Introduction to Electronics Signals Signals contain information about

Introduction to Electronics Signals § § Signals contain information about what is happening in the physical world v Weather information can be contained in signals that represent w Temperature, humidity, pressure, wind speed, etc. v The voice of an announcer contains information about the game To extract information from a signal it needs to be processed in some manner, most conveniently by electronic means. v But what if the signal is not an electrical signal? v It has to be converted to a current or voltage. v A transducer converts the signal into an electronic form or from and electronic form w Sound waves from your voice can be converted to an electronic signal by a microphone (pressure transducer) w A wide variety of transducers exist w Transducers will be studied in other courses, in this course will be interested in how the electronic signals are manipulated © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -1

Introduction to Electronics Forms of the signal § The electronic signals can be represented in two different but equivalent forms v Thevenin (voltage) form w A voltage source and a series source resistance, Rs w Preferred when Rs is low v and the Norton (current) form w A current source and a parallel source resistance, Rs w Preferred when Rs is high © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -2

Introduction to Electronics A time-varying signal § § Information content is contained in the change in the magnitude of the signal over time In general it is not easy to mathematically describe an arbitrary looking waveform, but it is very important to have such a description in order to design the appropriate signal processing circuits to perform the desired functions or operations on the given signal Analog Signal Amplitude Continuously time-varying signal time © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -3

Introduction to Electronics The Frequency Spectrum of Signals § § An extremely useful way to characterize a signal or arbitrary waveform (time function) is in terms of its frequency spectrum The frequency spectrum description is obtained through the use of mathematical tools such as the Fourier series and Fourier transform (we will only introduce the topic here and come back to it in future courses) These tools provide the means for representing a voltage signal vs(t) or a current signal is(t) as the sum of sine-wave signals of different frequencies and amplitudes This makes the sine-wave a very important signal in the analysis and testing of electronic circuits so we will briefly review the properties of the sinusoid. T § Where Va denotes the peak value or amplitude in volts and w denotes the angular frequency in radians per second, that is w = 2 pf radians per second, where f is the frequency in hertz, f = 1/T Hz, and T is the period in seconds © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -4

Introduction to Electronics More on the sinusoid § § The sine-wave is completely characterized by three quantities v it’s peak value, Va v it’s frequency, w v it’s phase, f with respect to some arbitrary reference time. On the previous page the time origin was chosen so that the phase angle is zero. It is common to express the amplitude of the sine-wave signal in terms of its root-meansquare (rms) value, which is its peak value divided by the square root of two. Thus the rms value of the sinusoid on the previous page is Va divided by the square root of two. v For example a 120 V wall ac power supply is an rms value so that the peak value is actually 120 times the square root of two. © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -5

Introduction to Electronics A Square-wave signal and its Fourier Series § § The Fourier series representation can also be used in the specific case where the signal is a periodic function of time The waveform is described by the sum of sinusoids that are harmonically (multiples of some base frequency) related, for instance, take the symmetrical (both positive and negative amplitude) square wave shown below Fundamental (base) frequency T § It can be expressed as the following infinite Fourier series § Because the amplitude of the harmonics decrease with increasing frequency we can truncate the series at some point (make it a finite series) with little error © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -6

Introduction to Electronics The Frequency Spectrum § The sinusoidal components of the square-wave signal constitute the frequency spectrum of the this particular signal as shown below § The frequency spectrum of an arbitrary non-periodic waveform has a continuous distribution of frequencies (all frequencies are present in some degree) but much of the information is usually confined to a small part of the frequency spectrum which can be useful in signal processing applications. time © REP 11/2/2020 EGRE 224 frequency Introduction, Page 1. 1 -7

Introduction to Electronics Time domain vs. Frequency domain § The spectrum of audible sounds such as speech and music goes from roughly 20 Hz to 20 k. Hz (audio band) © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -8

Introduction to Electronics Exercises § 1. 1 - Find the frequencies f, and w of a sine-wave signal with a period 1 ms § 1. 2 - What is the period of sine waveforms characterized by frequencies of v a) f = 60 Hz v b) f = 10 -3 Hz v c) f = 1 MHz © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -9

Introduction to Electronics Exercise 1. 3 § When a symmetrical square-wave signal whose Fourier series is given by the following equation is applied to a resistor the total power dissipated may be calculated directly from the frequency spectrum using the relationship § § The total power can also be found by summing the contribution of each of the harmonic components, which can be found from using the RMS values, that is, P = P 1 + P 3 + P 5 + … What fraction of a square wave’s power is in its fundamental? § What fraction of a square wave’s power is in its first five harmonics (through 5 w 0)? © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -10

Introduction to Electronics Analog vs. Digital Signals Analog Continuous time-varying signal time Digital - only two discrete levels Analog Discrete time-varying signal time 1 0 1 © REP 11/2/2020 EGRE 224 1 0 0 1 0 Introduction, Page 1. 1 -11

Introduction to Electronics § § Binary (base 2) numbers, powers of 2 20=1, 21=2, 22=4, 23=8, etc. § § § A single binary digit (or bit as it is called) can be either a zero or a one Assign each bit a weighting value based on powers of two, for example 8’s place, 4’s place, 2’s place and 1’s place Binary numbers are listed from MSB to LSB MSB - Most significant bit (8’s place) and LSB - least significant bit (1’s place) § A decimal number (base 10) such as 12 has two digits one in the tens place and one in the ones place. Each digit can be 0 -9 in base 10. There are 100 or 10 2 (10 to the number of digits) possible combinations (0 -99). § A binary number has two possible values for each digit (0 or a 1) so a four bit binary number will have 24 or 16 combinations © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -12

