Designing a full-wave bridge rectifier

Discussion in 'General Electronics Chat' started by kaiosama, May 1, 2013.

  1. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    Design a full-wave bridge rectifier circuit to deliver 10V dc with less than 0.1V(pp) ripple into a load drawing up to 10mA. Choose the appropriate ac input voltage, assuming 0.6v diode voltage drops.

    The ac must output 10V + 0.6V + 0.6V = 11.2V RMS
    which is about 16V amplitude.

    Vripple = Iload / (2*f*C)

    Assuming f = 60hz, C = 0.83 mF

    Am I missing something?

    I tried simulating this circuit using LTspice, here is the circuit:
    [​IMG]
    probing the red node, I get this:
    [​IMG]

    Obviously something is wrong.
    Thank you very much.
     
  2. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Your positive will be the other side of the capacitor from your red dot.

    Which node did you select as ground reference? It should be the red dot!

    You have no load. Add a resistor in parallel to the capacitor (hint: try 1200 ohm).

    Your math is wrong. Convert 10v rms to peak first, then subtract for two diode drops.
     
  3. #12

    Expert

    Nov 30, 2010
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    the frequency is wrong, too.
     
  4. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    #12
    He has 60 waves per second on the fine lines, I just don't know why he has a 6 Hz wave laid over the top?
     
  5. #12

    Expert

    Nov 30, 2010
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    im talking about 60 hz rectified = 120 hz to the filter

    Sorry about the bad typing. Sister on the phone and I was typing one handed.

    11.2 volts peak is not 16 volts. It's 11.2/sqrt2 = 7.92 RMS
     
    Last edited: May 2, 2013
  6. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    Why should I convert 10 rms to peak first? The way I see it is, the load needs 10V dc to work. The 2 diodes each need 0.6V dc, so we need 11.2V RMS at the output of the signal source

    then sqrt(2)* RMS = amplitude

    why? the frequency can be arbitrary as long as the capacitor matches it. right?
     
  7. BillB3857

    Senior Member

    Feb 28, 2009
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    Move your ground to the common anodes of the diode bridge and take your measurement from the common cathodes. Right now, you have one end of the AC supply as a reference. Also, you show the positive lead of the cap at the top and negative at the bottom. That's not really wrong, but unusual.
     
    Last edited: May 2, 2013
  8. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    Thank you very much.

    Without a load, here is what it plots now:
    [​IMG]
    Seems like I get 15V DC.

    If I add a 1200 ohm resistor load in parallel with the cap, I get this:
    [​IMG]
    Around 14.2V DC

    Two questions:
    1. Why is there a difference in voltage with a load and without a load? What causes this?
    2. I need 10V DC but I get from 14 to 15V DC. This means either my math is wrong, either there's an error in my ltspice schematic.

    I repeat my computations:
    10V + 0.6V + 0.6V = 11.2V DC RMS
    so amplitude must be: 11.2*sqrt(2) = 15.84V, let's round up to 16.

    frequency can be anything, C is determined by:
    Vripple = Iload / (2*f*C)
    In my case, C = 0.83 mF
     
  9. richard.cs

    Member

    Mar 3, 2012
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    As others have said the capacitor charges to the peak value, and thus the diodes only conduct on the voltage peak. So if you start with a requirement for 10 Volts then you need a *peak* voltage of 10+(2*0.6) = 11.2 Volts and therefore an rms of 11.2/sqrt(2)= 7.8V.

    In reality when any significant current is drawn the diode drop will be more than 0.6V. Remember that the diode current is greater than the d.c. load current as it is drawin in short spikes. I suspect a 8.5V rms source (this is 12V peak) will be about right for you. Standard transformer secondaries are 7.5V and 9V rms, if you need a more accurate output voltage than that you would be better off with a regulated supply.
     
  10. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    I tried your suggestions and it worked. However I don't really understand the reasoning...why does the capacitor charge to the peak value and why do the diodes only conduct on the voltage peak?

    To me it makes more sense to say that the load needs 10V dc on average (and therefore RMS).

    Say at time t, voltage is +5V, this is greater than the forward voltage drop of the diode, shouldn't they conduct and contribute to charging the capacitor?

    Thinking of it I think I don't really understand this circuit. I know the role of the capacitor is to smooth out the oscillations of AC but how exactly? The AC charges the capacitor which acts as a battery to the load, is that right?

    If the diodes only conduct on the voltage peak then they only conduct at discrete or very very small intervals in comparison to the entire AC spectrum, which does not make much sense to me.
     
  11. richard.cs

    Member

    Mar 3, 2012
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    The diodes only conduct when the voltage from the source is greater than the voltage across the capacitor by at least 0.6 Volts. The absolute voltage on the input is unimportant, it's the difference between that and the capacitor voltage.
    Plot the voltage at each and the current in the diodes you will see these pulses of current.
     
  12. #12

    Expert

    Nov 30, 2010
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    Your generator is labeled 60Hz. After a full wave bridge, the filter frequency is 120 Hz.
    You can make the generator any frequency you like, but the filter frequency will always be twice that.
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    The 6Hz is almost certainly nothing more than aliasing of the visual display, which is a sampling of the signal at a rate corresponding to the horizontal pixel rate. You see this is DSO's all the time.
     
  14. WBahn

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    Mar 31, 2012
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    You need to remember that circuits are very much creatures of the moment. They don't care about averages, they respond to what is happening this instant. It's mostly humans that care about characterizing the behavior of a circuit over time and who come up with measurements, such as average and RMS, to do so.

    Also, keep in mind that individual devices are very short-sited. They care only about what is happening at THEIR terminals. So you can't look at the voltage on one side, say the anode, of a diode and, seeing that it is 5V, conclude that the diode is conducting. This is because the 5V is measured relative to some other node in the circuit, while the diode only cares about the voltage ACROSS the diode. So if the voltage on the other side of the diode is 4.7V, the diode isn't conducting.

    And you are correct that the current from the supply will be delivered in short bursts of high current. This isn't desirable, it's just the way it is. There are some things you can do about it -- such as putting a resistor or choke in series with the supply and/or filter -- but those have disadvantages that have to be taken into account.
     
  15. kaiosama

    Thread Starter New Member

    Dec 6, 2010
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    thank you very much for your answer. how do you know all that? what can I do to improve my practical knowledge of circuits (putting a resistor here, a capacitor here...that kind of thing). Im getting started with the art of electronics, that's where this problem comes from. What else can I do to gain intuition and more practical knowledge?
     
  16. WBahn

    Moderator

    Mar 31, 2012
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    You gain practical knowledge by practicing the art.

    Read a lot and build a lot of circuits and play with them. You will let plenty of smoke out of resistors and other things along the way and probably burn a finger or two for good measure. Those experiences, however, will tend to be the ones that you will learn the most from.
     
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