ENGR 3401 Electronics I Laboratory  EXPERIMENT 6  

Rectifier Circuits and Power Supply Filters ©

Objectives:

1. To understand how diodes can rectify AC voltage to obtain a pulsating DC voltage.

2. To build the three basic types of rectifier circuits and determine the advantages and disadvantages of each type.

3. To discover the effects of capacitors connected across the output of these rectifier circuits (Power Supply Filters).

4. To investigate the effects of different loads on power supply filter circuits.

5. To measure ripple voltage and calculate percent ripple from a power supply filter circuit.

Parts and Equipment required at lab table:

One 12.6 volt center-tapped transformer with fused line cord
4 ea.: 1N4004 diodes (or equivalent)
Electrolytic capacitors: 10 µF, 100 µF, 1000 µF
Resistors: 1 kOhm, 10 kOhm, 100 kOhm
Oscilloscope
Multimeter
Breadboard and wire

Textbook Reference:

Hambley, Allan R., "Electronics", 3rd, Prentice Hall, 2000, pp. 131-189, or Serda and Smith, " Microelectronic Circuits" 3rd, Oxford University Press, 1991, pp. 116-190, or Savant, Roden, and Carpenter, "Electronic Design," The Benjamin/Cummings Publishing Company, Inc., Redwood City, CA, 1991, pp. 17-25

Material Review:

The ideal diode acts as a "one way valve" to electricity and acts as a short (r approached 0) under forward bias (VD = 0+) and an open (r approached infinity ) under reverse bias (VD< 0); it is impossible to increase VD above 0+ in an ideal diode because to do so would require an infinite current since r approaches 0.  

Of course, a real diode only approximates the ideal diode with the small reverse saturation current (-Is) flowing under reverse bias and a small but nonzero dynamic resistance (dVD/dID) under forward bias:

(recall that the "average" or "DC" resistance is different from rD and equals VD/ID). Recall that for most small signal diodes, the diode appears to "turn-on" (i.e., ID rapidly increasing past the "knee" of the exponential curve) at a VD of 0.6 to 0.7 V for silicon diodes,  0.2 to 0.3 V for germanium diodes, and about 1.2 V for Gallium Arsenide diodes.

This ability to control/modulate the effective resistance of a diode allows a variety of useful signal processing applications, the simplest of which is RECTIFICATION.

Refer to Fig. 1. R represents the total load connected to the output of the rectifier.

Applying Kirchhoff's voltage law to the circuit above gives:

Both Equations 3 and 4 can be solved numerically (not in closed form) for iD(t) or vD(t), respectively, in terms of the other system parameters. One, of course, would find that for vs(t) < 0, iD(t) is approximately equal to -Is (but not exactly) in a real diode and zero in an "ideal" diode, vD(t) is approximately equal to vs(t) (actually v(t) + RIs) and vR(t) is approximately equal to 0. (actually -RIs). For vs(t) > 0, iD(t) is positive and fairly large in magnitude, vD is approximately equal to 0 (in an "ideal" diode) or 0.7 V (in a "real" Si diode), and vR(t) is approximately equal to vs(t) or vs(t)-0.7 V.

The net effect is that current is blocked (for all practical purposes) in one direction and passed in the other. Depending upon whether the output voltage is taken across the diode or resistor, one can pass only the positive or only the negative loops of, say, a sinusoidal input, depending upon diode polarity (Fig. 2). Note that the signal now has a nonzero average or "DC" value as opposed to the input sinusoid which has an average value of zero.

The above circuit is a simple HALF WAVE RECTIFIER. By utilizing more than one diode in proper configurations, one can implement a FULL WAVE RECTIFIER that passes either solely the positive or negative portions, respectively, of an AC waveform and actually negates/reverses the negative or positive portions, respectively, to yield a monopolar (all positive or negative, respectively) waveform (Fig. 3) with a nonzero average ("DC") value, larger than that of a half wave rectified signal that has half of the cycle near zero in value.

Fig. 3 exhibits two full wave rectifier circuits. Fig. 3a is a so-called bridge rectifier and passes the current in the same direction through the load resistor regardless of the input voltage polarity (mentally trace the currents). Except for the "small" voltage drops in the two forward biased diodes (2 x 0 = 0 in an ideal diode or approximately 2 x 0.7 =1.4 V in a real Si diode), the input voltage is passed to the load in a unipolar manner.

