LABORATORY 10
TIME AVERAGES, RMS VALUES AND THE BRIDGE RECTIFIER
Bridge Rectifier
Full-wave Rectification:
For many electronic circuits, DC supply voltages are required but only AC voltages are
available. Then, the required DC voltages are obtained from the AC voltages by rectification
and filtering. In the rectification process, an AC current is converted to a time-varying current
that flows in a single direction, and so it is a time-varying DC current. Subsequent filtering
“smoothes out” the variations to produce an almost constant dc current and voltage.
The diode bridge rectifier circuit is shown in Fig. 10.1. The operation can be understood
by considering the polarity of the applied voltage Vi at the top node a, with respect to the bottom
node b. When this voltage is positive, diodes D1 and D2 are forward biased and therefore
conduct, thereby causing the output voltage Vo to be equal to the input voltage Vi. (For
simplicity, the small voltage drops across the conducting diodes are neglected.) When, however,
Vi is negative, node b is positive with respect to node a and thus diodes D3 and D4 conduct,
making Vo = -Vi. Since, however, Vi is negative, Vo is positive. Thus, Vo is always positive and
is equal to the magnitude of Vi. That is, Vo = | Vi | .
Fig. 10.1 Diode Bridge Full-Wave Rectifier
Suppose that the input voltage is Vi = 10 sin ωt V. Then, if VT = 0, the input and output
voltage waveforms will be as shown in Fig. 10.2 by the dotted curve. Actually, though, for a real
bridge rectifier, the output waveform will be shifted down by 2VT because of the diode voltage
drops. This is shown by the solid curve in the bottom figure 10.2. Also, because of this shift, the
output waveform will be zero for a short time between the rectified half cycles. This type of
rectification is called full-wave rectification because both half cycles of a cycle of the input
contribute to the output.
Fig. 10.2
Now, suppose that in addition to a sinusoid, the input voltage contains a DC component,
as can be obtained from the sources shown in Fig. 10.3. What then is the output voltage Vo?
Fig. 10.3
Again, the key to understanding the circuit operation is the polarity of the voltage drop
from node a to b, which is
Vab = 10 sin ωt + 3 V.
When Vab is positive, diodes D1 and D2 conduct. When this voltage is negative, only diodes D3
and D4 conduct. Thus, the input and output voltage waveforms are as shown in Fig. 10.4.
Fig. 10.4
Finally, suppose in the circuit of Fig. 10.1, that a capacitor is placed across the load
resistor. What would the Vo waveform be then? The answer is that when Vi increases for the
first time to its positive peak, the capacitor would charge to this peak value minus 2VT. Then,
when Vi started to decrease, the capacitor could not discharge through the diodes because that
would require reverse diode currents. The capacitor would, however, start discharging through
the load resistor RL. Then, if the RLC time constant was much greater than the duration of a half
cycle of the sinusoidal input voltage, the capacitor voltage would not decrease as fast as the
sinusoidal voltage, and thus would reverse bias the diodes. The diodes would remain reverse
biased until the input sinusoidal voltage (minus 2VT) exceeded the capacitor voltage. In the
meantime, though, the load voltage would be the slow exponentially decaying capacitor voltage
as is shown in Fig. 10.5. Clearly, the resulting load voltage is smoother than a full-wave
rectified sinusoid.
Fig. 10.5
Procedure:
1.
Construct the circuit shown in Fig. 10.6. Be careful to connect the diodes so that
they conduct in the directions indicated. Diode casings usually have a stripe at
one end to designate the end corresponding to the bar in the diode circuit symbol
(the cathode end). The 2.2-kΩ resistor is included to limit the current to a safe
value. Apply a 5-V peak, 60-Hz sine wave and, using an oscilloscope, observe
the voltage across the a-b terminals. Examine the waveform, including amplitude
and time values.
Fig. 10.6
2.
Place a 0.1-μF capacitor across (in parallel with) RL, and reexamine the voltage waveform across the a,b nodes.
3.
Add a 1.5 V offset voltage to the function generator. This has the
same effect as the configuration in Fig. 10.7. Using the
oscilloscope, again observe the voltage waveform across the a,b
nodes.
