Objective
To familiarize you with the concept of capacitance and to illustrate the use of capacitors in circuits such as an integrator and a pulse counter and also to build RC circuits and observe the step response of the circuit. In doing so you will:
· 1. Familiarize yourself with the exponential waveform,
· 2. Learn parallel and series combinations of capacitors,
· 3. Observe the filtering action of an RC circuit.
Background
Capacitor:
We are introducing a new circuit element, i.e. a capacitor. A capacitor is a dynamic element which allows you to build circuits that remember. You will be able to build a whole range of new and interesting circuits, such as filters, oscillators, differentiators and integrators. You will learn more about filters and oscillators in other courses. In this lab we will concentrate on integrators and differentiators. An integrator is a circuit that responds to the average of a signal over time while a differentiator responds to changes of a signal. Before discussing some of these interesting circuits, let’s first get a better understanding of what a capacitor really is.
Let’s start by seeing what a capacitor looks like and what it does. You know from your physics and circuit classes that a capacitor can be constructed by two closely spaced metal plates separated by an insulator or dielectric, as shown in Figure 1.
Figure 1: A parallel plate capacitor.
When you put a voltage, V, over a capacitor, C, a charge, Q, will be stored on its plates:
Q=C.V
The larger the value of the capacitor is, the more charges one can store on it. Thus a capacitor acts as a reservoir of charges. Another interesting way of looking at a capacitor is to see how it behaves when the voltage over it changes with time:
I=C dV/dt
Where I, is the current that flows through the capacitor. The faster the voltage changes, the larger the current will be. Notice that there can only be current when V changes over time, otherwise the current will be zero.
It may be easier to get a better feel for a capacitor by looking at a hydraulic equivalent as shown in Figure 2a. We can compare a capacitor to a bucket filled with water. The amount of water stored in the bucket corresponds to the charges on the capacitor; the water level corresponds to the voltage and the size of the bucket is equivalent to the capacitance C. Now when you pour water into the bucket (or have current flow), the water level (voltage over the capacitor) will increase as shown in Figure 2b.
Figure 2: (a) bucket with water corresponding to a capacitor with charge Q; (b) water (current) flow in a bucket raises the water level (voltage)
Capacitor types and reading capacitor values:
There are many different types of capacitors depending on the type of dielectric material used between the two electrodes and the shape of the electrodes. The main types are electrolytic, tantalum, mylar, ceramic and polyester capacitors as shown in Figure 3.
Figure 3: Capacitor Types (not to scale)
a. Electrolytic capacitors:
These are large capacitors both in volume and in capacitance value (tens of microfarads). They look like a cylinder, as shown in Figure 3a. Electrolytic capacitors consist of a pair of aluminum plates separated by borax paste electrolyte. This provides a large capacitance value in a relatively small volume. Be careful when using these capacitors. They are polarized which means that the capacitor is not symmetrical and that you can destroy it when putting a voltage of the wrong polarity over it. As shown in Figure 3b, one of the terminals has a plus sign (with the largest leg) (or minus sign - shortest leg) indicating that his terminal always has to be more positive (negative) than the other one. If you violate this rule, the capacitor will be destroyed and may even explode! Also capacitors have voltage ratings. Never put a larger voltage over the capacitor than what is specified on the capacitor. This is also the case for all other capacitor types.
These are smaller capacitors and can have different colors and have the shape of a deformed oval. Tantalum capacitors are also polarized, as shown in Figure 3b. The positive terminal has the longest leg and is usually marked by a plus sign.
These are usually yellow cylinders. They are not polarized.
These can have circular shapes (usually orange) or look like little boxes (often blue) as shown in Figure 3d. Ceramic capacitors work well up to high frequencies in contrast to mylar capacitors.
These capacitors usually have a glossy cover and can have a square or oval shape. They are not polarized.
Reading capacitance values can be tricky and is generally more difficult than reading resistance values. Here are some general guidelines. The unit of a capacitance is Farad. One Farad is a huge capacitor (also in physical size). Capacitors are usually expressed in microfarads (x10-6F) for large capacitors and picofarad (x10-12F) for small capacitors. The largest capacitance you will be using ever is around 100 uF. Figure 3 gives a few examples to help you read capacitor values.
