Asst. Prof. Rajdeep Singh Rawat Natural Science Academic Group Room No. 07-03-108 Email- firstname.lastname@example.org Phone: 67903826
Experiments: a. Analog Electronics
1. Diodes: I-V characteristics, Rectification and power supply. 2. Transistors: Amplification 3. Operational Amplifiers: Inverting and non-inverting amplifier and other applications.
b. Digital Electronics
1. 2. 3. 4. Logic gates, Boolean algebra and combinational logic Karnaugh Mapping Binary Arithmetic: half-adder, full-adder and adder-subtractor RS and JK flip-flops and Counting circuits
Understanding Diode and its Application as Rectifier
Resistors, capacitors and inductors are called Linear Components as the current through these increases in direct proportion to the applied voltage in accordance to the Ohm's law. Components for which this proportionality does not hold are called NonLinear devices and they are the basis of all practical electronic circuits. Typical examples of nonlinear electronic components are diode, transistor, thyristor etc. Let us start with diode which is one of the most important nonlinear components and whose one of the basic function is rectification that is to convert a pulsating alternating current or voltage into direct current or voltage. The term "diode" has come from the fact that it has two terminals, or electrodes. The diode rectifier is nonlinear in that it passes a greater current for one polarity of applied voltage than other. If a rectifier is included in an ac circuit, the current is negligible whenever the polarity of the voltage across the rectifier is in the reverse direction. Therefore, only a unidirectional current exists and alternating current is said to have been rectified. A major application of diode rectifiers is in power supply which convert conventional 240 V, 50 Hz ac line voltage to dc potentials suitable for use in place of batteries. The most useful description of the electrical properties of a nonlinear component is the relationship between the current through the component and the potential across it. This relationship is conveniently displayed graphically in the current-voltage (or IV) characteristics of the nonlinear component. The appropriate current-voltage characteristics are useful in analyzing the performance of the nonlinear component in any circuit. It is convenient to employ graphical techniques in the analysis because the IV characteristic is not expressible in manageable mathematical form. Characteristics IV curves of commercial devices are always presented in manufacturer's data sheet and are widely available. In the present experimental exercise we aim to achieve the following goals 1. To study the current-voltage (IV) characteristic curves of a simple junction diode, and 2. Construction of full wave rectifier circuit with capacitor filter.
Various diode symbols: p n p n p n p n
Light Emitting Diode
Fig. 1: Symbols for different types of diodes
Terminal recognition by Diode Appearance: p n p n
Fig. 2: Physical appearances of different diodes
Activity 1: The IV (current-voltage) characteristic curves of a simple diode. Components and apparatus required: Diode IN4007, 1 k resistor, bread board, variable power supply (0-30 V), connecting wires and multi-meter.
Forward characteristic curve: 1. Arrange the following circuit on breadboard.
Vd VR R 1k
Fig. 3: Forward biasing circuit of diode for IV curves
V (0-30) V
2. Switch on the power supply and increase the supply voltage from initial zero volts to higher values in steps. 3. Use multi-meter (with knob being set to voltage measurement) to measure the voltages across diode (Vd) and load resistor (VR). Note down the observations in the following table
Id= VR/R Amp
4. Calculate Id and then plot the curve Vd versus Id with Vd along the x-axes and Id along the y-axes to get the forward characteristic curve of diode.
Reverse characteristic curve: Repeat the above experiment with the power supply being reversed, as shown in the following figure. Make similar observation table and draw the reverse characteristic curve of the diode on the same graph paper.
Vd VR R 1k
Fig. 4: Reverse biasing circuit of diode for IV curves
V (0-30) V
Id= VR/R Amp
Results: Simple diodes conduct heavily only when they are forward biased.
Activity 2: Construction of full wave rectifier circuit with capacitor filter. Components and apparatus required: 2 pieces of Diode IN4007, 6-0-6 V transformer, 10 k resistor, 100 F capacitor, bread board, and oscilloscope. Theory of the experiment: Activity 2 demonstrated that the diode conducts only when it is forward biased. Now, let us analyze the circuit shown in fig. 5, which has a junction diode in series with a sinusoidal ac source and a resistive load.