4.
Repeat step 2.
Fig. 10.7
Time Averages and RMS Values
In electrical circuits, one frequently encounters periodically varying current and voltage
waveforms. If the waveform x(t) repeats itself every T seconds, then
x(t) = x(t + NT)
where N is an integer and T is called the period. Figures 10.8 (a)-(d) show several periodic
waveforms.
Figure 10.8
(a) sinusoid, (b) triangular, (c) square, (d) half-wave rectified waveforms
The average value of a periodic waveform is defined mathematically as
Xave = < x(t) > = {∫0T x(t) dt}/T.
The integration represents the area under the curve during one period. Thus, it can be seen that
the average of a sinusoid is zero.
Prelab Question:
1.
Determine the time averages of waveforms in Figures 10.8(b), (c), (d).
The Root-Mean Square of a Periodic Waveform:
We saw earlier that the time average of one of the most important periodic waveforms, the
sinusoid, is zero. However, the average itself is of little importance. Since power absorbed or
provided by any element in a circuit depends on the square of the current or voltage, average
power can be computed by knowing the time average (mean) of the square of the periodic
waveform.
< x2(t) > = ∫0T x2(t) dt/T.
The root-mean square (RMS) value is simply the square root of the mean of the square.
Xrms = √ < x2(t) >
Note: AC voltmeters and ammeters measure the RMS value, not the peak value. The RMS value
is sometimes also referred to as the effective value.
Prelab Questions:
2.
Show that the RMS value of the sinusoid x(t) = A cos(ωt + θ) is A/√2.
Hint: Use trigonometric identity: 2(cosα)2 = 1 + cos2α
in evaluating the integral.
3.
Determine the RMS value of the waveform shown in Figure 10.8(d).
4.
What would be the RMS value of a full-wave rectified waveform?
Give your answer in terms of the amplitude of the input sinusoid.
Ripple Filtering:
The full-wave rectified waveform shown in the bottom curve of Figure 10.2 presents two
problems:
1. The time-average value (the DC value) of the output is much lower than the amplitude of the
input sinusoid.
2. The output oscillates around the DC value, an undesirable feature for most applications
(Imagine it running the DC motor in a portable shaver).
The full-wave rectified waveform can be considered as made up of two parts: the DC part and
the oscillating part. The oscillating part is called the ripple.
v0(t) = VDC + vripple(t).
where VDC = < vo(t) >
The ripple can be reduced (filtered) by using an appropriate capacitor connected across the load
resistor RL (refer to Figure 10.6 for the load resistor). This is explained in the text preceding
Figure 10.5. The filtered waveform is shown in Figure 10.5, along with the unfiltered waveform.
The filtered waveform is much smoother.
DC Value of the Filtered Waveform and Ripple Factor:
At t = T/4, the capacitor is fully charged (to the peak voltage of the input sinusoid, Vp).
Then, it discharges via the load resistor with a time constant τ = RLC. Since the capacitor is
chosen to ensure that τ >> T, the exponential decay of the capacitor voltage continues almost
until t = 3T/4. Therefore, in the interval T/4 ≤ t ≤3T/4, the waveform is given by
v0(t) = Vp exp{− (t − T/4)/τ}.
The minimum occurs at t = 3T/4 and is given by
Vmin ≅ Vp exp( − T/2τ) ≅ Vp (1 − T/2τ)
The ripple factor (rf) and the DC voltage are given by the following equations:
rf = (Vp − Vmin)/VDC
VDC = (Vp + Vmin)/2 = Vp(1 − T/4τ) = Vp(1 − 1/4fRLC)
Prelab Question:
5.
Determine the ripple factor for f = 60 Hz, RL = 1 MΩ, and C = 0.1 μF.
Procedure:
For the circuit in Fig 10.6 with a capacitor added in parallel to the load resistor:
1.
Measure the peak value and the minimum value in the rectified output.
2.
Measure the average voltage (VDC) of the rectified output using the Oscilloscope.
3.
Determine the ripple factor.
4.
Compare the DC voltage and ripple factor with the respective theoretical values.