Fig. 3b shows the transformer-based full wave rectifier with only two diodes (vs. the four in the bridge rectifier). Recall that the transformer secondary output voltage acts as an AC voltage source.  For an ideal transformer, v2(t) = (N2/N1) v1(t), where N1 and N2 are the number of wire turns in the windings on the primary and secondary, respectively, v2(t) is the (AC) secondary voltage, and v1(t) is the (AC) primary voltage.  A key feature of a transformer is the complete DC decoupling of the input and output since the induced secondary voltage waveform is proportional to the first derivative of the primary current.  This allows one to FLOAT the secondary and earth-ground any of the three terminals, including the center tap as with the full wave rectifier.  Trace the current direction with both transformer secondary polarities to verify that when the upper diode is "on", the lower one is "off" and vice versa, thus causing the current to flow in the same direction through the load resistor for both half cycles.  Again, except for the single diode voltage drop, the outer transformer terminal-to-center tap voltage is passed to the load (of course, assuming that the transformer and circuit wiring are perfect conductors). The smaller number of diodes/diode voltage drops and the ability to float the output are major advantages but the cost, bulk, and heating of the transformer tend to overshadow these advantages.

The diode voltage drops are generally not a problem in large voltage/current power conditioning applications converting AC to DC power. However, in communication applications in which the signals of interest may be less than 0.7 V in magnitude, the diode voltages can be prohibitive. In such small signal applications, active operational amplifier or transistor-based diode/rectifier circuits exist with true zero voltage drop in the forward direction.

Even though the rectified waveforms have a nonzero average value, they still contain a substantial fluctuating AC component with a frequency of fo (half-wave rectified) or 2fo (full-wave rectified), where fo is the input waveform frequency. This total rectified signal can, thus, be divided (via Fourier Series analysis and superposition) into a DC/average value (f = 0) and various sinusoidal "harmonics" at f = fo, 2fo, 3fo, 4fo ...; f = fo characterizes the "fundamental" which is usually the largest AC component for half wave rectification. For Full wave rectification 2fo will be the largest component.

The AC component(s) or RIPPLE can be greatly decreased by placing a capacitor in parallel with the load impedance (Fig. 4); summing currents, one obtains

a differential equation which can be solved numerically for vR(t). Upon solving it with a sinusoidal vs(t), one finds that the capacitor charges up to near the maximum value of vs(t) and allows then only a much smaller AC fluctuation or ripple about the average or DC value.  As long as the capacitor voltage (vR(t)) is less than vs(t), the diode is forward biased and conducts, hence, charging the capacitor until vR(t) reaches vs(t).  As vs(t), on reaching its peak value, drops below vR(t)=vs(t), vR(t) becomes greater than vs(t) and the diode shuts off. The capacitor then slowly discharges through R until vR(t) again drops below the then increasing vs(t) at which point the diode turns on and recharges the capacitor.

The numerical solution of Equation 5 would yield the exact behavior of vR(t) and, hence, its AC or ripple component. However, as an approximation when the ripple is very small,

where at steady state the charge gained during charging exactly matches that lost during discharging, vripple p-p is the small peak-to-peak ripple voltage, VDC is the "DC" voltage or average value of the output voltage.  The period, T, of the input sinusoid (T = 1/fo), is an approximation to the discharge time since the charge time is only a small fraction of the total cycle.

This leads to an approximation for the ripple voltage

showing that the ripple voltage or the ratio vripple / VDC decreases with increasing load resistance, increasing filter capacitance, and with increasing ac input frequency.

The same approximation is also good for a full wave rectified signal fed to a load resistance in parallel with a filter capacitance as long as vripple/VDC is small ( < 0.1). In this case the period is 0.5 T where T is the period of the input sinusoid and

half the ripple of the half wave rectified signal with f = fo = 1/T. More exact analyses lead to small corrections on these approximate equations.  For order of magnitude estimates these easily conceptualized approximations are sufficient as long as vripple p-p /VDC << 1 and the diode "on" resistance is small so that the charge time is a small fraction (< 0.1) of the total charge-discharge cycle.

In this experiment you will investigate the half wave rectifier and both the bridge and transformer-based full wave rectifier circuits with and without filtering. You should become familiar with exactly what is required to convert an AC voltage to a good DC voltage with negligible ripple.  This can also be used to transform a bipolar signal into a monopolar one for a variety of communication, control, and signal processing applications.

Procedure 1: Half-wave Rectification

Review the results of procedure 4 of the previous experiment.  These results will be used in answering some of the questions at the end of this experiment.