Fig. 5: Half-wave rectifier circuit.
It is easier to guess that when the polarity of the source is in forward direction, the diode conducts and produces a current in the load. On the reverse half-cycle the diode does not conduct and the current is zero. The output current is therefore a succession of half sine wave and the circuit is called a half-wave rectifier. The input sine wave has, therefore, been rectified as shown in fig. 6(b).
t V (c)
Fig. 6: (a) Input waveform, (b) half-wave rectified waveform and (c) full-wave rectified waveform
The half-wave rectifier is inactive during one-half of the input cycle and is therefore less efficient than possible. By arranging two diodes as in fig. 7 so that each diode conducts on alternate half-cycles, full-wave rectification results. This is accomplished by using a center-tapped transformer winding. Then, diode D1 conducts on one half-cycle while diode D2 is reverse-biased. The result is that we get the output voltage across load RL. On the alternative half-cycle, the conditions are interchanged i.e. D2 conducts while D1 is reverse-biased. So we still have output voltage across R L and that too with similar polarity. As a result, we get full-wave rectified output as shown in fig. 6(c).
i1 D1 V i2 i1+i2
t D2 i2 (a) (b) i1+i2
Fig. 7: (a) Full-wave rectifier and (b) waveforms.
Capacitor Filter: It is usually desirable to reduce the alternating component of the rectified waveform so that the output is primarily a dc voltage. This is accomplished by means of low-pass filters which are composed of suitably connected capacitors and inductances. We will be discussing and using only the simple capacitor filter. The filter reduces the amplitudes of all alternating components in the rectified waveform and passes the dc component. A measure of the effectiveness of a filter is given by ripple factor r, which is defined as the ratio of the rms value of the ac component to the dc or average value. That is
It is desirable to make ripple factor as small as possible. The capacitor filter is the simplest filter circuit with a capacitor in parallel to the load resistance Rl. The capacitor is charged to the peak value of the rectified voltage Vp and begins to discharge through load resistance Rl after the rectified voltage decreases from the peak value. The rate of decrease in the capacitor voltage between charging pulses depends upon the relative vlaues of time constant RlC and the period of the input voltage.The large time constant results in slower decrease and and hence smaller ripple component. The working of capacitor filter is illustrated in fig. 8.
Fig. 8: Output voltage of capacitor filter is dc voltage and small triangular ripple voltge. The ripple voltage is approximately triangular wave if time constant RlC is long compared with the period T of the output wave, RlC >> T. With ripple waveform in triangular form the ripple factor can be estimated to
V pT r Vrms Vdc
1 V pp ripple 1 Rl C 1 1 Vp 2 3 V p 2 3 2 3 fRl C
Ripple factor therefore can be adjusted to desired values by using suitable choices of Rl and C. The actual dc output voltage from the capacitor filter is equal to the peak capacitor voltage less the average ripple component,
V pT Vdc V p V pp
V pT Rl C Vp 2 2 Rl C
The simple capacitor filter provides very good filtering action at low currents and is often used in high voltage low-current power supplies. Because of its simplicity, the corcuit is also found useful in those high current supplies where ripple is relatively less important. The disadvantages of capacitor filter are poor regulation and increased ripple at large loads. The Experiment: 1. Assemble the following circuit. Please don't conect the capacitor, C, as shown in the circuit to observe the full-wave rectified output waveform, on oscilloscope, without being filtered
Fig. 8: Full wave rectifier circuit with capacitor filter.
2. Connect the capacitor for filtering the rectified output and observe the filtered output on oscilloscope. 3. Measure the Vpp of ripple waveform at suitable volts/div (lower volts/div is more suitable as it will increse the least count). 4. Measure Vdc of the rectified filtered output by pressing dc button on scope. 5. Estimate experimental and theoetical values of ripple factor using relation given earlier.
rexperimental= rtheoretical= 6. Repeat experiment for C=1000 mF and compare experimental and theoretical values of ripple factor.
