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Full-Wave Bridge : problems

  1. Feb 24, 2007 #1
    I am constructing a full-wave bridge with the following:

    - four D1N4148
    - Function generator (50Hz with 10 Vpp)
    - 1k ohm resistor
    - Oscilloscope

    I've been trying to construct the circuit, but every time I did it I always get a half-wave instead of a full-wave. All the diodes are working fine, so I don't know what is the problem.

    Below is my circuit:

    Help anyone?
  2. jcsd
  3. Feb 24, 2007 #2


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    Is your scope ground referenced to the same ground that output of your function generator is?
  4. Feb 25, 2007 #3
    Yes it is.
  5. Feb 25, 2007 #4
    My guess is that you've wired it the wrong way. If you've used breadboard it is always possible. I'm thinking that you're measuring voltage not between load and the ground, but rather over D1 or D3 (referenced to your picture) since they are the only diodes in this circuit which produces half-wave (D1 is producing negative cycles and D3 - positive).
  6. Feb 25, 2007 #5


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    How is the circuit wired? Soldered? Proto board?

    Double check your connections... Then check them again.

    Measure the waveform across each individual diode (keep a close eye on your ground!). Are they all working?
  7. Feb 25, 2007 #6


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    The reason why you are getting the results that you are is because you have basically shorted out D4 because you are sharing grounds.
  8. Feb 26, 2007 #7
    I tried the experiment with different apparatus this time where I am using:

    1. Step-down transformer (220V to 12V)
    2. Four 1N4148 diodes
    3. 10k ohm resistor.

    It is a bridge connection. Below are pictures of my circuit and its output . The output is 6V (amplitude for a peak) and there are something that look like clippings at the peaks. Theoretical output is supposed to be 10.6V (amplitude for a peak) but I got 6V instead.

    Thus why do I have a some sort of clipped peaks and an output of 6V (amplitude for a peak) which is obviously lower than expected?

    So that you know the oscilloscope's probe is connected across the resistor whereas the yellow wire is the +ve and the black is the -ve/ground and they are connected to the transformer.

    http://img410.imageshack.us/my.php?image=testbridgeji7.gif - Picture of how the circuit was constructed.

    http://img206.imageshack.us/my.php?image=testbridge1oc2.gif - This is the picture of the output where one box represents 2V.


    Another question, if I am to use the circuit that I showed here, am I supposed to get the same result if I use a function generator (giving 15Vpp and a 10kHz frequency) as the AC supply? I mean is the oscilloscope supposed to show a rectified full-wave? Because I keep on getting a half-wave instead.
    Last edited: Feb 27, 2007
  9. Feb 26, 2007 #8


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    The output is only 6 volts what? DC? Peak to peak? What kind of instrument did you use to meausure this? The waveform you have there looks right to me as far as I can tell without knowing any of your scope connections. I'd say it's wired correctly also.
  10. Feb 26, 2007 #9
    The output is is approximately 6V AC (all +ve polarity) after rectified. It is read from the oscilloscope where the probe was connected across the 10k ohm resistor.
    Last edited: Feb 27, 2007
  11. Feb 27, 2007 #10


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    Then that is exactly what I would expect.
  12. Feb 28, 2007 #11
    The transformer gives an input of 12Vpp (checked it with an oscilloscope), so the amplitude is 6V per each peak. When the input is rectified (all +ve polarity) won't the value drop by 1.4V since two diodes are forward-biased? Shouldn't the output be around 4.6V, but instead I got a 5.6V output. Why the difference?

    This is the very clear connection of the whole thing. As you can see the transformer is a 12-0-12. And i connected it to 12-0.
    http://www.ietucsi.com/layout/KABOOM.gif [Broken]
    Last edited by a moderator: May 2, 2017
  13. Mar 1, 2007 #12
    I re-did it again and followed the bridge in http://www.ee.unlv.edu/kevin/index_files/labs/320L.pdf (as in the picture at http://www.ietucsi.com/layout/KABOOM.gif [Broken], you can clearly see that it is a 12-0-12V transformer. This time I've connected the circuit to both 12V where I connect to the the right and left side but left the center alone.)

    I supplied 24Vpp as the input and my output is 11V. I am wondering as to why it is not 10.6V. Where did the 0.4V in 11V comes from?
    Last edited by a moderator: May 2, 2017
  14. Mar 1, 2007 #13


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    You cannot count on a diode dropping .7 volts no matter what. If the current through the diode is relatively small it may only drop .3 volts, or even less. Naturally, if the current is relatively high, some diodes may drop nearly a full volt. The 1N4148 is not a special diode, I would expect your results. Are you measuring EVERYTHING with a scope?
  15. Mar 4, 2007 #14
    No, I measured the readings with a digital multimeter too. OK so let me get everything straight for myself ... a bit confused here.

    The transformer is a stepped-down transformer (230V to 12V). When the transformer says that it stepped-down the value from 230V to 12V it means that that 12V is the connection for either the right or left tap with the center tap right? So to get a 24V input I just need to use the right and left taps only, thus I will get a 24V input peak-to-peak. Am I right?

    Measuring the input using the digital multimeter I got a reading of 27V. Reading this value in the multimeter it is in RMS and is a peak-to-peak value right? If it is so then the input is approximately 19.1Vp in AC.

    Reading the voltage (in AC using the multimeter) across the load I got 11.94V and again it is in rms right? So that means I have 16.9Vp for the output cross the load in AC.

    Simulation in Pspice gave me an output of 17.5Vp (input of 19.1V).

    The simulation and practical output has a difference of 0.6V where simulation gives a higher output voltage compared to practical.

    What reasons other than the resistance in the jumper wires used explain the 0.6V difference?

    In addition to the above when I did the diode test, all the diodes gave a reading of 0.62V. Does this mean that 0.62V is the diode drop for each diode?
    Last edited: Mar 4, 2007
  16. Mar 10, 2007 #15


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    Sorry I've let this slip by. I was pretty busy for a couple of days and then I forgot about it. I'll post more later.
  17. Mar 11, 2007 #16


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    No. You will get 24 volts RMS. Peak-to-peak for a sine wave would be 24 * 2 * 1.414.

    No again. A 27 volt reading on your multimeter is 27 volts RMS.

    Ok. Here I'm not sure what you should get for a reading. It all depends on your meter. A good quality true RMS meter should read the same thing after the bridge rectifier as it did before the rectifier minus the loss across the diodes. In your case, you also have a DC offset which may throw the meter reading.

    Yes, that is what it means. How did you test for this?
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