Procedure 2A: Full-wave Rectification (unfiltered)

1. Build the following circuit (Fig. 6) using RL = 1.0 kOhm. Connect the oscilloscope to observe the voltage waveforms at points A and B of the circuit.  Connect the oscilloscope ground to the center-tap of the transformer.  Make sure both diodes are pointing toward the resistor.  Set the triggering source to line.  This is the best trigger source to use any time the circuit is being driven from the ac power line.  The triggering does not depend on the voltages being observed, but is synchronized with the ac power line voltage coming into the oscilloscope.  Since this is the same voltage that feeds the transformer it will automatically be synchronized with all the waveforms in the circuit.  You can adjust the position of the waveforms on the screen by adjusting the trigger level, but it is usually best to set the trigger level close to zero and use the horizontal position control to shift the time position of the waveforms.

2. Adjust the vertical sensitivity controls for maximum on screen display on both channels with zero at the center of the screen.  Print the waveforms and the measurements.  The ac rms voltage and the ac peak voltage will be needed.  Measure both of  these voltages  with the oscilloscope and also measure the rms voltage with the DMM.

3. Move one of the scope probes to point C.  You must use DC coupling to properly show the the output waveform, since it contains a large DC component.

4. Set the vertical zero of both channels at 1 division from the bottom of the screen(-3.0) and increase the vertical sensitivity for maximum on screen display of the positive part of the waveforms.  Set the horizontal sweep control for a display of two or three output pulses. Print the waveforms at point B and C and the measurements.  The input voltage at point B is used only as a reference for comparison with the output.  You can measure the peak value of the input voltage and the frequency or period, but any other measurements on this waveform will not be useful because the bottom of the waveform is off screen and therefore not available for measurement calculations.   How much voltage is lost across the diode?  You can measure this with the cursor mode.  It is also available as the difference between the peak voltage of the two waveforms.  Measure both ac rms voltage and the dc voltage across the load resister, RL, with the DMM.

5. Measure the period of and the frequency of the voltage waveform at point C.  Is it double the frequency of the ac input?

6.  Accurately measure the time that the load voltage is zero between the output pulses. 

Do this by centering one of the zero spaces on the screen and then expanding the time scale as much as possible while keeping a portion of the nonzero part on each side and using the time cursors to measure the time.  Increase the vertical gain so the nonzero portions form a steeper slope to make it easier to tell where it starts to deviate from zero.

Even steeper for better measurement accuracy.

7.  Leave the circuit intact for the next procedure.

Procedure 2B: Full-wave Rectification with Capacitor Filter

1. Disconnect the power to the transformer and add a 10 µF electrolytic capacitor, CF, in parallel with the resistor, RL, making sure that the correct polarity is observed (Negative to ground).

2. After all the connections are checked, connect the power to the transformer and observe the input and output waveforms.  Leave the zero reference at -3.00 divisions on the oscilloscope so the output waveform can be observed in more detail. 

3. Print the load voltage waveform and measurements.  Include the input AC waveform for reference.   Record the Vmax and Vmin of the output voltage.  The peak output, Vmax, should be the same as without the capacitor.  Show the correct voltage and time scales on the waveform graph.  DC coupling must be used on the oscilloscope to obtain the actual voltage levels of the waveform.  Use the multimeter to measure the DC voltage and the rms AC voltage across RL.

4. Draw a horizontal line on the waveform printout at the level of the average DC voltage which was previously measured using the multimeter.  Remember your zero volt level should be at -3 divisions on the screen,

Step 5 and 6 will be repeated using each of the following resistor-capacitor combinations:

A. RL = 1 K, CF = 10 µF.
B. RL = 1 K, CF = 100 µF.
C. RL = 1 K, CF = 1000 µF.
D. RL = 10 K, CF = 10 µF.
E. RL = 100 K, CF = 10 µF.
Measure the capacitor values using the LCR meter set at 120 Hz since that is the frequency at which you will be using them.  Make sure to put the negative terminal of the capacitor at the low potential terminal of the LCR meter.

5.  Measure both the AC and DC voltage across RL with the DMM. 

6.  Observe the ripple waveform across the load using AC coupling with the zero reference level set to the middle of the screen, adjust the vertical sensitivity control for maximum on screen display.  Print the ripple waveform and show the correct voltage and time scales on the printout.  Remember that on ac coupling the center of the screen (zero reference level) is actually the dc voltage level not zero volts.  Measure the values of  Vmax and Vmin on the ripple waveform.  The delta measurement, (Vmax - Vmin), will equal the total peak to peak ripple voltage.  The actual total voltage at Vmax is the same value as the peak rectified voltage measured in step 3 above.  It can also be found by adding the Vmax measured with ac coupling to the measured DC voltage.  Vmin is found by adding the negative value found for Vmin to the dc level or by subtracting Vripple p-p from Vmax.

