Bridge Rectifier Circuit Operation and Output Waveform Explanation

[STEMucator]

Homework Statement

The output of the transformer is a sinusoidal AC signal with $10.6V$ RMS amplitude, and acts as an input to the bridge rectifier.

Explain the operation of the following bridge rectifier circuit.

Then sketch the output waveform $V_L$ assuming the $0.7V$ drop model for the silicon diodes.

Homework Equations

The Attempt at a Solution

I want to make sure I understand the operation of this circuit correctly.

The output of the the transformer acts as the input to the bridge rectifier circuit, call the input $v_s$. The input will have a maximum amplitude given by $V_p = (10.6 V)\sqrt{2} = 15 V$.

During the positive half cycles of the input voltage, $v_s$ is positive, and current is conducted through $D_1, D_L, R_L$ and $D_4$. The diodes $D_2$ and $D_3$ are reversed biased during this time. The output voltage $V_L$ will be lower than $v_s$ by two $0.7V$ diode drops and a $1.5V$ LED drop (I've been told to assume LEDs have a 1.5 V drop). So the maximum amplitude of $V_L$ would be $12.1 V$.

During the negative half cycles of the input voltage, $v_s$ is negative, so $-v_s$ is positive. Current will be conducted through $D_2, D_L, R_L$ and $D_3$. The diodes $D_1$ and $D_4$ are reversed biased during this time. The output voltage $V_L$ will be lower than $v_s$ by two $0.7V$ diode drops and a $1.5V$ LED drop. So the maximum amplitude of $V_L$ would be $12.1 V$.

Does this sound okay? Thank you for your help in advance.

 

[Svein]

Does this sound okay?

Yes.

 

[Merlin3189]

Your logic sounds fine except for the LED drop. It may be I am misreading your diagram, but V_L seems to include the LED and the R_L because the arrows point to the top and bottom conductors and the + and - signs are placed at those points, not at the top and bottom of R_L.
Either way, you have the correct understanding.

 

[STEMucator]

Your logic sounds fine except for the LED drop. It may be I am misreading your diagram, but V_L seems to include the LED and the R_L because the arrows point to the top and bottom conductors and the + and - signs are placed at those points, not at the top and bottom of R_L.
Either way, you have the correct understanding.

So the max amplitude of $V_L$ should actually be lower by just the two diode drops, i.e 13.6 V.

 

[Merlin3189]

That's what I thought.
Since you correctly show how you got the result, if it's clear where you are measuring V_L then you can't be wrong.