The value of the electrolytic capacitors in Figure 3a are 100 uF and 6.8uF (sometimes the unit u may be omitted; however, we know that it cannot be picofarad as electrolytic capacitors usually have large values). The values of the capacitors in Figure 3b are 22uF (with 25V rating) and 4.7uF (35V). An equivalent way of indicating the value is 226 (=22uF). The third digit indicates 106 x pF (or 226M = 22 x 106 x 10-12 = 22 uF). The M in 226 does NOT stand for Mega or milli but refers to the tolerance (M = +/- 20%). The value of the capacitor in Figure 3c is 0.2uF (again, M stands NOT for mega but for the tolerance). The value of the capacitor in Figure 3d is 20nF (the 3rd digit is the power of 10: 203=20 x 103 x 10-12 = 20nF) with a 25V rating. Finally the value of the capacitor in Figure 3e is 1uF (45 V).
TABLE 1: Tolerance code of Capacitors
Tolerance Code Value (+/- %)
M 20
K 10
J 5
G 2
F 1
D 0.5
C 0.25
B 0.1
A 0.05
Z 0.025
Now you learned that capacitors can store charges and act as short-term memory elements (that is the basis for the operation of DRAM’s in your PC). You also know that when one charges a capacitor with a constant current, the voltage over the capacitor increases linearly with time, as shown in Figure 4a. This is a direct result of the expression,
I = C. dV/dt.
Indeed, one can solve the above equation for V,
(1)Thus, the voltage increases linearly with time. The larger the current the faster the voltage over the capacitor will increase, as is illustrated in Figure 4b.
We will now consider the case when the capacitor is charged up through a resistor and a constant voltage source, as is shown in Figure 5. This circuit is a little more difficult to understand but is more common in electric circuits. Let’s get an intuitive understanding of what happens in this case.
In the beginning when no charge is stored on the capacitor the voltage V over the capacitor is zero. The current I with which the capacitor is charged up is then, according to Ohm's law, equal to: I= (Vcc-0)/R. Now, when the capacitor charges up, the voltage over the capacitor increases, as seen in Figure 5b. As a result, the current that charges the capacitor will decrease: I= (Vcc-V)/R. The capacitor will charge up at a slower rate. The larger the voltage V becomes the slower the capacitor charges, as illustrated in Figure 5b.
Figure 5: (a) A capacitor charged up by a constant voltage source through a resistor R; (b) voltage over the capacitor.
In class you derived the exact expression for the voltage over the capacitor (by solving the first order differential equation). The expression can be written in general form,
(2)where Vi is the initial voltage (at t=0+), Vf is the final value (at time t=infinite) over the capacitor, and t is the time constant RC. For the example of Figure 5a, the expression for the case of charging the capacitor is equal to,
(3)For the case of discharging the capacitor, the exponential waveform is equal to,
(4)The above expressions represent an exponential waveform which, together with sinusoids, are one of the most important waveforms in electronic circuits. It is important that you understand the exponential waveform well. Here are some of the more important features:
The slope of the waveform at the origin (t=0) is equal to Vcc/t, as shown in Figure 6. Thus the time constant can be found by extrapolating the initial slope with a straight line until it intersects the final voltage level over the capacitor.
Figure 6: Exponential waveform and time constant: (a) charging of C, and (b) discharging of C.
When t= t, e-1 = 0.37 and 1- e-1 = 0.63. Thus after one time constant the capacitor will be charged up to 63% of the final value, as illustrated in Figure 6a above. The table below shows how the voltage varies after one, two, etc., time constants. It is interesting to notice that (only) after 5 time constants the voltage over the capacitor had reached 99% of the final value. The increase (or decrease) of the voltage gets slower and slower when the voltage reaches its final value, as we expected, because the current gets smaller (as discussed above).
TABLE II
Time t | e-t/t | 1-e-t/t |
1t | 0.37 | 0.63 |
2t | 0.14 | 0.86 |
3t | 0.05 | 0.95 |
4t | 0.02 | 0.98 |
5t | 0.01 | 0.99 |
6t | 0.003 | 0.997 |
You don't need to memorize these numbers. However, the value for t=t and t=5t are very handy to know.
1. Read the textbook ("Basic Engineering Circuit Design" by D. Irwin), sections 5.1 and 5.7 on capacitors. Also review section 6.2 in the textbook (Basic Engineering Analysis, 5th Ed., by D.Irwin, Prentice Hall).
