The output of the rectifier circuits discussed in the last lecture is pulsating significantly with time. Hence, it’s not useful as the output from a DC power supply.
One way to reduce this ripple is to use a filtering capacitor. Consider the half-cycle rectifier again, but now add a capacitor in parallel with the load:
We expect that as soon as we turn on the source, the capacitor will charge up on “+” cycles of v_{I} and discharge on the “-” cycles.
To smooth out the voltage, we need this discharge to occur slowly in time. This means we need to choose C large enough to make this happen, presuming that R is a given quantity (the Thévenin resistance of the rest of the circuit). The output voltage vO will then be a smoothed-out signal that pulsates with time:
Notice the diode current and the capacitor voltage. They display behavior much different than what one would find in an AC circuit.
Analysis of Peak Rectifier Circuits
We’ll require that RC T ô= , which means that the time constant of the RC circuit must be much greater that the period
of the input sinusoidal signal:
Now, our quest is to approximately determine the ripple voltage
v_{r}assuming t>>T
When D is off, and assuming it is an ideal diode
[If D is not ideal then v_{0}(t)~(v_{0}-0.7)e-t/r.]
At the end of the discharge time, td, the output voltage equals
Substituting for vO from (1) at this time td leads to
This equation has the two unknowns Vr and td, assuming ô isknown. If we can determine td, then we can find Vr. Finding tdcan be done numerically by equating (1) to the expression forthe input voltage
and solving for the time td when the two are equal as
This needs to be done numerically since (5) is a “transcendentalequation.”Alternatively, if Ät is small compared to T (true when T ô , asassumed), then from (3)
Again, because T ô then we can truncate the series expansionof the exponential function to two terms (see Lecture 4) giving
This simple equation gives the ratio of the ripple voltage to the peak voltage of the input sinusoidal signal for the half-cyclerectifier. It’s worth memorizing, or knowing how to derive.Often R and T are fixed quantities. So from (7)
to obtain a small ripple voltage we need a large C in this case.
Conduction Interval
Lastly, the conduction interval Ät is defined as the time intervalin which the diode is actually conducting current. This timeperiod is sketched in the preceding two figures. The diode conducts current beginning at time td and ending at T,within each period. Using equation (4) at time td
We expect the conduction interval to be small. So truncating theseries expansion of cosine to two terms, (9) gives
The factor t ùÄ is sometimes called the conduction angle, è. For V_{r} V_{p} this conduction angle (and conduction interval) will besmall, as expected.
Discussion
To reiterate, the objective of the peak rectifier is to charge theshunt C when D is on, and slowly discharge it during thosetimes when D is off.When does D conduct? During the Ät periods in the previousfigure. Also seeNote that this peak rectifier is not a linear circuit. iD is a very complicated waveform and not a sinusoid, as seen earlier in
There are no simple exact formulas for the solution tothis problem. The text only shows approximate solutions for peak iD:
Example N8.1 (similar to text example 3.9). A half-cycle peakrectifier with R = 10 kÙ is fed by a 60-Hz sinusoidal voltagewith a peak amplitude of 100 V.(a) Determine C for a ripple voltage of 2 Vpp. From (8):
or c=83.3uf
For a “factor of safety” of two, make C twice as large. Remember, a bigger C translates to smaller ripple. (b) Determine the peak diode current. Using (11):
When specifying a diode for your circuit design, you would need to find one that could safely handle this amount of current.Example N8.2. A half-cycle peak rectifier with R = 10 kÙ is fedby a 60-Hz triangula voltage with a peak amplitude of 100 V.
(a) Determine C for a ripple voltage of 2 Vpp. If you go back and look at the derivation of (8) you’ll find that there were no approximations made that required a sinusoidal waveform. Consequently, (8) applies to this triangular waveform as well, provided T ô . Hence, as before 83.3 C = ìF. (b) Determine the diode conduction time, Ät. Referring to this sketch of the region near the positive peak voltage for vI:
Because the rising portion of the waveform is a straight line:
To find Ät, equate
Therefore, for a triangular waveform
In this particular case,
Compare this time to a sinusoidal waveform:
This time is much longer than for the triangular waveform. Consequently, we would expect max D i for D to be much larger for the triangular waveform than for the sinusoid!
Full-Cycle Peak Rectifiers
In a similar fashion, we can also add a shunt C to full cycle and bridge rectifiers to convert them to peak rectifiers. For example, for a full-cycle peak rectifier:
The output voltage has less ripple than from a half-cycle peakrectifier (actually one half less ripple).
The “ripple frequency” is twice that of a half-cycle peak rectifier. Using the same derivation procedure as before with thehalf cycle, but with 2 T T . gives from (7)
Lastly, it can be shown that the i_{d}|max for the full-cycle peakrectifier:
is approximately one-half that of the half-cycle peak rectifierwhen V_{r} V _{p} .