0
\$\begingroup\$

I have this question: enter image description here

This is how I went about answering this question:

In the above question, it's telling us to presume the ideal diode model. Therefore the D4 (diode 4) is in forward bias when Vs is negative. So this would mean that D4 is on when Vs is negative. But according to the diode model, when the diode is on doesn't this mean that although it allows an infinite current, the voltage through the diode is 0? Therefore, during the region that Vs is negative Vd4 is 0.

When Vs is positive, D4 is is the negative bias state (i.e. "Off state"). Therefore, voltage in the direction of anode to the cathode of D4 is negative.

All in all I'd get a graph like the following: enter image description here

However, I believe that the actual solution is: enter image description here

Could someone please tell me where I am going wrong in my logic?

Thank you

\$\endgroup\$
1
  • \$\begingroup\$ No error, the red-blue image is ok. The voltage doesn't flow, it is and has a direction. Fix that to your text. Write "voltage direction is from X to Y" instead of "voltage flows from X to Y". Term "flow" refers to some kind of movement. Voltage causes movement like a pressure difference. \$\endgroup\$
    – user136077
    Oct 28, 2017 at 15:46

2 Answers 2

1
\$\begingroup\$

The question is a little difficult because neither side of the AC is ground referenced. With the bridge rectifier in place the ground keeps switching from side to side of the AC source as each is "grounded" by D3 or D4 on alternate half-cycles. That means that your nice blue sine graph has to be on a separate axis to the diode waveforms.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The original circuit for simulation.

enter image description here

Figure 2. The results of the simulation (using real diode models).

In Figure 2 you can see the blue and orange side of the bridge get alternately "grounded" or clamped at zero volts.

schematic

simulate this circuit

Figure 3. (a) Positive half-cycle equivalent circuit. (b) Negative half-cycle equivalent circuit.

Splitting the circuit as shown in Figure 3 may help with your analysis.

  • On the V1 positive half-cycle D4 (not-shown) will be reverse biased so it will have the positive half-cycle voltage applied to it's cathode. But, since the question has a '+' on the anode it shows that voltage is to be measured anode to cathode so the sign is reversed.
  • On the V1 negative half-cycle D4 is forward biased so the voltage is zero.

This means that your large red and blue graph is correct.

\$\endgroup\$
0
\$\begingroup\$

Arguably, it depends on your definition of \$V_{D4}\$.

enter image description here

For the sake of this argument let us choose it to mean the voltage from the positive terminal, \$V_{D4+}\$ to the negative terminal \$V_{D4-}\$.

So \$V_{D4} = V_{D4+} - V_{D4-}\$

But whatever. You are overthinking this, and need to concentrate on the currents and when the diode is on.

Whenever the positive terminal of the sine wave generator is negative, D4 must be on, and since it's an ideal diode, that means \$V_{D4}\$ during negative half cycles must be zero.

There, half done already.

During positive half cycles D4 is off / open circuit, and the voltage on the negative pin, \$D_{4-}\$ must follow the AC voltage.

Meanwhile, \$D_{4+}\$ is now shorted to the negative pin of the source via D3. So D4 has the full source voltage across it, biasing it off.

Since we are measuring the thing sort of backwards that means \$V_{D4}\$ is a negative voltage.

enter image description here

Of course if you measure if with the more traditional method of having the red wire on the more positive side with respect to the indicated ground, the blue trace above is the other way up. That would be \$V_{D4} = V_{D4-} - V_{D4+}\$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.