Introduction to Electronics Analog to Digital Conversion Analog Digital Continuously varying signal time 1 volt resolution Discrete voltage varying signal 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Discrete time steps as well © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -13

Introduction to Electronics Amplifiers § § § Most transducers provide signals that are in the microvolt or millivolt range and have little energy These signals in their raw state are too weak for reliable processing Stronger signals are easier to process, therefore we need an amplifier Amplifiers should be linear so that the information being amplified is not being changed or distorted in an unwanted way A distortion free amplifier can be described by the following relationship vo(t) = Avi(t) Where vi and vo are the input and output signals and A is a constant which represents the amplifier gain. The previous equation describes a linear amplifier © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -14

Introduction to Electronics Voltage Gain § § Triangle of the symbol points in the direction of signal flow Signals can each have different reference level or share a common reference output input Transfer Characteristic Voltage gain (Av ) = i. I v. I(t) i. O RL v. O v. I Av + v. O(t) 0 1 v. I © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -15

Introduction to Electronics Power Gain and Current Gain § § § We are trying to increase the signal power (its ability to do work) The additional power is usually externally supplied The power is to be delivered to the Load ( RL) i. O = v. O / R L Power gain (Ap ) = Load Power PL Input Power PI Current gain (Ai ) = v. O i. O v. I i. O i. I Ap = Av Ai © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -16

Introduction to Electronics Expressing Gain in Decibels § § Voltage and Current Gain are defined as a ratio of quantities with the same dimension so they may be written as unit less or as Sedra and Smith prefer, V/V or I/I for emphasis In many instances a logarithmic expression is preferred Voltage gain in decibels = 20 log Av Current gain in decibels = 20 log Ai Recall that P=V 2/R or I 2 R , the squared term causes the power gain in decibels to be 1/2 Power gain in decibels = 10 log Ap § § Absolute values are used because in some cases the gain may be negative Negative gain simply means that there is a phase change of 180 degrees Attenuation (weakening) of a signal is indicated by a gain of less than one Amplification (strengthening) of a signal is indicated by a gain of more than one © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -17

Introduction to Electronics Amplifier Efficiency and Power Supplies i. I v. I(t) i 1 i 2 V 1 V+ i. O V+ RL -V 2 + v. O(t) - DC Power delivered to the amplifier is Pdc = V 1 I 1 + V 2 I 2 Some power is delivered into the circuit by the power supplies and some is delivered by the input signal Pdc + PI Some power is consumed by the load and some is dissapated by the amplifier circuit. The Efficiency is defined as © REP 11/2/2020 EGRE 224 Pdc + PI = PL + Pdissapated PL x 100 h= Pdc Introduction, Page 1. 1 -18

Introduction to Electronics Example 1. 1 § § Consider an amplifier which uses +/- 10 V supplies. The input signal is a 1 V peak sinusoid while the current is 0. 1 m. A peak. The output delivers a 9 V peak sinusoid into a 1 k. W load. The amplifier draws 9. 5 m. A from each power supply. Find the voltage gain, the current gain, the power drawn from the dc supplies, the power dissipated in the amplifier, in the load and the amplifier efficiency © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -19

Introduction to Electronics Amplifier Transfer Characteristic, Linear to Saturation § § v. O L+ is the positive output saturation level L- is the negative output saturation level Output peaks are clipped or distorted L+ Undistorted Output Range v. I 0 The steeper the slope of the transfer characteristic the higher the gain LAv © REP 11/2/2020 EGRE 224 L- Safe input swing range L+ Av Introduction, Page 1. 1 -20

Introduction to Electronics Nonlinear Transfer Characteristics and Biasing § § § Bias the circuit so that it operates in a “linear” region of the non-linear characteristic for small input signal levels The dc operating point or quiescent operating point (Q point, from audio, where we have quiet or “off” levels) The gain is given by the slope of the transfer characteristic at the Q point L+ Slope=Av Q LAv = © REP 11/2/2020 EGRE 224 dv. O dv. I at Q Introduction, Page 1. 1 -21

Introduction to Electronics Biasing Example § A transistor amplifier has the following transfer characteristic § § The characteristic applies for Find the limits (L- and L+, and the corresponding values of v. I. ) v The L- limit is obviously 0. 3 V based on the range restriction given above v By substitution into the transfer characteristic equation and solving for v. I we get 0. 69 V v The L+ limit is determined by where v. I =0, and thus 10 § Also find the value of the dc bias voltage VI that results in VO=5 V and find the voltage gain at the corresponding operating point. v To bias the device so that VO=5 V we use the equation and solve for v. I. 5 w v. I =0. 673 V v We can find the gain by taking the derivative at v. I =0. 673 V 0. 3 v Note the negative gain which indicates that we have an inverting amplifier © REP 11/2/2020 EGRE 224 0. 673 0. 69 Introduction, Page 1. 1 -22