2. Consider the integrator of Figure 7. The input signal is a square wave of 10 Vpp and frequency of 5 kHz, as shown in Figure 7, calculate and sketch the output waveform Vo as a function of time.
Figure 7: Integrator Circuit
3. Do the same as in question 2 but for an input waveform shown in Figure 8. Calculate and sketch the output Vo.
Figure 8: Input waveform of the integrator of Figure 7.
4. You will design a traffic light counter that keeps track of the number of vehicles passing over a narrow bridge (one lane only) as shown in Figure 9. Every time a car passes over the bridge the sensor gives a 5V short pulse (0.2 s). This pulse will be added to the previous counts. When 10 cars have passed in one direction, a red light goes on indicating that no more cars coming from that direction can cross the bridge.
Figure 9: Traffic light controlled by a sensor and counter
You will use an integrator, as shown in Figure 10, to add the pulses coming from the sensor. Design the circuit so the integrator's output decreases by 1V for each car crossing the bridge. When 10 cars have crossed a red LED should go on.
· Find the value of the resistor R1 that will give a 1V decrease for each pulse.
· What circuit will you use that switches the LED on when the 10th car has passed? Design and draw this circuit. From previous labs we know that the LED has about 1.7V over its terminals when it is on. Also, the current in the LED should not be larger than 60 mA. Design the circuit so that the LED draws about 20mA when it is on, assuming that the output voltage of your circuit is 15V (so you will have to use a resistor to limit the current). You can use an Op Amp 741, a potentiometer and resistors to build the circuit. (Hint: you have used a similar circuit in previous labs.)
A manual switch will be used to reset the integrator (discharge the capacitor) 10 pulses after the red LED has gone on.
Figure 10: Integrator circuit for counting number of cars
5. It is often easier to measure the rise and fall time of a waveform. The rise and fall times are defined as the time it takes for the waveform to rise from the 10% to the 90% (or fall from the 90% to the 10%) level, as shown in Figure 11. Prove that the relationship between the rise or fall time and the time constant is as follows,
(5)
Figure 11: Definition of rise time (similarly for fall time)
6. Consider the simple RC circuit of Figure 12a. Find the initial voltage V (0+) over the capacitor, the final voltage and the time constant. Give the expression v (t); sketch v (t) carefully and label the axis; also indicate the value of the time constant.
Figure 12: RC circuits with (a) single capacitor and (b) two capacitors in series.
7. The circuit in Figure 12b is similar to the one in Figure 12a except for the two capacitors in series. Find the new value of the time constant of the circuit. What is the expression of v (t)? Sketch v (t) on the same graph as the response of Figure 12a above.
8. Consider the circuit of Figure 13. Find the expression of the voltage over the capacitor v (t): first find the initial value of v (0+), final value vf, and the time constant.
Figure 13: RC circuit with two resistors
9. Next you will design a timer circuit can be used in cameras with a self-timer. We will be making use of the time it takes to charge up a capacitor through a resistor. The circuit is shown in Figure 14. A buzzer will go on during the delay time for which the circuit has been programmed.
Figure 14: Self-timer circuit
We will be making use of a relay to switch on the buzzer. Relays are useful devices which are used in many applications varying from telephone switches, power stations to pinball machines. A relay is an electrically controlled switch. It consists of a coil that pulls down an armature when a current flows through the coil as is schematically shown in Figure 14. When the relay closes, the buzzer is powered and will start to buzz.
When you press the switch S1 over the capacitor, the voltage V will be zero (see Figure 14). The capacitor will charge up slowly. As long as this voltage V (t) is below the voltage on the non-inverting input of the comparator, the output voltage of the comparator will be high. This will energize the relay and close the switch S2 which causes the buzzer to go on. The diode is put in series with the relay so that only current in one direction can flow (and thus the relay will not be energized when the comparator output is negative).
We want the buzzer to go on for 3 seconds after you discharge the capacitor (or push the switch S1).
Find:
a. The value of the resistor R that will give you the required delay of 3 seconds, using the values of the elements shown in Figure 14.
b. Assume that you want to make the delay variable. You could use a variable resistor R (potentiometer) or change the voltage at the non-inverting input. Let’s choose the latter option. What voltage would you need to apply on the non-inverting input of the comparator in order to make a delay of 10 sec using the value of R calculated above? In practice we can use a potentiometer to adjust the voltage at the non-inverting input.