Transistor: Familiarization, Characteristic Curves and Amplification
Amplification is one of the most basic needs in electronics. Consider this, the radio or TV signal received by an antenna is so weak that it cannot adequately drive the loudspeaker or TV tube. We need to amplify this weak signal to make it useful for an electronic application. This amplification is usually achieved through semiconductor device called transistor, invented in 1951 by Shockley at Bell Labs. The transistor has led to many other inventions including the integrated circuits (IC) used by you in your digital lab. The transistors have the advantage of being operative at low supply voltage and using less power as compared to vacuum tubes. Being semiconductor device they can last indefinitely. Moreover, they can be made very small leading to substantial miniaturization. There are many types of transistors. We will be using bipolar transistor in our lab for our experiments. Let me just give a brief introduction to transistor.
(commonly called as BJT) is uses both free electrons and holes for flow of current through the device. That is why it is called so. A Bipolar Transistor
essentially consists of a pair of PN Junction Diodes that Fig.1: The bipolar junction transistor and its symbol. The emitter lead is with arrow on it. are joined back-to-back,
refer fig.1. This forms a sort
of a sandwich where one kind of semiconductor is placed in between two others. There are therefore two kinds of Bipolar sandwich, the NPN and PNP varieties. The N-type region contains free electrons which are negative carriers. The P-type region contains free holes which are positive carriers. The three layers of the sandwich are conventionally called the Collector (C), Base (B), and Emitter (E). A schematic of typical transistor structure is shown in fig.2. The emitter is very rich in current carriers as it is a heavily doped region. Its job is to send its carriers into the base region and then on to the collector. If the transistor is NPN then the emitter is rich in electron and therefore electrons act as majority carrier. . If the transistor is PNP then the emitter is rich in holes and therefore holes act as majority carrier. The base is a lightly doped (with opposite, opposite to that of emitter and collector, polarity carriers) very thin region between the emitter and the collector and it controls the flow of charge carriers (electrons or holes) from emitter to collector i.e. it may allow none, some or many carriers from emitter to collector. The collector collects the carriers coming from the emitter.
Very lightly doped. This “collects” the current carriers.
Base P/N N/P
Very lightly doped. This region “controls” the flow of current carriers. Heavily doped. This region “emits” the current carriers.
Fig.2: The schematic of transistor structure.
Please note that there are two junctions in a transistor: one is between the emitter and the base called emitter-base junction and another between the base and the collector called the collector-base junction, refer fig.3. A depletion layer (depleted of charge carriers) is formed at these junctions as the free electrons from N-region diffuse across the Collector
junction (leaving the parent pentavalent atom in N-region short of one negative
charge, making it positive ion) and recombine with holes in the P-region (leaving the parent trivalent atom in P-
Fig.3: Transistor junctions.
region short of one positive charge,
making it negative ion). The electric field is thus created across the depletion layer between the positive ions in N-region and negative ions in P-region. The polarity of this field is such that it opposes the diffusion of electrons. The strength of this field increases with every diffusing electron till the equilibrium is reached. This means that electric field eventually stops the diffusion of electrons across the junction. The electric field between the ions is equivalent to the difference of potential called the barrier potential. At 25oC, the barrier potential equals approximately 0.3 V for germanium and 0.7 V for silicon. So each of the depletion layer at these two junctions is having a barrier potential of 0.7 V as we will be using silicon transistors.
The Biased Transistor:
The two junctions must be properly biased properly for
transistor to work properly. This is why one cannot replace an NPN transistor with PNP transistor. The properties would be wrong. The transistor biasing is shown in fig.4. Rule: The emitter-base junction must be forward-biases whereas the collector-base junction should be reversed biased for the proper operation of the transistor.
Collector Base Emitter
This collector-base junction is required to be reversed biased. This emitter-base junction is required to be forward biased.
Fig.4: Biasing of transistor junctions.
Fig.5: The CE connection.
Let us consider a practical circuit shown in fig.5. Please note that I will be making discussion with reference to the NPN transistor throughout. Similar analysis can be carried out for PNP transistor with corresponding proper biasing and hole being the charge carrier rather than electrons. The portion of the circuit on the left shows the forward biasing of the emitter-base junction using base voltage source VBB through base resistor RB. The right collector voltage source VCC reverse biases the collector base junction. This configuration with emitter common to both input and the output circuits is known as common-emitter (CE) connection.