Introduction to Electronics Symbol Convention § Total instantaneous quantities (have a dc and an ac component or are arbitrary) are denoted by a lower case symbol and an upper case subscript § Direct-current (dc) quantities are denoted by an upper case symbol and an upper case subscript. Say to yourself, if they are both uppercase it is a dc quantity § Incremental or ac quantities are denoted by a lowercase symbol and a lower case subscript. Incremental ac value total value © REP 11/2/2020 EGRE 224 Peak ac value dc value Introduction, Page 1. 1 -23

Introduction to Electronics Some more Exercises § An amplifier has a voltage gain of 100 V/V and a current gain of 1000 A/A. Express the voltage and current gain in decibels and find the power gain in d. B. § An amplifier operating from a single 15 V power supply provides a 12 V peak-to-peak sine -wave signal to a 1 k. W load, and draws negligible input current from the signal source. The dc current drawn from the 15 V supply is 8 m. A. What is the power dissipated in the amplifier and what is the amplifier efficiency? © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -24

Introduction to Electronics Small-signal approximation Exercise § Lets look at the limitation of the small signal approximation. Consider amplifier with the following transfer characteristic § 10 5 output swing Input swing A positive input signal of 1 m. V is superimposed on the dc bias voltage VI. § Find the corresponding output for the following two situations 1) Use the transfer characteristic of the amplifier 2) Assume the amplifier gain is linear about the bias point (gain of -200) 1) Let v. I=VI+vi and v. O=VO+vo=5+vo thus, 0. 3 0. 673 0. 69 2) © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -25

Introduction to Electronics Small-signal approximation Exercise continued § § Repeat the calculations for input signals of 5 m. V and 10 m. V 5 m. V v Using the approximation v Using the full equation § 10 m. V v Using the approximation v Using the full equation © REP 11/2/2020 EGRE 224 The larger the signal the larger the error when using the approximation Introduction, Page 1. 1 -26

Introduction to Electronics Circuit Models for Voltage Amplifiers § § § Simple but effective. Input resistance accounts for the power drawn from the signal source a the output resistance accounts for changes in output voltage as current is supplied to the load. The model might internally represent from one to to 20 or more transistors Model values determined by analysis and calculation or by direct measurement Ro + vi § § § Ri +- Avovi + vo - A voltage controlled voltage source which has a gain factor of Avo An input resistance of Ri that accounts for the fact that the amplifier draws an input current from the signal source. An output resistance Ro that accounts for the change in output voltage as the load resistance (hence the load current supplied by the amplifier) is changed © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -27

Introduction to Electronics Using Amplifier Circuit Models with a Signal Source and a Load § § § The non-zero output resistance causes only a fraction of the Avovi to appear across the output (load) as shown on the right below. It follows that in order to reduce the losses in coupling the amplifier to the load, the output resistance, R o should be as LOW as possible (much lower than RL). If there is no load, RL is infinitely large (open circuit) then the voltage gain A V becomes AVo or as it is known, the open-circuit voltage gain. When you specify the voltage gain of an amplifier you must also specify the value of the load resistance the gain is measured at. Using voltage division In order not to lose signal strength at the input, Ri should be infinitely large or much, much greater than Rs vo = Avovi is vs Rs © REP 11/2/2020 EGRE 224 Ro + vi - Ri +- Avovi RL RL + R o io RL + vo - Introduction, Page 1. 1 -28

Introduction to Electronics Comment on the Voltage Amplifier § The overall voltage gain is given by § A buffer amplifier, has high input resistance and low output resistance. A buffer has minimal voltage gain but can have a significant POWER gain (see exercise 1. 8). © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -29

Introduction to Electronics An Example of Cascaded Amplifier Stages Signal source Stage 1 Stage 2 Stage 3 Load ii 100 k. W vi 1 § vi 2 1 MW vs § 1 k. W 10 vi 1 100 k. W 100 vi 2 10 W vi 3 10 k. W 100 W 1 vi 3 The amplifier shown is a cascade of three separate stages. It is fed by a signal source and delivers its output to a load resistance of 100 W. Find the fraction of the source signal appearing at the input terminals of the amplifier v Use voltage division © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -30

Introduction to Electronics Example continued Signal source Stage 1 Stage 2 Stage 3 Load ii 100 k. W vi 1 vi 2 1 MW vs § § 1 k. W 10 vi 1 100 k. W 100 vi 2 10 W vi 3 10 k. W 100 W 1 vi 3 Find the gain of each of the three stages and the three in cascade, For the first stage consider the input resistance of the second stage to be the load resistance of the first stage © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -31

Introduction to Electronics Example continued Signal source Stage 1 Stage 2 Stage 3 Load ii 100 k. W 1 k. W vi 1 vi 2 1 MW vs 10 vi 1 100 k. W 100 vi 2 10 W vi 3 10 k. W 100 W 1 vi 3 § Find the gain from source to load. Here we simply realize that some of the signal is lost due to voltage division between the source and the first stage and we already calculated that only 90. 9% gets to the input, so § Find the current gain and the power gain © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -32