(the lab is divided in 2 weeks. Part I (complete Pulse Counter Circuit) is due the week of 11/28, Part II is due the week of 12/6)
A. Equipment:
· 1. HP function generator/waveform generator (HP 33120A)
· 2. HP digital oscilloscope HP54600
· 3. HP triple-output power supply (HP E3631A)
· 4. Protoboard
· 5. RLC meter (Philips PM6303)
· 6. Two HP Scope Probes
· 7. Resistors: Two 1 MOhm , 1 and 2 KOhm and one 10KOhm potentiometer
· 8. Capacitors: one 1 nF ,three 1 uF and one 22 uF
· 9. Two 741 Op-Amps
· 10. DIP relay <<<< only for Part II
· 11. Buzzer <<<< only for Part II
· 12. Diode 1N4004 <<< only for Part II
· 13. One Red LED
· 14. Blue box with cables and connectors
· 15. PC
· 16. Multisim software on PCs (optional)
1. Build and measure the following integrator (Figure 15). The resistor R2 has been added to prevent the Op Amp from going into saturation (as a result of the offset voltage). Don't worry about this right now; you will learn more about this in EE216. Make a sketch in your notebook of the protoboard layout before building the circuit.
Figure 15: Integrator circuit
V1 is a square wave with frequency of 5 kHz and peak-to-peak voltage of 10V (notice that this requires a setting of 5Vpp on the function generator due to the 50 Ohm output resistance; you can change the output impedance to a "high Z" using the menu functions). Verify the peak-to-peak voltage and frequency of V1 on the oscilloscope (see waveform in Figure 7). Display both the input voltage V1 and the output voltage Vo on the oscilloscope. Measure the output voltage and make a print-out. Compare with the calculations of the Prelab.
2. Change the input voltage V1 to a triangular waveform and measure and sketch the output voltage. Experiment with some other waveforms. Change the amplitude or the frequency.
3. Select a sinusoid for V1 and measure the output voltage. Is this what you expected? Verify this by measuring the phase difference between the input voltage V1 and the output voltage Vo. Measure the phase difference on the HP oscilloscope (go to menu for time measurements). Record the procedure and results in your lab notebook.
a. Build the circuit of Figure 16 (repeated below) including the circuit to switch on the LED. Have your circuit checked by the lab instructor before building it. A switch shown in Figure 16 is a pair of wires whose ends can be touched momentarily. Use the switch to reset the integrator (discharge the capacitor). Notice that the value of the capacitor is different from the one used in the previous circuit.
Figure 16: Pulse counter circuit
b. After building the circuit, apply a voltage Vs that simulates the output of the sensor as shown in Figure 16. Use a 1 Hz square wave with a 20% duty cycle with a voltage that varies between 0 and 5V (you will need to apply an offset voltage; also take into account the 50 Ohm output resistance of the function generator). Verify the waveform on the oscilloscope.
c. Reset the switch (touch the end of wires ) and observe the output voltage Vo (of the integrator) and Vs on the oscilloscope.
In order to see these slowly varying signals use the "Roll" mode (press the MAIN/DELAY key first, then push the Roll key at the bottom of the display). The ROLL mode is handy to display slowly varying signals.
When you display the output of the integrator after resetting the capacitor, there is a good chance that the output of the integrator is not constant between 2 pulses in contrast to what we expect as shown in Figure 17a. The slope of the output can be upwards or downwards depending on the OpAmp. This is due to the offset voltage of the OpAmp. You can compensate for this by subtracting an offset voltage from the input signal Vs. Instead of having Vs go between 0 to 10V, adjust the lower level (using the offset control on the function generator) flat as in Figure 17b (this level can be positive or negative depending on the slope of graph in Figure 17a). The output voltage of the integrator should be flat between two pulses as the solid (blue) line in Figure 17a
Figure 17: (a) output of an ideal integrator (solid line) and an integrator with offset voltage (stippled line); (b) adjust lower level of Vs until curve in (a) is flat between two pulses.
Verify that for each pulse at the input the output decreases by 1V (use the cursors). You can stop the waveform by pressing the STOP key on the top right of the panel. To continue, push the RUN key.
Also verify that after 10 pulses the LED goes on. Adjust the time scale so that you can see the 10 sensor pulses (channel 1 or 2) on the display as well as the output of the integrator (channel 2 or 1).
