Can't understand Clipper

Discussion in 'General Electronics Chat' started by macman, Dec 18, 2014.

  1. macman

    Thread Starter New Member

    Dec 18, 2014
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    [​IMG]

    That is the input and output graph of clipper. I have gone through several video lectures and all of that they say that:

    When V(IN) equals the barrier potential or V(ref) then V(out) becomes a straight line until again V(in) gets below barrier potential.
    Ok but why? Isn't V(in) increasing continously..... i mean in an AC source voltage increments to the peak...
    So suppose if V(ref)=V(in) at 0.7 volts then even after that V(in) will continously increase so how does V(out) become constant(straight line)??? Isn't total energy in=total energy out?
     
  2. MikeML

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    What is the element(s) used to create the clipping? Back-to-back Zeners? What is the ratio of the dynamic resistance of a Zener compared to the series resistor?

    249.gif
     
    Last edited: Dec 18, 2014
  3. Papabravo

    Expert

    Feb 24, 2006
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    You're only looking at part of the picture, namely the voltage waveform. The other part of the picture is what happens to the current. A clipper is made with diodes that have a non-linear V-I characteristic. At the fixed voltage V(ref) the diode will conduct large amounts of current and the excess voltage will be dropped by other circuit elements. You should give us a schematic so we know precisely what we are talking about.

    With respect to your question about energy in and out the answer in no. That is not one of the rules of circuitry. In electronic terms we often ask about power in and power out. In all real circuits the power will always be less, sometimes much less, than the power in. Where does the power(energy) go? It is usually lost as heat in a resistive load.
     
  4. kubeek

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    Sep 20, 2005
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    What does energy have to do with all this?
    Anyway the difference between Vin and Vout will most likely be spread across a resistor. Show your circuit.
     
  5. macman

    Thread Starter New Member

    Dec 18, 2014
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    In clippers....when V(in)>V(barrier) then V(out)=V(barrier). This is what they say right considering a simple clipper circuit?

    But my doubt is that V(in) continues to grow...that is suppose it's peak is 5v then it will go to 5v and only then decrease. But V(out) stays fixed at the value of V(in) when it equaled V(barrier).
    So how come the graph of V(out) is straight line, isn't it should overlap with the graph of V(in) ? How does voltage gets clipped?
     
    Last edited: Dec 18, 2014
  6. MikeML

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    The expression (for the positive half-cycle) should be:

    If (V(in)<=V(barrier)) then V(out)= V(in) else V(out) = V(barrier)
     
  7. #12

    Expert

    Nov 30, 2010
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    Fill in the rest of the circuit. If the impedance of the supply voltage is zero, then the circuit will not clip because the clipper diodes will smoke or explode. Attach a resistor to the input and the clippers will clip with the excess voltage being used up by the resistor.
     
  8. macman

    Thread Starter New Member

    Dec 18, 2014
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    Lots of websites give conflicting information. At the one hand they graph what your logic says but on the other hand they say that if V(in)<V(barrier) then it's reverse biased and nothing flows ....so what right?
     
  9. #12

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    They are both right. Fill in the rest of the circuit! There must be some resistance in series between the input signal and the clipper circuit or the clipper will not work.

    Post #2. Mike's R1. What do you think it's there for?
     
    Last edited: Dec 18, 2014
  10. Papabravo

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    It's all correct.
    A reverse biased diode has little or no current flowing in it. If no current flows in the diodes then no current flows in the resistor. No current in the resistor means no voltage drop and V(out) = V(in)
    A forward biased diode will draw a fixed amount of current and that current will correspond to the zener voltage in MikeML's circuit. The resistor will determine the amount of zener current that flows as ((V(in) - V(barrier))/R
     
  11. macman

    Thread Starter New Member

    Dec 18, 2014
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    I will go through my thoughts and you can check if i am wrong.

    [​IMG]

    (Couldn't get another pic so consider V(dc) polarity reversed and D inverted)

    Suppose AC source provides a peak of 10V.
    When V(in)>V(dc) then voltage drop across R is 10-V(dc) and thats why V(out) stays constant at V(dc) even though V(in) continues to climb to peak?
     
  12. #12

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    Adding the battery messes up the concept. A clipper wastes the excess current into ground pin.
    Take the battery out and reverse the diode.
    When Vin is less than 0.6 volts, the diode is not conducting so the output voltage tracks the input voltage.
    When the input voltage is greater than 0.6 volts, the diode conducts and all current that would support more than 0.6 volts at the output is shunted to ground (the place marked - ).
     
  13. Papabravo

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    You're asking us to do a lot of mental gymnastics to try and answer your question. As it is your drawing is worse than useless! Instead of continuing to confuse us and yourself, try and learn to draw a proper schematic.
     
  14. MikeML

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    Yes, and the ones that do are wrong! When it comes to electronics, most of the stuff posted out there is crap, posted by newbies that know more about how to create a web site than practical or theoretical electronics...

    Look up the current vs voltage (IV) curve of a Zener diode. If a clipper uses just one Zener (unlike the circuit I posted which had two), then the Zener effectively has two intrinsic barrier voltages, one for either polarity. In the forward direction (anode positive with respect to cathode), the Zener acts as a standard diode, so its V(barrier) is about 0.65V. In the reverse direction, V(barrier) is its rated Zener voltage V(Z), so if V(in) is greater than V(Z), current will flow on both positive and negative half-cycles of the AC waveform.
     
  15. ian field

    Distinguished Member

    Oct 27, 2012
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    Vin must be greater than the limiting device, or it won't clip.

    Vin must be fed via a series resistance to limit the current when the device clips - if the limiting device is a forward biased diode, a standard silicon type will clamp at about 0.7V, up to that voltage only a tiny leakage current flows, at a certain point around 0.7V you hit the "knee curve" of the diode forward characteristic, at that point the current rises sharply which is what causes the clamping effect in a clipper circuit.

    With no series resistor between Vin and the clipper - the sharp rise in current at the knee curve will rise high enough to destroy the device.
     
  16. Kishor Tiwari

    New Member

    Dec 25, 2014
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    The Diode used in the clipper only conducts in the forward bias and open circuit in the reverse bias that is why some portion of the input wave is cut off and output wave is clipped wave with some of the portion removed.
     
  17. atferrari

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    Hola macman,

    Aren't you able to draw a circuit by hand? Putting yourself in the necessity to do it right, it is a good exercise that could even help you to understand how it works. And then, to post it here, there is little more needed.

    The LTSpice simulator is a quick way to verify things, starting with the simplest ones. (Why not Ohm's law?)

    Question to all contributing already: I got surprised by the "barrier" expression. Probably more than correct but not usual, right? Willing to learn myself.
     
  18. WBahn

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    Mar 31, 2012
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    At least, after repeated requests, we got a schematic, even though it isn't the one you want us to consider and neither matches the graph you gave in the first post.

    Instead of scouring the internet looking for a schematic that is exactly what you want, either draw one from scratch or edit one that is close. The following took less than one minute to prepare from your diagram and explanation using nothing more than Paint:

    Edit_2014-12-25_1.png

    You want to consider the circuit in two pieces. When Vo would "like" to be more than Vdc+Vd (Vd is the forward-biased voltage drop of the diode which is 0V for an ideal diode and about 0.7V for a silicon diode), then the diode is "turned on" and the circuit behaves very similar to the following:
    Edit_2014-12-25_1_fwd.png
    And when Vo would "like" to be less than Vdc+Vd, then the diode will be "turned off" and the circuit behaves very similar to the following:
    Edit_2014-12-25_1_rev.png

    Does that help?
     
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