Introduction to Electronics Exercise to Determine the Benefits of Using a Buffer § 1. 8 - A transducer produces 1 Volt rms and has a resistance of 1 MW and its output is to drive a 10 W load. If the transducer is connected directly to the load what voltage and power levels are seen at the load (draw a circuit diagram, it always helps). ii 1 MW Signal source (transducer) vout 10 W vs § If a unity gain buffer amplifier (AV=1) with a 1 MW input resistance and 10 W output resistance is connected between the transducer and the load, what are the output voltage and power levels now? Find the gain (in d. B) from source to load and the ac power gain (also in d. B). Buffer ii 1 MW Signal vs source (transducer) 10 W vi 1 © REP 11/2/2020 EGRE 224 vout 1 MW 10 W 1 vi 1 Introduction, Page 1. 1 -33

Introduction to Electronics Exercise Continued © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -34

Introduction to Electronics More Exercises § 1. 9 - The output voltage of an amplifier is found to decrease by 20% when a total load resistance of 1 k. W is connected. What is the value of the amplifier output resistance? v We can view this as a voltage division problem between the amplifier output resistance and the load resistance § 1. 10 - An amplifier with a voltage gain of +40 d. B, an input resistance of 10 k. W, and an output resistance of 1 k. W is used to drive a 1 k. W load. What is the value of the voltage gain AV? Find the value of power gain in d. B. v Since the load resistance is equal to the output resistance the voltage gain will be the same as the amplifier gain © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -35

Introduction to Electronics Other Amplifier Types Current Amp Voltage Amp Ro + vi - Ri +- io Avovi + vo - Ri R o io Ri +- © REP 11/2/2020 EGRE 224 R mi i Aisii Ro + vo - Transconductance Amp Transresistance Amp ii io + vo - + vi - Ri Gm v i Ro + vo - Introduction, Page 1. 1 -36

Introduction to Electronics Voltage Amplifier § Voltage in and voltage out, V/V is unit-less Voltage Amp Ro + vi - © REP 11/2/2020 EGRE 224 Ri +- Avovi io + vo - Introduction, Page 1. 1 -37

Introduction to Electronics Current Amplifier § Current in and current out, I/I is unit-less Current Amp io ii Ri © REP 11/2/2020 EGRE 224 Aisii Ro + vo - Introduction, Page 1. 1 -38

Introduction to Electronics Transconductance Amplifier § Voltage in and current out, I/V=G Transconductance Amp io + vi - © REP 11/2/2020 EGRE 224 Ri Gm v i Ro + vo - Introduction, Page 1. 1 -39

Introduction to Electronics Transresistance Amplifier § Transresistance Amp Current in and a voltage out, V/I =R ii R o io Ri © REP 11/2/2020 EGRE 224 +- R mi i Introduction, Page 1. 1 -40 + vo -

Introduction to Electronics Bipolar Junction Transistor Example § A bipolar junction transistor (BJT) is a three-terminal device with the following symbol. C B i. B + v. BE § § - E i. C + v. CE i. E The terminals are called the emitter (E), the collector (C) and the base (B). The device is basically non-linear (relationships between the terminal voltages and terminal currents). We can use the small-signal linear approximation about a quiescent bias point as we discussed in section 1. 4. The total instantaneous terminal voltage and current quantities can be expressed using the conventions presented earlier. v. BE = VBE + vbe i B = IB + i b i E = IE + i e i C = IC + i c v. CE = VCE + vce § The device itself may be represented by one of several equivalent circuit models © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -41

Introduction to Electronics Two Equivalent (p) Circuit Models for Bipolar Junction Transistors § Notice that the components of the two models are the same but that they use different dependencies for the current source (one a current and the other a voltage). ib = vbe rp ib B + vbe - rp Current Amplifier bib ie E B ib rp i. C + vbe - gmvbe E © REP 11/2/2020 EGRE 224 ie = ib+ ic=(b + 1)ib ic = gmvbe C gm = ie ic = b ib C i. C Trans-conductance Amplifier b rp P model name comes from the fact that the circuit configuration looks like an upside down greek letter p Introduction, Page 1. 1 -42

Introduction to Electronics Why so many different models of the same device? § § Different models are used in different circuit configurations in order to make the derivation of the circuit behavior easier. For example, the p models on the previous page are easy to use when the input signal source resistance appears in series with rp in the model since they can be combined by adding. Other circuit configurations become easier to analyze when other models are used It takes a little experience to pick the right models to simplify your design work. The alternative is to use computer algorithms to solve the complex circuits regardless of the choice of models, but the designer loses some of the intuitive feel for the behavior of the circuit when they do that. Another issue is conversion between model types. What if someone gives me parameters for one model and I find that a different model simplifies the analysis? v Equations for each model can usually be linked and solved such that the parameter values for one model can be found in terms of the other model parameters allowing conversion between models © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -43

Introduction to Electronics Low Frequency “T” Model for common base configuration § B ib i. C + vbe - re a ie C ie = vbe re But what are a and re? From the p model we can write ic = ie E Therefore + v 1 re ic = a ie ie=ib +ic therefore ib= (1 -a )ie ie E We can derive expressions that allow us to convert from one model form to another T gmv 1 i. C B © REP 11/2/2020 EGRE 224 C b b +1 b a = b +1 ie = b +1 rp re = vbe rp b +1 The “T” model is useful in the Common base configuration Introduction, Page 1. 1 -44

