is to find by yourself, but if you can't within 24 hours, I will supply the (well, one of the possible) answers.

by vanderghast
March 22, 2023

Note: Be careful if you use an oscilloscope powered up by the domestic electric distribution network: the ground of the probe that you will use MUST BE the same as the ground of the scope, in this circuit, ELSE, you likely kill your oscilloscope or, at best, its probe. Not problem with differential probe (such as the ADLM2000, but you are limited to +5/-5 volt in this specific case), or with isolated oscilloscope (hand held, or otherwise uncoupled).

Sure, a simulator like this one is without danger ... to the oscilloscope. :-)

by vanderghast
March 22, 2023

4 Answers

Answer by klh1228

Thank you for taking the time to reply so thoroughly, @vanderghast. And also thanks for leading me in the right direction, but making me work a bit to find an answer. The added node names and pointing out the negative AC voltage to ground reference issue helped quite a lot.

I ended leaving the ground symbol on the DC circuit and plotting the voltage difference across the plus side and negative side of the AC source (each component of which I think Circuit Lab measures relative to the ground I have placed on the DC side). See updated circuit and plot below.

On the topic of measuring using a wall powered oscilloscope, I confess that I wasn't aware of the need to make sure the ground of the probe match the ground of the scope. I will follow-up with some internet research on this topic, which intuitively makes sense. I am grateful to be warned! Thanks again.


+1 vote
by klh1228
March 22, 2023

Googling around I found this suggestion for how to safely and properly measure the voltage across the load. I think this advice aligns with what you warned me about @vanderghast.

Use two oscilloscope probes. Connect the probe tips to either side of Rload. Connect both of their black alligator clips to circuit ground, which should be the place where the ground lead from the function generator is also connected. On the scope, display the difference between the two probes. Try pressing that button labeled "Math" – Elliot Alderson May 5, 2022 at 9:44

by klh1228
March 22, 2023

Answer by RandyPearson

Thanks for explaining. I am a college student and I am making a final project and you helped me a lot. If I want to know more, I will message you.

+1 vote
by RandyPearson
April 18, 2023

For those who may have not found a solution:

Take a look at the description section of the circuit here up.

Note the position of the ground, and compare it to the original post. That is the main difference.

You can run the simulation to get the three voltages: at Node1, at Node2 and their difference, V(Node1)-V(Node2)

by vanderghast
April 18, 2023

Answer by SharonHarris

When the AC input voltage is positive, diodes D1 and D3 are forward-biased, allowing current to flow through them and into the load resistor. At the same time, diodes D2 and D4 are reverse-biased and do not conduct current.

+1 vote
by SharonHarris
April 22, 2023

Answer by betterwound

Diodes D1 and D3 are forward-biased when the AC input voltage is positive, allowing current to pass through them and into the load resistor. doodle jump

+1 vote
by betterwound
May 15, 2023

Your Answer

You must log in or create an account (free!) to answer a question.

Log in Create an account

Go Ad-Free. Activate your CircuitLab membership. No more ads. Save unlimited circuits. Run unlimited simulations.

Search Questions & Answers

Ask a Question

Anyone can ask a question.

Did you already search (see above) to see if a similar question has already been answered? If you can't find the answer, you may ask a question.

About This Site

CircuitLab's Q&A site is a FREE questions and answers forum for electronics and electrical engineering students, hobbyists, and professionals.

We encourage you to use our built-in schematic & simulation software to add more detail to your questions and answers.

Acceptable Questions:

  • Concept or theory questions
  • Practical engineering questions
  • “Homework” questions
  • Software/hardware intersection
  • Best practices
  • Design choices & component selection
  • Troubleshooting

Unacceptable Questions:

  • Non-English language content
  • Non-question discussion
  • Non-electronics questions
  • Vendor-specific topics
  • Pure software questions
  • CircuitLab software support

Please respect that there are both seasoned experts and total newbies here: please be nice, be constructive, and be specific!

About CircuitLab

CircuitLab is an in-browser schematic capture and circuit simulation software tool to help you rapidly design and analyze analog and digital electronics systems.