1 Equipment familiarization: multimeter, breadboard, and oscilloscope
3.6 Exploring rectification
Next we take up the basic principles of rectification. Almost all electronic equipment requires power from a steady voltage source, i.e., a DC power supply. For portable equipment, the power is supplied by batteries. How-ever, the most convenient power source is the 120 V 60 Hz AC line.1(120 V is in fact the r.m.s. value of the sinusoidal voltage, the amplitude being V0=√
2× 120 V = 170 V, (3.10)
and the peak-to-peak voltage is, of course, twice this, or Vp−p= 340 V, as you can easily verify from the definition of the root-mean-square by integrating over the sine wave.)
Within most electronic equipment using the AC line, there is a power transformer that steps down the 120 V AC to a more convenient voltage, a rectifier that converts the alternating voltage from the transformer to a DC voltage, and a regulator that maintains the output voltage at the desired level.
Caution: In usinga power transformer, bear in mind that an especially large transient current sometimes flows when the line cord is first plugged in.
You will probably blow fewer fuses if you leave the power transformer plugged in at all times. Attach banana-plug leads to the transformer’s sec-ondary only after you are sure your circuit will not damage any of the equipment. Do not permit powered lines to dangle loosely; when reconfig-uring your circuit, it is safest to disconnect the leads at the transformer, not at the breadboard.
Set up the circuit shown in Fig. 3.6(a) using a 10 k resistor as the load, RL. Observe the sinusoidal voltage waveform across RL. Measure the amplitude V0and the r.m.s. voltage. Check the relation
√2Vrms= V0. (3.11)
You will probably find Vrms> 25 V. Since the windings of the transformer have some ohmic resistance, the transformer’s output voltage depends on
1 In North America, the supply voltage from a standard wall socket is 120 V, and the supply frequency is 60 Hz; the discussion is equally valid for other values, which may be substituted according to your local supply voltage and frequency.
41 3 Diodes
Center
Tap RL Vout
Transformer
Vrms
Vp--p V0
V0 fuse
120
(a) VAC
(b)
Fig. 3.6. (a) Power transformer supplies Vout≈ 25 V r.m.s.; (b) waveform produced by the circuit in (a).
the current drawn, and its 25.2 V r.m.s. nominal output voltage is for sub-stantially higher current than is drawn by the 10 k load.
Add a 1N4001 diode to give the half-wave rectifier of Fig. 3.7(a) with RL = 10 k.
Observe and record the voltage waveform. Measure the amplitude V0
using the oscilloscope (due to the rectification it is now equal to the peak-to-peak voltage).
Compare the amplitude of the half-rectified waveform with the amplitude of the unrectified waveform measured above. By how much has the amplitude decreased? Is this the amount you expect? Explain.
Measure the average voltage Vav across the load with a DC voltmeter.
Check that for a half-wave rectifier Vav= V0
π. (3.12)
Add a filter capacitor in parallel with the load as shown in Fig. 3.8(a).
Caution: The 100 F electrolytic capacitor is polarized -- be careful not to hook it up backwards! The negative terminal should be labeled with a ‘−’ sign.
Vout RL
fuse 120
VAC
Transformer
Vav
Vp--p V0 Rectifier Diode
(a)
(b)
Fig. 3.7. (a) Power transformer with half-wave rectification; (b) waveform produced by circuit shown in (a).
Vout RL
Transformer
Rectifier Diode
+ 100 Fµ
1 60s fuse
120
(a) VAC
(b)
Ripple Voltage
1 60s
(c)
Ripple Voltage
Fig. 3.8. (a) Half-wave rectifier with filter capacitor; (b) waveform produced by circuit shown in (a); (c) simple approximation to waveform produced by circuit shown in (a).
43 3 Diodes
What is the voltage rating of your capacitor? Make sure it is sufficient for the voltage that will be applied!
Observe and record the output-voltage waveform across the load resis-tance RLand measure the peak-to-peak ‘ripple voltage’ (i.e., the amount by which the output voltage is varying; see Fig. 3.8(b)).
Here is that rare situation – measuring accurately a small AC signal on top of a large DC offset – in which you should use the AC-coupling feature of the scope’s vertical menu. If you are troubled with noise, you may want to trigger the scope on ‘line’ and employ signal averaging.
A simplified analysis approach for predicting the expected output wave-form, illustrated in Fig. 3.8(c), is to assume that the capacitor charges up to the peak voltage instantaneously and discharges at a uniform rate dQ/dt, equal to the average load current. The average current through the load can be determined using the known resistance RLand the average DC voltage across it as measured with a voltmeter. Using these assumptions, and the fundamental capacitor equation Q = CV ,
Calculate the output-voltage droop in each cycle, and compare with the peak-to-peak ripple voltage as measured. Is the observed percentage discrepancy within the tolerances of your components? Explain.
Replace the 100F electrolytic capacitor with a 1000 F capacitor. Do you expect the ripple voltage to increase or decrease? Explain.
Measure the ripple voltage and compare with your expectations.
Full-wave rectification should decrease the ripple by about a factor of two. This can be accomplished using two diodes and the transformer’s center tap or by using a diode bridge. Diode-bridge rectifiers are available as a single encapsulated unit, making the bridge-rectifier approach partic-ularly convenient. These bridges have four diodes within. The terminals are labeled: ‘∼’ marks the two terminals that should be connected to the transformer secondary, and ‘+’ and ‘−’ denote the positive and negative outputs (see Fig. 3.9).
~
~ + _ Diode Bridge
Fig. 3.9. An example of how to insert a diode bridge into a breadboard.
Vout RL
Transformer
Diode Bridge
+
_
~
~ fuse
120 VAC
Fig. 3.10. Full-wave rectification using a diode bridge.
Caution: A defective bridge element can blow the power transformer fuse -- check it be-fore placingit in service. It should show essentially infinite resistance between the terminals marked ‘∼’. When using an ohmmeter to check the resistance, remember to measure it for both orientations of the terminals -- since you are dealingwith diodes, the resistance could be different in each direction.
Set up the bridge rectifier circuit of Fig. 3.10 with RL= 10 k. (Insert the rectifier package straddling the central groove of a breadboard socket unit, with the long dimension of the package running along the groove, as shown in Fig. 3.9.) Four rectifier diodes can be used if a bridge rectifier is not available. As before, observe and record the voltage waveform across RL.
Add a filter capacitor to the full-wave rectifier as shown in Fig. 3.11(a) – again, be careful not to connect the capacitor backwards. This form of power supply is very common. Measure the average output voltage across the 10 k load. Record the peak-to-peak ripple voltage.
Repeat these measurements for RL= 1 k. Caution: What power rating must the 1 k resistor have, and why? Use socket adapters to handle the fat leads of the 2 W resistor.
Repeat the ripple voltage calculations for these two values of RL, keeping in mind that the filter-capacitor discharge time is now one-half of the AC cycle.
To make this circuit into a ‘complete’ power supply, one would want to regulate the output, that is, employ feedback to make the output voltage and ripple less dependent on the load resistance. This could be done using Zener diodes, but a more effective technique is a transistorized regulating
45 3 Diodes
Vout RL
Transformer
Bridge Rectifier
+ 1000 Fµ
1 120s
+
_
~
~
Ripple Voltage fuse
120
(a) VAC
(b)
Fig. 3.11. (a) Full-wave rectification with filter capacitor; (b) waveform produced by circuit shown in (a).
circuit. Integrated three-terminal voltage regulators (such as the 7800 and 7900 series) have made this particularly simple.