Introduction to Electronics Input Resistance in an active device (BJT) § 1. 14 - Find the input resistance between the base and ground in the circuit shown below. The voltage vx is a test voltage and the input resistance is defined as Rin = vx/ix. ix ib B rp vx i. C + vbe - E Rin G © REP 11/2/2020 EGRE 224 Using KCL at the emitter node we get § Now using KVL around the input loop we get C bib ie § RL ie=ib +ic or ic= (b+1 )ib= (b+1 )ix vx=vbe+v. E=(ixrp)+ (b+1 )ix. Re or vx/ix= Rin=rp + (b+1 )Re Re § Note: We can’t “see” any resistances on the other side of the current source since the current source output is constant no matter what voltage appears across it. Introduction, Page 1. 1 -45

Introduction to Electronics Frequency Response of Amplifiers § § § We can make up any signal from a sum of different frequency and amplitude sinusoidal signals (Fourier series), so it is important to know how our amplifier responds to various different sinusoidal frequencies. Such a characterization of amplifier performance is called the amplifier frequency response. Various test instruments can assist in the characterization process as we shall see in lab. How can we apply a signal and measure the response? linear amplifier + - § § § The gain is given by Vo/Vi and the phase difference or phase angle is given by f. The gain and phase angle measured can change depending on the test frequency w used. An oscilloscope can be used to measure these quantities (see next page) The amplifier transmission or transfer function, T(w) versus frequency is found by making measurements over a wide range of frequencies. © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -46

Introduction to Electronics Measuring Frequency Response with an Oscilloscope § Low Frequency f input § output w= w 1 § Higher Frequency w= w 2 f © REP 11/2/2020 EGRE 224 input output § If the output is one half cycle behind the input then the phase difference is 180 degrees In this example it appears that the gain (Vo/Vi) goes up with increasing frequency and the phase difference (referenced to the input) goes down (gets smaller If we took data at many more frequencies we could construct a detailed plot of these two responses versus frequency This is a tedious process so the Gain-Phase Analyzer was invented to do this for us and plot the results Introduction, Page 1. 1 -47

Introduction to Electronics Single Time Constant (STC) Networks § § Lets consider a Low Pass filter The transfer function T(s) of an STC (single time constant) low-pass circuit can always be written in the form of the complex frequency variable, s § For physical frequencies, where s=jw, we get § § Where K is the magnitude of the transfer function at w=0 (dc) and w 0 is defined being 1/t. Tau is the time constant, which in this case is RC. On the next page we will look at the magnitude response of the transfer function for various frequencies and plot the results © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -48

Introduction to Electronics Sample Calculations Let R=100, 000 W and C equal 0. 1 m. F, t=RC is 0. 01, w 0=1/t=100 so the transfer function is Lets pick a frequency and calculate the magnitude of the response, LET w=1 Plot the denominator as a complex number +jw -Re The magnitude of the denominator is given by the length of the complex vector. Using trigonometry we know that c 2 =a 2+b 2 so that c is 0. 01 1 +Re The phase angle of the denominator is given by the angle the complex vector makes with the positive real axis. Using trigonometry we know that this angle is given by the inverse tangent of imaginary part over the real part -jw © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -49

Introduction to Electronics Continuing with the example § § We looked at the denominator so the total magnitude response is the magnitude of the numerator divided by the magnitude of the denominator, or 1/1 =1 The total phase response is the phase angle on the numerator (zero) minus the phase angle of the denominator (since it is division), or -f Now lets consider a different frequency w=10. Remember w 0 is 100. And w=100 +jw -Re 1 +Re 0. 1 1 +Re -jw © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -50

Introduction to Electronics and at higher frequencies § Finally lets consider a frequency of w=1000. Remember w 0 is 100. -jw 10 -Re 1 +Re -jw § What we are observing is this capacitor gradually changing its behavior from an open circuit at low frequencies (w<<w 0) to a short circuit at higher frequencies (w>>w 0) which shorts out the output response © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -51

Introduction to Electronics The General Expression (Low Pass) Thus the low pass magnitude response in general is given by The general low pass phase response is given by © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -52

Introduction to Electronics Plots of the Frequency Response Magnitude 1 0 1 10 100 1 k Radian Frequency Phase Angle of the transfer function © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -53

Introduction to Electronics Bode (Bo-duh) Plots of the Frequency Response (Low Pass) Normalized Magnitude § § A decade is a factor of 10 and an octave is a doubling Note § and 0 -10 -20 -30 0. 1 1 10 Normalized Frequency 0. 1 © REP 11/2/2020 EGRE 224 1 10 Introduction, Page 1. 1 -54

Introduction to Electronics Single Time Constant (STC) High Pass Network § § Lets consider a High Pass filter now The transfer function T(s) of an STC (single time constant) low-pass circuit can always be written in the form of the complex frequency variable, s § For physical frequencies, where s=jw, we get § Where K is the magnitude of the transfer function at w=0 (dc) and w 0 is defined as being 1/t. Tau is the time constant, which in this case is 1/RC. § © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -55

Introduction to Electronics The General Expression (High Pass) Thus the high pass magnitude response in general is given by The general high pass phase response is given by In this case the capacitor is and open at low frequencies and the output is zero, but at higher frequencies the capacitor becomes a short circuit and the output is finite © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -56

Introduction to Electronics Bode (Bo-duh) Plots of the Frequency Response (High Pass) Normalized Magnitude 0 -10 -20 -30 0. 1 1 10 Normalized Frequency slope 0. 1 1 © REP 11/2/2020 EGRE 224 10 Introduction, Page 1. 1 -57

