Bridge rectifier output voltage

[i_am_stupid]

Hi, I'm trying to figure out why the output voltage waveform of this circuit would not include a DC offset of E. My understanding is that D1 becomes forward biased when Vs > E, and so that is when it should conduct. Every waveform drawing I've seen, on the other hand, begins conduction at 0 volts.

Also, what would be the difference between this circuit and one that replaces two diodes with thyristors? I've seen the latter referred to as half-controlled.

Thanks!

 

[mfb]

I didn't analyze the circuit in detail, but don't forget the coil: At Vs close to 0 you have the fastest voltage changes -> current changes in the resistor -> the coil will have some induced voltage.

 

[skeptic2]

At which nodes is the output voltage taken?

 

[256bits]

Take the inductor out of the circuit and you have a diode clipping circuit, where the diode will not conduct until the input waveform Vs > Vd + E. You can see the if you remove the R as shown and instead add an R to the anodes of D1 and D3, to obtain the textbook diode clipping circuit, where the output is taken anode of D1 to the bottom of the circuit. Of course, you have another diode in there, D2, but that is just another 0.7v ( for silicon ).

There is no DC bias voltage in a clipping circuit in the normal sense of where the output voltage for a clipping circuit is taken. Textbooks will add in a capacitor for DC bias and show the output waveform at the appropriate connections.

Then you can add in the inductor to see how that affects the circuit.

Skeptic is correct. One has to specify the nodes to properly identify the output waveform.

 

[alphy]

Hello, thank you for your question.

First, let's discuss the output voltage waveform of a bridge rectifier circuit. The output voltage waveform will not include a DC offset of E because of the way the diodes are arranged in the circuit. Each diode conducts only when its anode is at a higher voltage than its cathode. In a bridge rectifier, the diodes are arranged in such a way that the anode of one diode is always at a higher voltage than the cathode of the other diode. This means that the diodes will only conduct during the positive half of the input voltage, resulting in a pulsating DC output.

Now, onto your question about the difference between a bridge rectifier circuit and one that replaces two diodes with thyristors. As you mentioned, the latter is referred to as a half-controlled circuit because only two of the four diodes are replaced with thyristors. This allows for more control over the output voltage, as thyristors can be triggered to conduct at specific points in the input waveform. This can result in a smoother output voltage compared to a bridge rectifier. However, thyristors are more expensive and require more complex control circuits, so a bridge rectifier is often the preferred choice for simpler applications.

I hope this helps clarify the differences between these two circuits. Let me know if you have any further questions.