The load voltage out of a rectifier is pulsating rather than steady. For instance, look at Fig. 4-11. Over one complete output cycle, the load voltage increases from zero to a peak, then decreases back to zero. This is not the kind of dc voltage needed for most electronic circuits. What is needed is a steady or constant voltage similar to what a battery produce. To get this type of rectified load voltage, we need to use a “filter.”
Half-wave Filtering
The most common type of filter is the capacitor-input filter shown in Fig. 4-12. To simplify the initial discussion of filters, we have
represented an ideal diode by a switch. As you can see, a capacitor has been inserted parallel with the load resistor. Before the power is turned on, the capacitor is uncharged; therefore, the load voltage is zero. During the first quarter cycle of the secondary voltage, the diode is forward-biased. Ideally, it looks like a closed switch. Since the diode connects the secondary winding directly across the capacitor, the capacitor charges to the peak voltage, Vp.
Figure 4–12 Capacitor-input Filter
Just past the positive peak, the diode stops conducting, which means the switch opens. Why? Because the capacitor has Vp. volts
across it. Since the secondary voltage is slightly less than Vp, the
diode goes into reverse bias. With the diode now open, the
capacitor discharges through the load resistance. But here is the key idea behind the capacitor-input filter: by deliberate design, the discharging time constant (the product of RL and C) is much greater
than the period, T, of the input signal. Because of this, the capacitor will lose only a small part of its charge during the off time of the diode as shown in Fig. 4-13a.
Figure 4–13
When the source voltage again reaches its peak, the diode conducts briefly and recharges the capacitor to the peak voltage. In other words, after the capacitor is initially charged during the first quarter cycle, its voltage is approximately equal to the peak
secondary voltage. This is why the circuit is sometimes called a peak detector.
The load voltage is now almost a steady or constant dc voltage. The only deviation from a pure dc voltage is the small ripple caused by charging and discharging the capacitor. The smaller the ripple is, the better. One way to reduce this ripple is by increasing the discharging time constant, which equals RLC.
Full-Wave Filtering
Another way to reduce the ripple is to use a full-wave rectifier or bridge rectifier; then the ripple frequency is 120 Hz instead of 60 Hz. In this case, the capacitor is charged twice as often and has only half the discharge time (see Fig. 4-13b). As a result, the ripple is smaller and the dc output voltage more closely approaches the peak voltage. From now on, our discussion will emphasize the bridge rectifier driving a capacitor-input filter because this is the most commonly used circuit.
Brief Conduction of Diode
In the unfiltered rectifiers discussed earlier, each diode conducts for half a cycle. In the filtered rectifiers we are now discussing, each diode conducts for much less than half a cycle. When the power switch is first turned on, the capacitor is uncharged. Ideally, it takes only a quarter of a cycle to charge the capacitor to the peak
secondary voltage. After this initial charging, the diodes turn on only briefly near the peak and are off during the rest of the cycle. In terms of degrees, the diodes turn on for only a couple of degrees during each cycle (half a cycle is 1 8 0 ) .
An Important Formula
Whether you are troubleshooting, analyzing, or designing, you have got to know how to estimate the size of the ripple. Normally, the ripple is small compared to the peak secondary voltage. For most applications, the ripple is considered small when it is less than 10 percent of the load voltage. For instance, if the load voltage is I5 V, the ripple in most filtered rectifiers will be less than 1.5 V peak- to-peak.
Here is the formula for ripple expressed in terms of easily measured circuit values:
V I
f
R
C (4-8)
where VR = peak-to-peak ripple voltage
I = dc load current
f = ripple frequency
C = capacitance
The proof of Eq. (4-8) is too lengthy and complicated to show in this book. But the derivation assumes that the peak-to-peak ripple is less than 20 percent of the load voltage. Beyond this point, you cannot use Eq. (4-8) without encountering a lot of error. But as was already discussed earlier, the whole point of the capacitor-input filter is to produce a steady or constant dc voltage. For this reason, most designers deliberately select circuit values to keep the ripple less than 10 percent of the load voltage. In the circuits you
encounter, you will find that the ripple is usually less than 10 percent of the load voltage.
DC Voltage
To be successful in electronics, you have to learn the following basic idea: approximations are the rule, not the exception. Why? Because electronics is not an exact science like pure mathematics. The idea that you must always get exact answers is a false idea, a left-brain trap. For most of the work in electronics, approximate answers are adequate and even desirable.
The situation is like an artist painting a picture. The best artist starts with the largest brush when beginning a painting. The artist then switches to a medium-sized brush to improve the picture, and, finally, may use the smallest brush to get the finest detail. No good artist ever uses a small brush all of the time.
The three diode approximations are like an artist’s brushes. You should start with the ideal diode to get the big picture. In many cases (trouble shooting, for instance), this will be all you need. Often, you will want to improve your analysis by using the second approximation (a lot of everyday work is done with this one).