Introduction to Electronics Bode Plots of Factored Transfer Functions z 2 z 1 p 1 z 2 p 3 I expected everyone to do the problem graphically and by calculation p 1 T(jw) in d. B p 2 83. 5 80 43. 5 40 p 3 43. 5 1 z 1 © REP 11/2/2020 EGRE 224 10 102 103 104 105 106 107 108 z 2 p 3 p 1 p 2 Introduction, Page 1. 1 -58

Introduction to Electronics Example 1. 5 § The circuit below shows a voltage amplifier having an input resistance Ri, an input capacitance Ci, a gain factor m, and an output resistance Ro. The amplifier is fed with a voltage source Vs having a source resistance Rs, and a load resistance RL is connected to the output. Rs Vs § Ro + Vi - Ri Ci +- m. V i RL + Vo - Derive an expression for the amplifier voltage gain Vo/Vs as a function of frequency. From this find expressions for the dc gain and the 3 -d. B frequency. © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -59

Introduction to Electronics Example 1. 5 cont’d Using the voltage divider rule: , Thus, Low pass STC network At the output side of the amplifier: Amplifier transfer function: © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -60

Introduction to Electronics Example 1. 5 cont’d § By inspection, we see that the input circuit is a STC network. We can find the time constant by reducing Vs to zero, resulting in the resistance seen by Ci is Ri // Rs. (from previous page) dc gain: 3 -d. B frequency: © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -61

Introduction to Electronics Example 1. 5 cont’d § Calculate the values of the dc gain, the 3 -db frequency, and the frequency at which the gain becomes 0 db (i. e. , unity) for the case Rs = 20 k. W, Ri = 100 KW, Ci = 60 p. F, m = 144 V/V, Ro = 200 W, and RL = 1 k. W. V/V d. B rad/s § Since the gain falls off at a rate of -20 db/decade, the gain will reach 0 d. B in two decades (starting at w 0). Unity-gain frequency = 108 rad/s or 15. 92 MHz © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -62

Introduction to Electronics Example 1. 5 cont’d § Find vo for vi = 0. 1 sin 102 t, V V § Find vo for vi = 0. 1 sin 105 t, V V § Find vo for vi = 0. 1 sin 106 t, V V © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -63

Introduction to Electronics Classification of Amplifiers based on Frequency Response § Capacitively coupled amplifier decibels (d. B) mid-band frequencies bandwidth § Directly coupled amplifier decibels (d. B) bandwidth § Tuned or bandpass amplifier decibels (d. B) bandwidth © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -64

Introduction to Electronics Frequency Dependent Gain § § § INTERNAL device (transistor) capacitances (in th p. F range) cause the gain of amplifiers to fall off at high frequencies. The gain at low frequencies usually falls off at low frequencies due to coupling capacitors. Coupling capacitors are used to couple (connect) one amplifier stage to another. Coupling capacitors simplify the design of each stage by allowing the designer to separately bias each stage. The coupling capacitors are usually quite large (0. 1 m. F - 50 m. F typically). This value range is chosen so that the coupling capacitors will have a low resistance (be short circuits) at the frequencies of interest. The internal capacitance will still be open circuits at midband frequencies. Coupling capacitors cause the gain to be zero at dc § A directly coupled amplifier has gain down to dc and, integrated circuit fabrication techniques do not easily allow for the fabrication of the large value capacitors required for capacitively coupling amplifiers, so most are direct coupled. § Tuned amplifiers are often needed in the design of radio and television circuits and are used in the tuner part of the receiver (allows you to select the channel). The center frequency is adjusted to match the broadcast frequency. Other frequencies (channels) outside of the pass band are not amplified or received. © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -65

Introduction to Electronics The complex frequency plane § § § § Complex frequency is a unifying concept. It ties together v resistive circuit analysis v steady state sinusoidal analysis v transient analysis v forced response v complete response and the analysis of of circuits excited by exponential forcing functions and exponentially damped sinusoidal forcing functions v They all become special cases of one general technique which is associated with the complex frequency concept Sigma is the damping level (real axis), + is growing and - is decreasing Omega is the frequency sigma = 0 and omega = 0 is a dc signal sigma + and omega = 0 is an increasing exponential sigma - and omega = 0 is a decreasing exponential See page 420 of Hayt and Kemmerly © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -66

Introduction to Electronics The Inverter § § § An inverter is the fundamental circuit in digital logic as the amplifier is in analog circuitry, in fact, an inverter is an amplifier We have looked at amplifier transfer characteristics and determined that by proper biasing (choosing a Q point) and use of a small ac signal we can use a non-linear characteristic to create a linear amplifier. In digital logic we make use of the gross nonlinearity of the characteristic since we are only interested in the two (binary) saturation levels and not in the transition region in between. One saturation region is called a “ 1”, or high or true. The signal is usually a voltage level but current signals are sometimes used as well The other saturation region is called a “ 0”, or low or false In positive logic a 1 is a more positive voltage than a 0, in a negative logic system the more negative voltage is defined to be a 1. Negative logic is commonly used with logic circuits made up of only p-channel metal oxide semiconductor field effect transistors (MOSFETs). The voltage transfer functions of several inverters are shown on the next page. Some key points to observe are: v The slope at any point on the VTC is the gain of the inverter v If the slope is greater than one (steep) the gain is high and if the slope is less than one we have attenuation v A slope of -1 on the VTC is a 45 degree line (provided both axes are scaled the same). The minus sign come from the fact that it is an inverter. v The point at which the output voltage is equal to the input voltage is called Vinvert and is of interest in digital logic (ideally Vinvert should be 1/2 the supply voltage). © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -67