Let us see that how this (a) circuit works. For that we present the circuit in fig.5 in simpler equivalent
version with transistor symbol replaced by a physical transistor
structure, as shown in fig.6. (b) At the instant that
forward bias is applied to the emitter-base junction, electrons in the emitter have not yet entered the base region, as shown in fig.6a. If VBB is greater than the barrier potential, emitter electrons will enter the base region as shown in fig.6b. These free
electrons can flow in either of two directions. First, they can flow to the Fig.6: (a) Biased transistor. (b) Electron enter base. (c) Electrons enter collector. left and move out of the base passing through RB on the way to the positive source terminal. Second, the free electron can flow into the collector. The question is which way do most of the free electrons go? Most will follow the second path as the base is lightly doped (providing free electrons longer life time in base) and moreover it is very thin. To flow out of the base, to the left, the free electrons must first recombine with holes in the base. Then, as valance electrons, they can flow to the left until they leave the base and enter the external connecting wire. Since the base is lightly doped and very thin, very few electrons manage to recombine and escape into the external base lead.
Almost all the free electrons go into the collector as shown in fig.6c. Once they are in the collector, they feel the attraction of the VCC source voltage. Because of this, the free electrons flow through the collector and through RC until they reach the positive terminal of the collector supply voltage. In most transistors, more than 95 percent of the emitter electron flow to the collector; less than 5 percent flow out of the external base lead.
Currents in Transistor:
There are three different currents in a transistor: emitter
current IE, base current IB, and collector current IC, shown in fig.7 for NPN transistor. Because the emitter is the source of the electrons, the emitter current is the largest of three currents. As almost all the emitter electrons flow into the collector, the collector current approximately equals the emitter current. The base current is very small compared to these two currents. Using Kirchoff’s current law, we can write Fig.7: Current flow in NPN transistor. As, IE = IC + IB IB<< IC (1)
One can safely assume
IC I E
It is important for us to realize that though the base current is quite small but very important. Suppose, for example, that the base lead of the transistor is open. Then the base current will be zero. Opening the base lead for zero base current removes the forward biasing of the emitter-base junction which leads to no emitter current and hence no collector current. The fact, the low base current controls much higher currents in the emitter and collector is very important. This shows the transistor is capable of good current gain. The current gain
β dc of a transistor is defined as the collector current divided by the base current.
For most low-power transistors the current gain is typically 100 to 300.
Activity 3: Common-emitter amplifier: dc biasing, voltage gain and frequency response.
Components and equipment required: 2N2222A (or BC547) transistor, the general purpose kit with breadboard, 560 , 1 k, 8.2 k, 18 k resistors, Two 25 F (16V) (or 22F) and one 100 F (16V) capacitors, 9V supply (adjust the one on the kit), functional generator (on the kit), multi-meter and a oscilloscope. Circuit diagram:
G Fig.8: Experimental common emitter amplifier circuit.
Procedure: This activity is sub-divided into three parts. Part A: dc bias check. 1. Set the variable supply on the kit to +9V to be used in this experiment. 2. Connect the circuit shown in dashed rectangle only. But don’t switch on the power supply. 3. Before switching on the supply let us analyze the circuit using simple rules and approximation. 4. Calculate the voltage drop across R2. This is also called the base voltage, or C (or VBG). The two base resistors form a voltage divider across the supply VCC. The voltage divider equation is:
R2 VCC R1 R2
So, V B =……….V. 5. Assume a 0.7 V drop from base to emitter for silicon transistor and a 0.2 V drop for germanium transistors. Why we assume that? Calculate VE by subtracting this drop from VB:
VE VB 0.7
So, V E =……….V. 6. Calculate the emitter current using the Ohm’s law:
So, I E =……….mA. 7. Assume that the collector current equals the emitter current:
IC I E
So, I C =……….mA. 8. Calculate the voltage drop across the collector resistor RC (which is also called as load resistor) using the Ohm’s law:
V RC I C RC
So, VC =……….V. 9. Calculate the collector-to-emitter drop using Kirchoff’s law:
VCE VCC V RC V E
So, V CE =……….V. 10. This six-step (4-9) process is not exact, but is accurate enough for practical work. 11. Now switch on the power supply and using multi-meter measure the various voltages
V B , V E , VC , V CE , and VEB . Records these voltages.
