0
\$\begingroup\$

In my class, we are using Proteus 8 to simulate simple rectifier circuits. We are not familiar with analog electronics yet. Since the class is all about using engineering software, there are not many explanations for the theories.

In this particular class, we tried 3 different rectifiers including half-wave, center tap, and bridge rectifier. I applied the same signal and resistor value to all of the circuits, 6V,10Hz, RL = 500ohm, 1:1 ratio for all transformers, and the same setting for both the oscilloscope and signal generator.

I was able to figure the half-wave and the bridge rectifier's output voltages. The half-wave rectifier's output voltage is 5.25V (5.3V) because of the diode's voltage drop 6V-0.7V, while the bridge rectifier got 4.65V(4.6V) because the current goes through 2 diodes in each cycle 6V-0.7V-0.7V.

Half-wave rectifier:

enter image description here enter image description here enter image description here

Bridge rectifier:

enter image description here enter image description here

Now for the real question. From what I understand, since the voltage is divided across the secondary winding, each cycle has 3V. If current only passes through 1 diode in each cycle, shouldn't the amplitude be around 2.3V?

enter image description here enter image description here

I know it should be more complicated. From what I understand, it should include something like Vrms, Vavg which includes integrals, or something like that, but I turned the output channels to DC instead of AC as per my lecturer instructed so that I can only calculate it as in DC analysis, so I'm confused right now. I'm not sure if this all makes sense at this point, hope anyone would help me understand better, thank you for reading this far.

\$\endgroup\$
5
  • \$\begingroup\$ I note a little problem with the transformers. In the first picture, it is a TRAN-2P2S. In the second, TRAN-2P2S. In the last, it is TRAN-2P3S. Have you changed the transfo ? \$\endgroup\$
    – Antonio51
    Aug 7, 2021 at 14:00
  • \$\begingroup\$ Have you changed the parameters accordingly ? In the first two cases, the primary and secondary value of the selfs (1H, 1H) are the same, ok. For the last case, the number of globals turns at secondary is twice the values as in the first cases. So here, the primary self is 1H, the total secondary self is 4 H. \$\endgroup\$
    – Antonio51
    Aug 7, 2021 at 14:12
  • \$\begingroup\$ So the center tap transformer shouldn't be 1H to 1H?Hmm, if the secondary is twice the primary.. Then shouldn't it be 1H to 2H?. \$\endgroup\$
    – Syahrul
    Aug 8, 2021 at 16:37
  • \$\begingroup\$ No. L is proportional to N^2. \$\endgroup\$
    – Antonio51
    Aug 8, 2021 at 18:22
  • 1
    \$\begingroup\$ Your question title says "center tapped rectifier". Shouldn't that be "center tapped transformer"? You can't tap a rectifier. \$\endgroup\$
    – Transistor
    Aug 8, 2021 at 18:36

1 Answer 1

1
\$\begingroup\$

Review the tran-2p3s in the library.

enter image description here

The 3 windings are identical. If primary has N turns, each secondary has N turns. Proteus lib for trans-2P3S is done so that the 2 secondaries are taken together into account for defining the numbers of turns of secondary line and inductance in the library. So, if N turns for primary give 1H, secondary has 2*N turns ... this lead to inductance of 2^2= 4 H.

\$\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.