Introduction to Electronics The Voltage Transfer Characteristic (VTC) of an Inverter § The gain is the change in Vout divided by the change in Vin or the slope of the VTC VOUT (Volts) 5 HIGH, “ 1” 5 V BJT Common. Emitter Inverter 4 3 Ideal Inverter GAIN=SLOPE 2 LOW, “ 0” 0 V CMOS Inverter 1 0 0 LOW, “ 0” 0 V © REP 11/2/2020 EGRE 224 1 2 3 VIN (Volts) 4 5 HIGH, “ 1” 5 V Introduction, Page 1. 1 -68

Introduction to Electronics n-channel MOS Transistors § Enhancement Mode Transistors. A normally open switch. At zero volts on the gate no current flows ( a positive Voltage must be applied to the gate to enhance a channel of electrons) Source Drain Gate n p substrate Substrate n Source n-channel Source - where electrons come from (-) Drain - where electrons flow to (+) IDS Channel is enhanced (resistive) § Depletion Mode Transistors. A normally closed switch. At zero volts on the gate a current flows ( a negative Voltage must be applied to the gate to deplete the channel of electrons) -VGS Channel is off +VGS 0 Source Gate Drain Gate n n p substrate IDS n-channel Substrate Source - where electrons come from (-) Drain - where electrons flow to (+) Channel is made stronger Channel is depleted (Off) -VGS © REP 11/2/2020 EGRE 224 0 +VGS Introduction, Page 1. 1 -69

Introduction to Electronics p-channel MOS Transistors § Enhancement Mode Transistors. A normally open switch. At zero volts on the gate no current flows ( a negative Voltage must be applied to the gate to enhance a channel of holes) Source Drain Gate p Substrate p Source p-channel n substrate 0 -VGS Source - where holes come from (+) Drain - where holes flow to (-) +VGS Channel is enhanced § Depletion Mode Transistors. A normally closed switch. At zero volts on the gate a current flows ( a positive Voltage must be applied to the gate to deplete the channel of electrons) IDS Source Drain Gate p p p-channel n substrate -VGS Substrate Source - where holes come from (+) Drain - where holes flow to (-) 0 +VGS Channel is depleted IDS © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -70

Introduction to Electronics Inverter operation § § § The figure on the left shows that when the switch is open (transistor is off) there is no connection between Vout and ground and there is no path for current flow through the resistor. If there is no current flow through the resistor then both ends must be at the same potential and Vout must be five volts. When the switch is closed (transistor is on) a low resistance path is formed between Vout and ground through the transistor. The output will be pulled towards ground since the on resistance of the transistor is less than the other resistance value and Vout will be close to zero volts. A current will flow from the 5 volt supply to ground when the transistor is on. Resistors take up too much space (area) +5 Vin=0 V Logic 0 +5 Vout=5 V Logic 1 open © REP 11/2/2020 EGRE 224 say R = 10 k Ohms Vin=5 V Logic 1 Vout ~ = 0. 5 Volts Logic 0 closed say Ron = 1 k Ohms Introduction, Page 1. 1 -71

Introduction to Electronics The Complementary MOS (CMOS) Inverter § § Complementary means both n. MOS and p. MOS transistors are used A polite “tug O’War”! Only one device pulls at a time v A High Voltage on Vin turns On the NMOS device and turns Off the PMOS device v A Low Voltage on Vin turns off the NMOS and turns On the PMOS v Power is dissipated only when the output is switching from low to high or high to low +5 V gate Vin source of holes p channel (n well) drain for holes Vgsp= Vgp-Vsp = 5 -5 = 0 +5 V g=5 s=5 open Vin = + 5 V Vout = 0 V Vout gate drain for electrons n channel (p wafer) source of electrons The source is where charge carriers come from and the drain is where they flow to, holes come from the higher voltage and flow towards a more negative terminal, electrons come from the more negative terminal and flow towards the positive © REP 11/2/2020 EGRE 224 closed g=5 s=0 Vgsn= Vgn-Vsn = 5 -0 = 5 +5 V Vgsp= Vgp-Vsp = 0 -5 = -5 g=0 s=5 closed Vin = 0 Vgsn= Vgn-Vsn = 0 -0 = 0 Vout = + 5 V g=0 open s=0 Introduction, Page 1. 1 -72

Introduction to Electronics Noise Margins § § § § In positive logic a “high” is more positive than a “low” v High is 5 V or High is 3. 3 V or 12 V, etc. v Low is zero In negative logic a “high” is more negative than a “low” v High is -10 V v Low is zero In a real binary (2 level) logic system we have to assign a “high” to a range of voltages and a “low” to another range of voltages v For example, let a high be 4 V < Vout < 5 V and v let a low be 0 V < Vout < 1 V The output of logic gates must fall within these ranges in order to clearly (solidly) convey to the outside world what value the gate is producing If Vout is 4. 2 V it is a high. If Vout is 0. 3 V it is a low. If Vout is 2. 1 V then the output is said to be in an illegal or undertermined logic state. Another problem is that we would like to be able to transmit these “solid” high and low signals to remote locations. In the process these signals might get degraded. So, we must be able to accept these degraded inputs (within reason) and still recognize them. v For example, A 4. 05 V signal generated by a logic gate is a solid high, but if it degrades to a 3. 7 V signal when it arrives, we must still recognize it as a “marginal” high. Noise margins are ranges adjacent to strong signal levels, levels which are also recognized as having the same level. If 92 and above is a strong “A”, a grade of 88 might be a marginal “A”. © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -73