12. How close are they to theoretical values? If they are nearly the same that means that your circuit connection are correct and you can now switch on to the next part.
Part B: Voltage gain at mid frequency (1 kHz). 1. Setting function generator: Connect the function generator to the oscilloscope using the probes provided. Set the function generator output at low voltage (about 50 mV peak to peak, sinusoidal) and at 1 kHz frequency. Switch off the function generator after setting it at desired level of voltage and frequency.
2. Now connect the rest of the circuit shown in fig.10. Be careful about the polarity of the capacitors. 3. Connect the probes, one at input (across the –ive end of capacitor C1 and the ground) and the other at the output (across the –ive end of capacitor C2 and the ground), to the channel 1 and 2 of the oscilloscope to measure the input and the output. 4. Switch on the power supply VCC. 5. Observe the output signal. Is it showing any sign of saturation in terms of clipping? Slowly increase the input function generator voltage (without changing its frequency), you can observe the clipping of the output at some input voltage. That is where you are at saturation. We don’t measure voltage gain of this distorted clipped output waveform as it will not give the correct value. Slowly reduce the input voltage and adjust the output for maximum undistorted sine wave. 6. Record the input and output peak-to-peak voltages and calculate the voltage gain.
Part C: Frequency response of transistor amplifier. 1 This part is just the extension of previous part to see how the voltage gain of a transistor affected with the change in the frequency of the input signal. 2 Reduce the frequency of the input signal to 10 Hz and record the input and output voltages. 3 Gradually increase the frequency of the input signal and keep recording the corresponding input and output voltages. 4 5 Calculate and record the voltage gain at each frequency. Plot a Vgain v/s log10(f) graph.
The diode and transistors circuits that we have dealt so far are called as discrete circuits. The word discrete means separate or distinct. It refers to using separate resistors and transistors in building a circuit. The discrete circuit is one where all the components have been soldered together or otherwise mechanically connected. The integrated circuit (IC) was first developed in 1960s to do away the need to mechanically connect discrete components. The internal
components in an IC are not discrete; they are integrated. The IC is a device that includes its own resistors and transistors. One of the first ICs to be manufactured was the operational amplifier, commonly called as op amp. At one time, the op amps were built as discrete circuits. The operational amplifier is called so as such it can be used to perform mathematical operations of summing, integration, and differentiation. But to be more precise, op amp is a high gain, direct-coupled differential linear amplifier whose response characteristics are externally controlled by negative feedback from the output to the input. Op amps are also used as video and audio amplifiers, oscillators and so on, in communication electronics.
The symbol for an op amp is shown in fig.9. Note that the op amp is basically a differential amplifier, an amplifier that amplifies the voltage difference at its two inputs. The op amp, therefore like differential amplifier, has two inputs marked (-) and (+). The minus is the inverting input. A signal applied to the minus terminal will be shifted in phase by 180o at the output. The plus input is the non-inverting input. A signal applied to the plus terminal will appear in the same phase at the output as at the input. Because of the complexity of the internal circuitry of an op amp, shown in fig.10, op amp symbol is used exclusively in the circuit diagrams.
Fig.9: Symbol for Op Amp.
Fig.10: Simplified schematic for op amp 741.
General Characteristics of Op Amp
It not possible to list all the characteristics of op amp. The following list represents some general characteristics for a typical op amp, like that of most commonly used 741C. 1. High Voltage gain: 200,000 2. High Input impedance: 2 M 3. Low output impedance: 75 4. Offset adjustment range: 15 mV
Input Bias current
For the circuit, shown in fig.10, to work one need to connect the supplies VCC and VEE . If the inputs (inverting and non-inverting), which are base of Q1 and Q2, are grounded through equal resistances R B as shown in fig.11, the input current flowing into the
Fig.11: Input bias current.
two bases may not be same. The slightly unequal base currents flowing through the external resistance
produce small differential input voltage or unbalance; this represents a false input signal. When amplified, this small input unbalance produces an offset in the output voltage. The input bias current shown on data sheets is average of the two input current. It tells approximately what each input current is. The smaller the input bias current, the smaller the possible unbalance. The 741 has a worst-case input bias current of 500 nA.
