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Bypass capacitor

  1. Feb 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Hello! I would like to check on my knowledge of bypass capacitor which bodders me a bit

    2. Relevant equations
    Is that true?

    3. The attempt at a solution
    Bypass capacitors are use to filter AC ripples. It does it in the following way: Lets imagine an RC amplifier. When the friquency is high, Xc is low and it acts like low resistance path. So the Ac signal gets shortened to ground and instead of riding on the DC through the resistor RE, it bypasses it and so the voltage at the emitter remains unchanged. But witjout the capacitor, the AC would pass through RE with the DC signal and couse VE to rise and so eventualy it causes some distorsions.
     
  2. jcsd
  3. Feb 15, 2016 #2

    BvU

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    Yes, that's a way of putting it.
     
  4. Feb 15, 2016 #3

    Hesch

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    I don't know what is meant by "RC", "RE", "VE", but have a look at this circuit:

    upload_2016-2-15_12-52-48.png

    To the right, the circuit could be connected to a load like a bulb.
    To the left, the circuit is connected to an AC-outlet.
    The AC voltage is rectified, creating a DC voltage with a lot of ripple.
    To remove (some of) the ripple, a smoothing capacitor is connected between the DC voltage and "ground".

    I wouldn't compare the capacitor to a "low resistance path", but to a battery that is charged/discharged, so that whenever the rectified AC voltage is higher than the voltage across the capacitor, the capacitor/battery is charged and vica versa.

    Now, the current discharging the battery is limited to the current consumed by the bulb. The current has nowhere else to go, because the rectifier is a one-way path. Hence the steepnes of the rectified AC voltage going toward 0V is limited, so before the voltage reaches 0V, a new rectified AC pulse will occur and the battery will be recharged. In this way the ripple will be decreased because the peak to peak voltage is decreased.

    So instead of regarding the capacitor as an AC short circuit, I would regard it as a back up battery, supplying power when needed. You can calculate the needed capacity ( in Ah ) to provide power to the bulb within, say 0.01 s.
     
  5. Feb 15, 2016 #4

    gneill

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    Based upon the OP's referring to RE and the term "bypass capacitor", the context would appear to be that of a capacitor paralleling the emitter resistance in a common emitter transistor amplifier circuit. The concept of treating it as a low(er) impedance path for AC signals is a good one. The parallel combination of R and C form a high pass filter for the emitter current.

    In the common emitter layout the voltage at the emitter has two effects. The first is the role of the DC component in the bias network for the transistor, establishing the operating point of the transistor (no input signal condition). The second is that the AC voltage developed across RE acts as a negative feedback for the circuit, reducing the gain for AC input signals by reducing the forward base-emitter bias as the potential at the emitter follows the input signal at the base. The result of bypassing RE is to reduce the size of the AC signal developed at the emitter, thus reducing the negative feedback and increasing the gain of the circuit for higher AC frequencies.

    Whether or not distortion develops depends upon the operating point of the transistor and nature of the input signal. At a single frequency the gain is fixed with or without bypass, and distortion depends on the linearity of the region over which the transistor is driven by the input. If the input contains a range of frequencies (such as an audio signal, for example) then the bypass will increase the range of operation in a frequency dependent way, so any non-linearities will affect different frequencies to different extents. Having frequency-dependent gain is a form of distortion, too. This is sometimes intended -- consider tone controls in an audio system.

    If the choice of bypass capacitor size is such that Xc << RE for all frequencies of interest in the input signal then the gain of the circuit will be maximized across the whole range. But this choice comes with certain design issues, one of which is that gain of the circuit will be tied to the inherent gain of the transistor (β) which varies from component to component even in the same manufacturing run. This is not a good thing if the aim is a circuit which is to be mass produced and all all expected to behave identically. Reducing the gain by the use of feedback can add stability and reproducability to a design.
     
  6. Feb 15, 2016 #5

    Hesch

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    My misunderstanding is due to
    and
     
  7. Feb 15, 2016 #6

    gneill

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    Hi Hesch. Understood. Different people have different ways of looking at things and using analogies to understand the effects. Whatever works for you is fine so long as you can do what math is required to analyze the circuit when necessary.
     
  8. Feb 15, 2016 #7
    Thanks all for youre answers!
     
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