Introduction to Electronics How are Noise Margins Determined? § § The slope of the voltage transfer characteristic of an inverter is the gain. There are three key points on the gain plot v The point at which the magnitude of the gain is first equal to unity (one, 45 degrees) v The point at which the magnitude is maximum v The point second point at which the gain is again equal to unity VOH VOUT 4 3 (Volts) 2 VOL |slope| = 1 5 |slope| = Maximum = 5 |slope| = 1 1 0 |gain| slope 5 1 © REP 11/2/2020 EGRE 224 1 VIL 2 3 VIH 4 5 VIN (Volts) max VIN (Volts) Introduction, Page 1. 1 -74

Introduction to Electronics What Noise Margins really mean § § On the previous page VIL was equal to 1. 2 V and VIH was equal to 3 V VOL was equal to 0. 7 V and VOH was equal to 4. 9 V § The Noise Margins are defined as follows NML = VIL - VOL in our case = 1. 2 - 0. 7 = 0. 5 Volts NMH = VOL - VIL = 4. 9 - 3. 0 = 1. 9 Volts What the output produces High VOH = (5 - 4. 9) V “solid” signals Low VOL = (0. 7 - 0) V © REP 11/2/2020 EGRE 224 What the input accepts solid high VIH = 3 V and up VIL = 1. 2 V solid low +5 V Marginal High NMH = (5 -3)-(5 -4. 9) =1. 9 V Marginal Low NML = 1. 2 -9. 7=0. 5 V 0 V Introduction, Page 1. 1 -75

Introduction to Electronics How does an Inverter (with gain) restore a “poor” signal level? § § Assume that we have two identical inverters in series and that they both have the same voltage transfer characteristic given below. Lets say that the input to the first inverter is 3. 1 Volts, which is about as marginal a high signal as will be recognized as a high by the inverter. The output of the first inverter will be 0. 65 volts. If we take that as the VOH input to the second inverter the output of the second inverter will be 5 volts which is a solid VOUT high, and is much improved over the 3. 1 volts we saw on the (Volts) original input signal. The inverter supplies (+5 and ground) and the gain VOL drive illegal and marginal signals towards solid Output of levels inverter # 1 = 0. 65 V © REP 11/2/2020 EGRE 224 5 0. 65 3. 1 5 4 3 2 |slope| = 1 1 0 2 1 VIL Input to INV #2 3 VIH 4 VIN (Volts) 5 Input to inverter # 1 = 3. 1 V Introduction, Page 1. 1 -76

Introduction to Electronics Capacitors § A capacitor is a circuit element that stores charge. The energy is not stored chemically as in a battery but rather the energy is stored in the form of an electric field. Partially charged + capacitor Electric field § § + + - - Fully charged capacitor Stronger electric field ++++ ---- The Electric Field (Volts per cm) is proportional to the voltage across the capacitor’s terminal. The charge stored is proportional to the voltage (and the area), and the proportionality constant is called the capacitance v Q=CV d. Q/dt = C DV/dt but d. Q/dt is current, I so that v I = C (d. V/dt) the units of capacitance are Farads = Coulomb/Volt § The bathtub analogy. The voltage across a capacitor can not change instantaneously just like the water level in a bathtub can not change instantaneously, instead the rate at which it changes depends on the value of the current flowing into or out of the capacitor. I = C (d. V/dt) Note: the level can change pretty fast if the current is large enough © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -77

Introduction to Electronics The RC time constant § Consider the following circuit with a battery, switch, resistor and capacitor. The capacitor initially has no charge stored on it. v Charging a capacitor VC t=0 V 0 = 10 V R + - i C VC = 0 VC = 10 V = V 0 (1 - e-t/t) 2. 3 RC time VC t=0 i 6. 3 V RC v Discharging a capacitor R 10 V 90% 10 V C 3. 6 V V = V 0(e-t/t) 10% RC § § § 2. 3 RC time Note RC has units of time. Ohms are Volts divided by Coulombs per second and capacitance is Farads which are Coulombs per Volt, which leaves seconds. We use the greek letter t (tau) to represent the RC product. The behavior of the circuit is found by solving a differential equation. The solution uses the base e = ~ 2. 72 e-1 = 0. 37~37% e-2 = 0. 135~14% e-3= 0. 05~5% © REP 11/2/2020 EGRE 224 Introduction, Page 1. 1 -78

Introduction to Electronics Propagation Delay and Rise and Fall Times of a signal v. I Input Signal VOH 90% of (VOH-VOL) 50% 1/2(VOL+VOH) VOL v. O 10% of (VOH-VOL) tr tf t. PHL Time t. PLH VOL © REP 11/2/2020 EGRE 224 t. THL Output Signal t. TLH Time Introduction, Page 1. 1 -79