Another handy way to verify the operation of the circuit is to display both the output of the integrator (e.g. on channel 1) and the output of the circuit that drives the LED (channel 2). Verify that indeed the circuit switches on after the 10th sensor pulse has been integrated. Have the instructor check your circuit (give him a demo) and have him sign your lab notebook. Make a printout of the output of the integrator showing that the LED circuit switches after 10 pulses (after resetting the capacitor).
The counter designed above is a good illustration of how a capacitor can remember the previous states, in this case up to 10 pulses, that arrived at an earlier time. By resetting (shorting) the capacitor one erases its memory and one can start off fresh.
-------------------------------------End of Lab Part I-------------------------------------------------------------------------------------------------------
-------------------------------------End of Lab Part I-------------------------------------------------------------------------------------------------------
5. You will be using scope probes (two per station) to measure and display the waveforms on the scope.
First check to see if the propes are properly compensated using the reference square wave on the scope front terminal underneath the display. Remember that a probe is adjusted when the square wave has a flat top and bottom. Make sure you set the scope for a 10x probe so that you get the proper readings.
6. a. Build the circuit of Figure 18a. When building the protoboard don't put elements in the air. Keep the legs and wires short. The circuit on the protoboard should reflect the topography of the circuit.
Figure 18a : RC Circuit
b. For the input voltage Vs, use the function generator and select a 5Vpp square wave of 50Hz as shown in Fig. 18a. Check the amplitude, offset voltage and frequency of the waveform with the oscilloscope.
Display the output V of the capacitor C on the other channel of the scope, using the scope probe. Observe the shape and sketch it in your lab notebook. Measure its characteristics by using the voltage and time functions of the scope: voltage level (low and high), rise and fall times, and period. Determine the time constant using equation (5). Make a print out for use in your report using Benchlink.
7. Next replace capacitor C with two capacitors in series as shown in Figure 18b. Use the same voltage Vs as for the task above. Display both voltages V and V1 on the scope. Measure the time constants and voltage levels for both curves. Is this what you would expect? Sketch the waveforms in your notebook. Make a printout as well. Explain what you measured: time constant and voltage levels in your report.
Figure 18b : RC circuit with two capacitors in series
8. Build the circuit of Figure 19. Use the same value for Vs as before. Display both Vs and V on the scope and measure the characteristics. Make a print out. Does the waveform correspond to what you calculated in the pre-lab?
Figure 19: RC circuit
9. Timer circuit.a. Build the actual circuit of Figure 20 on your protoboard using a value for R as close as possible to the one calculated. Measure the actual resistance and record the value in your notebook. You will be using an electrolytic capacitor of 22 uF. Remember that these are polarized and must to be connected with the right polarity. If not, there is chance that the capacitor will explode! Be careful! Measure the exact value of C with the RLC meter (Philips PM 6303) and record the value in your lab notebook.
Figure 20 : self-timer circuit
Use a 10 kOhm potentiometer to adjust the voltage at the non-inverting input of the comparator to 2.5V. Measure this voltage with the multimeter.
The relay comes in a DIP (dual in line pin package) as shown in Figure 20.
b. First display the voltage over the capacitor and check at what time it reaches 2.5V (press the push button switch and observe the waveform). The signals are very slow. The best way to display them on the scope is to use the Roll mode. To measure the time accurately, set the voltage cursor at 2.5V and see when V(t) crosses this level. Push the STOP button to hold the display. You can now use the time cursor to measure the time.
Based on the measured values of R and C, at what time should the capacitor reach 2.5V? Compare the calculation with the above measurement.
Does the buzzer switch on at that time? If not, debug your circuit.
c. Now display both the voltage over the capacitor and the output of the comparator on the scope. Check that the comparator switches around 3 sec; record this in your lab notebook. Make a printout.
d. Change the potentiometer so that the buzzer is on during 10 sec after pushing the switch. Measure the voltage at the non-inverting amplifier and compare this with your hand calculations from the pre-lab.
1. J.D. Irwin, "Basic Engineering Analysis," 5th ed., Prentice Hall, Upper Saddle River, NJ, 1996.
2. T. Hayes and P. Horowitz, "The Art of Electronics - Student Manual", Cambridge University Press, Cambridge, MA, 1989.
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