Bridge rectifier output voltage

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Discussion Overview

The discussion centers on the output voltage waveform of a bridge rectifier circuit, specifically addressing the presence or absence of a DC offset and the comparison of this circuit with one using thyristors. Participants explore the implications of circuit components, such as inductors and resistors, on the output characteristics.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant questions why the output voltage waveform does not include a DC offset of E, suggesting that D1 should conduct when Vs > E, while noting that typical waveform drawings start conduction at 0 volts.
  • Another participant mentions the role of the inductor, indicating that at Vs close to 0, rapid voltage changes can induce current changes in the resistor, affecting the output.
  • A participant asks for clarification on the nodes from which the output voltage is taken, implying that this is crucial for understanding the waveform.
  • Another participant proposes that removing the inductor results in a diode clipping circuit, where conduction occurs only when Vs > Vd + E, and emphasizes the importance of specifying output connections for accurate waveform identification.
  • It is noted that the presence of another diode (D2) adds a voltage drop, but does not contribute to a DC bias in the clipping circuit unless additional components, like a capacitor, are included.

Areas of Agreement / Disagreement

Participants express differing views on the output voltage characteristics and the role of circuit components, indicating that multiple competing perspectives exist without a clear consensus.

Contextual Notes

Participants have not fully analyzed the circuit details, and there are unresolved aspects regarding the specific nodes for output voltage measurement and the effects of different components on the waveform.

i_am_stupid
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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.

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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!
 
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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.
 
At which nodes is the output voltage taken?
 
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.
 
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