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In general, you can consider a diode to be open-circuit when the voltage on its anode is less than that on its cathode.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Positive half-cycles. (b) Negative half-cycles.

The bridge rectifier situation is easy! Remove the reverse biased diodes on alternating half cycles. The 'dot' in the captions refers to the dot on the transformer secondary winding.

In each of the other circuits you need to work out at what input voltage the diode becomes reverse biasedaffects the circuit. Draw that line on the input waveform and you'll be most of the way to your answer.

Load the image into an image editor, mark up the positions as I've described then mark up the resultant output voltage. (Just overlay both on the blue sinewave that is in the question.) Post your effort into your question and we'll see if you are understanding it correctly.

Example for (a):

enter image description here

*Figure 2. (a) When the input voltage exceeds 12 V the diode will conduct and current will flow into the battery. The output will be clamped at 12 V. At all other times the output will follow the input.

Now, you do the same for the other three and post your results in your question.

In general, you can consider a diode to be open-circuit when the voltage on its anode is less than that on its cathode.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Positive half-cycles. (b) Negative half-cycles.

The bridge rectifier situation is easy! Remove the reverse biased diodes on alternating half cycles. The 'dot' in the captions refers to the dot on the transformer secondary winding.

In each of the other circuits you need to work out at what input voltage the diode becomes reverse biased. Draw that line on the input waveform and you'll be most of the way to your answer.

Load the image into an image editor, mark up the positions as I've described then mark up the resultant output voltage. (Just overlay both on the blue sinewave that is in the question.) Post your effort into your question and we'll see if you are understanding it correctly.

In general, you can consider a diode to be open-circuit when the voltage on its anode is less than that on its cathode.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Positive half-cycles. (b) Negative half-cycles.

The bridge rectifier situation is easy! Remove the reverse biased diodes on alternating half cycles. The 'dot' in the captions refers to the dot on the transformer secondary winding.

In each of the other circuits you need to work out at what input voltage the diode becomes affects the circuit. Draw that line on the input waveform and you'll be most of the way to your answer.

Load the image into an image editor, mark up the positions as I've described then mark up the resultant output voltage. (Just overlay both on the blue sinewave that is in the question.) Post your effort into your question and we'll see if you are understanding it correctly.

Example for (a):

enter image description here

*Figure 2. (a) When the input voltage exceeds 12 V the diode will conduct and current will flow into the battery. The output will be clamped at 12 V. At all other times the output will follow the input.

Now, you do the same for the other three and post your results in your question.

Source Link
Transistor
  • 180.3k
  • 14
  • 193
  • 417

In general, you can consider a diode to be open-circuit when the voltage on its anode is less than that on its cathode.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Positive half-cycles. (b) Negative half-cycles.

The bridge rectifier situation is easy! Remove the reverse biased diodes on alternating half cycles. The 'dot' in the captions refers to the dot on the transformer secondary winding.

In each of the other circuits you need to work out at what input voltage the diode becomes reverse biased. Draw that line on the input waveform and you'll be most of the way to your answer.

Load the image into an image editor, mark up the positions as I've described then mark up the resultant output voltage. (Just overlay both on the blue sinewave that is in the question.) Post your effort into your question and we'll see if you are understanding it correctly.