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Gre4nikov [31]
3 years ago
12

Explain the necessity of smoothing the output voltage before applying it to a transistor amplifier

Physics
1 answer:
Sveta_85 [38]3 years ago
4 0

Answer:

  smoothing avoids "hum" in the output of the amplifier

Explanation:

We presume the question is why half-wave or full-wave rectified AC needs to be smoothed before using it to supply power to a transistor amplifier.

Most transistor amplifiers have output voltage that is a function of both the signal input voltage and the power supply voltage. Half-wave or full-wave rectified power goes from full value to zero once or twice in each cycle of the AC power from which it is derived. That modulation of the power voltage can show up as a modulation of the output signal from the amplifier. Such modulation is usually undesirable.

Some amplifier designs have a significant factor of "power supply rejection", meaning large changes in power supply voltage only produce small changes in amplifier output voltage. However, when the supply voltage goes to zero, the amplifier output necessarily goes to zero also. That is, power supply rejection is never perfect and is rarely sufficient to tolerate full-wave or half-wave rectified power without producing some modulation of the amplifier's output.

It is necessary to smooth the power supply voltage in order to reduce or eliminate amplifier output "hum."

__

<em>Caveat</em>

If the power supply frequency and its harmonics are well-separated from the amplifier's signal frequency band, modulation on the power supply may be of little or no consequence. Often power supplies operated at an audible frequency, so power for audio amplifiers needs to be quite "clean." There are other power and signal frequency bands for which the results will vary.

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\sum d_{x} =  - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\  { \underline{d _{x} =  -  5.702 \: m}} \\  \\  \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0  \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y}  =  - 11.476 \: m}}

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When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the
KonstantinChe [14]

Answer:

The ratio is  KE : TM  =  0.75

Explanation:

from the question we are told that

  The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical  energy of the mass is mathematically represented as

        TM  =  \frac{1}{2}  *  k  *  A^2

Here  k is the spring constant  ,  A is the total displacement of the  the mass  from maximum  compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

        TM  =  KE  + PE

=>     KE = TM  - PE

Here the potential  energy of the mass is mathematically represented as

     PE   = \frac{1}{ 2}  *  k *  [ x ]^2

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is  

      x = \frac{A}{2}

So

     PE   = \frac{1}{ 2}  *  k *  [ \frac{A}{2}  ]^2

So

      KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{2}  *  k  *  [\frac{A}{2} ]^2

=>    KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{8}  *  k  *  A ^2

=>    KE =  0.375  *  k  *  A^2

So the ratio of  KE :  TM is  mathematically represented as

       \frac{KE}{TM} =  \frac{0.375  k A^2 }{0.5 k A^2}

=>    \frac{KE}{TM} = 0.75

3 0
3 years ago
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