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Facebook Clipping circuits (also known as limiters, amplitude selectors, or slicers), are used to remove the part of a signal that is above or below some defined reference level. We’ve already seen an example of a clipper in the half-wave rectifier – that circuit basically cut off everything at the reference level of
zero and let only the positive-going (or negative-going) portion of the input waveform through.
To clip to a reference level other than zero, a dc source (shown as a battery in your text) is put in series with the diode. Depending on the direction of the diode and the polarity of the battery, the circuit will either clip the input waveform above or below the reference level (the battery voltage for an
ideal diode; i.e., for Von=0). This process is illustrated in the four parts of
Figure 3.43:
Without the battery, the output of the circuit below would be the negative portion of the input wave (assuming the bottom node is grounded). When vi
> 0, the diode is on (short-circuited), vi is dropped across R and vo=0. When vi <0, the diode is off (open-circuited), the voltage across R is zero and vo=vi. (Don’t worry; we won’t be doing this for all the circuits!) Anyway, the reference level would be zero.
Figure has the battery with the same orientation as in part (a), but the diode has been flipped. Without the battery, the positive portion of the input waveform would be passed (i.e., a reference level of zero).With the battery, the diode conducts for vi < VB. This means that the reference level is shifted to +VB and only
vi > VB appears at the output.
zero and let only the positive-going (or negative-going) portion of the input waveform through.
To clip to a reference level other than zero, a dc source (shown as a battery in your text) is put in series with the diode. Depending on the direction of the diode and the polarity of the battery, the circuit will either clip the input waveform above or below the reference level (the battery voltage for an
ideal diode; i.e., for Von=0). This process is illustrated in the four parts of
Figure 3.43:
Without the battery, the output of the circuit below would be the negative portion of the input wave (assuming the bottom node is grounded). When vi
> 0, the diode is on (short-circuited), vi is dropped across R and vo=0. When vi <0, the diode is off (open-circuited), the voltage across R is zero and vo=vi. (Don’t worry; we won’t be doing this for all the circuits!) Anyway, the reference level would be zero.
With the battery in the orientation shown in Figure (and below), the diode doesn’t turn on until vi
> VB (If this looks strange, revisit the definition of forward bias). This shifts the reference level up and clips the input at +VB and passes everything for vi < VB.
vi > VB appears at the output.
Again referencing part (a), the diode is in the original position but the polarities on the battery have been switched in Figure . The discussion follows the same logic as earlier, but now the reference level has been shifted to –VB. The final result is that vo = vi for vi < -VB.
Finally, Figure 3.43d behaves the same as part (b), but the polarity on the battery has been switched, shifting the reference level to -VB. The signal that appears as the output is vi as long as vi > - VB.
To accommodate a practical diode, the turnon voltage (VON=0.7V for silicon) and forward resistance (Rf
) are included, along with the ideal diode, in the model (as shown in Figure, reproduced to the right).
The effective reference level will either have a magnitude of VB+VON or VB-VON, depending on the relative polarities of the two sources (review combining voltage sources in series if necessary). Including Rf
in the diode path creates a voltage divider when the diode is forward biased. The result of this slight drop
across Rf (remember that the forward resistance is generally pretty small), is a slight distortion in the output waveform – it is no longer strictly “limited” or “clipped” to the reference level, as is illustrated in Figure 3.44b in your text. The four possible configurations of Figure 3.43 are still valid, with the effective reference level ranging in magnitude from 0.7V (if VB=0) on up
Parallel-biased clipper
A parallel-biased clipper is a circuit that clips the positive and negativegoing portions of the input signal simultaneously. This is designed by using two parallel diodes oriented in opposite directions – note that it is very important that the diodes are oppositely oriented (think voltage sources in parallel – a big no-no!). Just as in our previous discussion, the path containing diode D1 will provide the upper limit with reference level V
B1+VON (with the VB1 polarity shown) and the path containing D2 will provide the lower limit with reference level VB2+VON (with the VB2 polarity shown). An example of this type of clipper, with the resulting output waveform is shown below (Figure of your text, where it looks like they assumed Rf
was negligible):
series-biased clipper
A series-biased clipper involves placing a battery in series with the input.The result of this modification is that the input signal is no longer symmetric about the zero axis, but instead shifts by an amount defined by the
magnitude and polarity of VB. In Figure, the four possible permutations and the resulting output waveforms are shown for an ideal diode. When you look at this figure, keep in mind that the input signal is swinging between +2 and –2 volts and the battery magnitude VB is 1 volt. This may avoid some confusion when looking at the output waveforms – the (2V+VB) is just 3 volts and the –(2V-VB) may be replaced by –1 volt. The series placement of the battery is not changing the input waveform in any way, it is simply affecting when the diode turns on. To include the effects of a practical diode, include VON and RF in the diode path and crunch the math..
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