Procedure 3: Full-wave Bridge Rectifier

1. Build the following circuit (Fig. 7) using RL  = 1 k Ohm. Connect the oscilloscope to observe the waveform across the load resistor.   You should be using DC coupling, the triggering should still be set to AC Line, and the zero reference should be set to -3.00 div. 

2. Observe and print the waveform observed across RL and measurements from Wavestar.   Note the ground symbol at one end of the resistor.  This point should be used as the reference node in this circuit.  This is where the ground clips from the oscilloscope will be connected. Print the waveform.  Use the multimeter to measure the DC and AC rms voltages across RL. 

3. Accurately measure the time that the output voltage is zero between the adjacent output pulses, as you did in procedure 2A step 6.  This time will be a bit longer than in the previous measurement by about 200 ms.

4. Use the multimeter (make sure multimeter is set on "AC") to measure the AC RMS voltage on the input to the bridge (points A to B).

5. Use the oscilloscope to observe the AC voltage at points A and B relative to the reference node.  Print waveforms and measurements.  Compare the waveforms to the waveform across RL.  Measure the Vmax  and Vmin of each waveform and compare them to the waveform across RL.   How does the peak to peak voltage of these waveforms compare to the transformer voltage measured in previous procedures?  Is it the same as VS max?  You can reconstruct the input ac waveform by subtracting one of these voltages from the other.  This can be done by exporting your waveform to a CVS file and then import the data into a spreadsheet where you can subtract V1 from V2 and plot the waveform.

CALCULATIONS/GRAPHS

For nonfiltered rectification circuits:

1. Make a table showing the peak AC  input voltage (VS max), the peak output voltage (VL max), the difference between the max input and the max output, the DC output voltage (from the multimeter) for each of the three unfiltered configurations( including the half-wave rectifier from the previous experiment), and the ratio of DC output voltage to  AC peak input voltage for each case.  Express the ratio as a percent of the peak AC input voltage.  Use the maximum values from Wavestar or from the cursor mode on the oscilloscope.

2. Explain the differences between the half-wave rectifier, the full-wave rectifier, and the full-wave bridge rectifier.

3. How much voltage is lost due to the diodes in each type of rectifier configuration? (compare the peak input to the peak output as measured by Wavestar).

4.  How much did the time the voltage was zero between pulses change between the full wave and the full wave bridge circuits?  Why do you think this happened?

For rectification circuits with filter capacitor:

1. Calculate the percent ripple factor (measured peak to peak ripple voltage as a percent of the measured DC voltage) for each (RL, CF) combination.

2. On a log-log graph, plot the percent ripple as a function of CF for RL = 1 k Ohm.  Click to see sample of a log-log graph and how to make one.

3. On another log-log graph, plot the percent ripple as a function of RL for CF = 10 µF.

4.  On a third graph plot percent ripple as a function of  f  for RL = 1 kOhm and CF = 10 µF.  Remember the output frequency for the full wave circuits was double that of the half wave circuit.

4. Verify that percent ripple varies inversely with R, C, and f. Recall that from both Eq. 7 and 8 in the Material Review.  If  %ripple can be represented in terms of a constant multiplied by the resistance and capacitance each raised to some integer power then the exponent values can be determined by plotting the log of %ripple verses the log of R with constant C and constant output pulse frequency, and the log of %ripple verses the log of C with constant R and constant pulse frequency, and the log of %ripple verses the log of output pulse frequency (60 Hz for the half wave circuit and 120 Hz for the full wave circuits) with constant R and constant C, and checking the slopes.  See the derivation  below.  You do not have to actually take the log of the values.  You can use a log-log scale graph and then count the number of decades the line drops or rises per decade change in R, C, or frequency to get the slope.  Since the points for frequency are less than a decade apart, you will have to extrapolate the line to cover a full decade of frequency.

 

 Thus, your log-log plot of (% ripple) vs. R, C, or f with the other variables held constant should have a slope of  -1 decade/decade.  Read the slopes from your graphs and round to the nearest integer value.

QUESTIONS:

1. How does the average or DC output voltage vary with the change of RL,   CF, and output pulse frequency?

2. How does the percent ripple vary with the change of RL,    CF, and output pulse frequency?

3. Which of the three unfiltered rectifier circuit configuration appears to be the most efficient at converting AC voltage to DC voltage?  Check the table made in Calculations 1.

4. Comment on and try to explain any practical difficulties or unanticipated observations experienced with this Experiment.

Conclusions:

What can you say about rectifier circuits based on your experimental observations?

This page last updated 09/25/2007