Input offset Voltage
Ideally, the output voltage should be 0 when the voltage between the inverting and non-inverting inputs is 0. In reality,
Fig.12: Output offset voltage.
the output voltage may still have a slight offset or unbalance. This output offset is caused by internal mismatches,
tolerances and so on. In other words, even if one short-circuit the inverting and non-inverting inputs together to eliminate the effect of input bias, as shown in fig.12, the output may still have a slight offset from 0. The input offset voltage is the differential input voltage between the inverting and non-inverting inputs needed to null or zero the quiescent output voltage.
Operational Amplifier as Inverting and Non-Inverting Amplifier
Fig.13 shows the basic circuit, including the negative feedback loop of an op amp. Note that the output is fed back to the inverting input terminal in order to provide negative feedback for the amplifier. The input signal is applied to the inverting input. As a result the output will be inverted, that is, it will be out of phase by 180o. Such an amplifier is called as
Fig.13: Negative feed loop for inverting amplifier.
inverting amplifier. For such an amplifier the
output of the amplifier is given by the equation
RF Vin RR
The minus sign indicates that the sign of the output is inverted as compared to the input. The gain of inverting amplifier is therefore
Vout R F Vin RR
Fig.14: Non inverting amplifier.
It is possible to operate op amp as a non-inverting amplifier by applying the signal to the noninverting (+) input, as shown in fig.14. Note that the feed is still connected to the inverting input. The output voltage for a non–inverting amplifier is given by
R Vout F 1Vin R R
and the corresponding gain is
Vout RF 1 Vin RR
Activity 4: Measurement of gains of inverting and non-inverting amplifiers. Components and equipment required: IC 741C op amp, the general purpose kit with breadboard, Resistors of your own choice, 1 k pot (on the kit), +15 V and –15 V supplies (use the one on the kit), two multi-meters, function generator (on the kit), and oscilloscope.
Fig.15: Circuit for inverting amplifier
Procedure Inverting Amplifier
1. Connect the circuit of fig.15. Initially remove the input ac source (function generator) and ground the 1 k resistor. 2. Measure if there is any output offset voltage at terminal 6. 3. If output offset voltage is appreciable (or in any case), adjust it to null using a 1 k potentiometer connect across pin 1,4 and 5 of the op amp circuit (as explained by instructor, please ask if not explained). Leave the pot there after nullifying the output offset voltage. 4. For the circuit shown in fig.15. The voltage gain is 10. Therefore, at no instance the input voltage should exceed (15-2)/10= 1.3 V. Set input voltage accordingly. 5. Record output voltages for different input dc voltages and measure the gain. 6. Notice that the output is out of phase with respect to the input.
Choosing your own set of resistors construct a non-inverting amplifier with the gain of 40. Draw a neat circuit. Write down the detailed procedure. Clearly indicate the precautions to be taken while performing the experiment. Demonstrate your experiment, using dc input voltages only, to one of the lecturers in the class. How much is the error? How to minimize this error? Did you take enough measures to minimize it?
Op-Amp as an Integrator
Fig.16: Op-amp as an integrator.
1. The op-amp can be used as integrator using circuit shown if fig.16. 2. Set the function generator at 1 kHz square wave. 3. Connect the circuit shown in fig.16b with R1 R 10 k , C=0.2 F. 4. The shunt resistor R 2 is generally used to avoid the saturation of op-amp by the input offset. Typically this resistance is 5 to 10 times the input resistance R 1 . The shunt resistance has virtually no effect on the output, provided the input frequency is much greater (at least 10 times) than
1 2R2 C
5. Using above relation as the guiding rule, calculate the value of shunt resistance that you would like to use in your circuit. 6. Set the input square wave to op-amp at 10 volts peak-to-peak. 7. Verify that the output is triangular wave with peak-to-peak value of
Vout ( P P )
Vin ( P P ) 4 fR1C
8. Explain why you get triangular wave at the output? 9. Set the input to sinusoidal wave. What is the expected output? Observe the input and output and from the observed signals explain how it is in accordance with the expected results?