Finally, the third approximation may be best in some situations (if the circuit uses 1 percent resistors, for example).
First Approximation
With the foregoing in mind, here is how the diode approximations affect the value of the load voltage. For an ideal diode and no ripple, the dc load voltage out of a filtered bridge rectifier equals the peak secondary voltage:
Vd c Vp2
This is what you want to remember when you are trouble-shooting or making a preliminary analysis of a filtered bridge rectifier.
Second Approximation
With the second approximation of a diode, we have to allow for the 0.7 V across each diode. Since there are two conducting diodes in series with the load resistor, the dc load voltage with no ripple out of a filtered bridge rectifier is
Vd c Vp2 1 .4 V
Third Approximation
In the third approximation, two bulk resistances are in the charging path of the capacitor. This complicates the analysis because the diode conducts briefly only near the peak. Fortunately, bulk
resistances of rectifier diodes are typically less than 1 . Because of
this, they usually have little or no effect on the load voltage. Unless you are designing a filtered bridge rectifier, you will not need to consider the effect of bulk resistance. (If you are designing the circuit, you will need to use advanced mathematics because you have to deal with an exponential function. The alternative is to build the circuit and arrive at circuit values by experiment. The main rule here is to keep the load resistance as large as possible compared to the bulk resistance.)
There is one more improvement that we can use. We can include the effect of the ripple as follows:
Vd c(w ith r ip p le) Vd c w ith o u tr ip p le( ) VR
2
The idea here is to subtract half the peak-to-peak ripple to refine the answer slightly. Since peak-to-peak is usually less than 10 percent, the improvement in the answer is less than 5 percent.
A Basic Guideline
The resistors used in typical electronic circuits have tolerances of 5 percent. Sometimes, you will see precision resistors of 1 percent used in critical applications. And sometimes, you will see resistors of 10 percent used. But if we take 5 percent as the usual tolerance, then one guideline for selecting an approxima-tion is this: Ignore a quantity if it produces an error of less than 5 percent. This means we can use the ideal diode if it produces less than 5 percent error. If the ideal diode results in 5 percent or more error, switch to the second approximation. Also, ignore the effect of ripple when it is less than 10 percent of the load voltage. (Remember: the peak-to-peak ripple is divided by two before subtracting from the load voltage. Therefore, a 10 percent ripple produces only a 4 percent error in load voltage.)
The foregoing guideline will be of some help in deciding which approximation to use, but don't lean on this guideline too heavily. You may have a situation where a 5 percent guideline is not suitable, Remember the artist's brushes. The job may require a smaller or larger brush. It is impossible to give you a rule for every situation because real life is too messy and has too many
exceptions. But don't be discouraged. That's what makes electronics more interesting than accounting. Use the basic guideline given here, but be ready to abandon it if you feel it doesn't apply to your situation.
Example 4–7
Suppose a bridge rectifier has a dc load current of 10 mA and a filter capacitance of 470 F. What is the peak-to-peak ripple out of a capacitor-input filer?
Solution
Use Eq. ( 4-8) to get
VR 1 0 m A
1 2 0 H z ) ( 4 7 0 F
0 .1 1 7 V
( )
This assumes the input frequency is 60 Hz,. which is the normal line frequency in the United States.
Example 4–8
Assume we have a filtered bridge rectifier with a line voltage of 120 V rms, a turns ratio of 9.45, a filter capacitance of 470 F, and a load resistance of 1 k . What is the dc load voltage?
Solution
Start by calculating the rms secondary voltage:
V2 1 2 0 V
9 .4 5
1 2 .7 V
This is what you would measure with an ac voltmeter connected across the secondary winding.
Next, calculate the peak secondary voltage:
Vp2 1 2 .7 V
0 .7 0 7
1 8 V
With an ideal diode and ignoring the ripple, the dc load voltage equals the peak secondary voltage:
Vd c 1 8 V
This answer would be adequate if you were troubleshooting a circuit like this. The dc load voltage is the approximate value you would read with a dc voltmeter across the load resistor. If there were trouble in such a circuit, the dc voltage probably would be much lower than 18 V.
The second approximation improves the answer by including the effect of the two-diode voltage drops:
Vd c 1 8 V 1 .4 V 1 6 .6 V
This is more accurate, so let us use it in the remaining calculations. To calculate the ripple, we need the value of dc load current:
I k
1 6 .6 V
1 6 .6 m A
1
V F R 1 6 .6 m A 1 2 0 H z ) ( 4 7 0 0 .2 9 4 V ( )
This is the peak-to-peak ripple and is what you would see if you looked at the load voltage with the ac input of an oscilloscope. This ripple has little effect on the dc load voltage:
V ( w i t h r i p p l e ) = 1 6 .6 - 0 .2 9 4 V 2
1 6 .5 V d c
This gives you the basic idea of how to calculate the dc load voltage and ripple.