Full Wave Rectifier with Diode Bridge

Discussion in 'Homework Help' started by PsySc0rpi0n, Mar 25, 2014.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Starting a new thread to try to keep things cleaner.

    Ok, I'll start drawing a circuit with a diode bridge. Then we will discuss it...

    In the first screen I have just built a circuit to see the diode voltage drop.
    We can see it by the difference existing between the peak of the Vin curve and the peak of the Vout curve...

    Please, correct me if I'm wrong!
     
  2. MaxHeadRoom

    Expert

    Jul 18, 2013
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    Looks about right for two diodes in series.
    Max.
     
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    What I'm doing next is to place a resistor between Vout and the ground and increase the resistor value...

    I can see that the diode Vdrop decreases, meaning that the drop of 2*0.7 is changing to something like 2*0.66 or 2*0.62 or so.

    Why does this happens?
     
  4. MaxHeadRoom

    Expert

    Jul 18, 2013
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    What would be Your first guess?
    Max.
     
  5. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I really don't know but that would depend on the specification of the diode 1N4001, I think. It has 0.75 of Vf.
     
  6. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    Have a look through the datasheet for the 1N4001 diode. (1N400x series).

    There are some charts there that might give you an idea. :)
     
  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I have this in standard.dio

    Code ( (Unknown Language)):
    1. .MODEL 1N4001 D(IS=2.55E-9 RS=0.042 N=1.75 TT=5.76E-6 CJO=1.85E-11 VJ=0.75 M=0.333 BV=50 IBV=1E-5 )
    So I think it's VJ=0.75 that should give me the Vdrop...
     
  8. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi Psy,
    Recall the earlier posts on the other thread of yours, when we LTS plotted the effect on the diode charging current when using different values of C1 smoothing capacitor.

    E
     
  9. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    This is a clip from the 1N4001 datasheet, study the curve of I inst versus V forward diode drop.

    What do you make the diode current to be at say Vfwd of 0.75V.???

    EDIT:
    Added a diode pdf for you to read.
     
  10. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    That graph looks like to have a logarithmic scale at Y-axis... So, I would say that until around 0.7V~0.8V the current goes up to 0.1A. Then in only 0.4V, the current goes up to 10A...

    This is what we say to be the Vf, I guess... Is where the current as a considerable increase for a small amount of voltage!
     
  11. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    Did you look back at the diode current plots in the other thread.?

    As a rough guide we assume a Vfwd of 0.7V for a general purpose silicon rectifier.

    This image shows the I/V curve for a silicon diode.

    Whats the next step in your experiment.?
     
  12. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Yes, I know...

    We also assume that value of 0.7V for silicon diodes, but the definition in standard.dio says vf=0.75. When we have a rectifier with a bridge of diodes, that will produce a 1.5V rather than 1.4V as we should have assuming 0.7V
     
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