Basic Electronics Introductory Lecture Course for Technology and Instrumentation in Particle Physics 2011 Chicago, Illinois June 9-14, 2011 Presented By Gary Drake Argonne National Laboratory Session 2
Session 2 Time-Varying Signals & Circuits Basic Electronics Special Lecture for TIPP 2011 2
Time-Varying, Periodic Sources Periodic Time-Varying Sources Steady State Sinusoidal Ramp t t Square Wave Pulse Train t t Triangle Wave Comb Function t t Can be Voltage Sources or Current Sources Depends on.. Source Impedance! Basic Electronics Special Lecture for TIPP 2011 3
Sinusoidal Sources Sinusoidal Sources Characterized by frequency f), phase, period (T = 1 / f), and peak amplitude V p t v O (t) = sin(t + ) - T Is the average value important? T V AVG = 1/T f v O (t) dt = 1/T f sin(t) dt = 0 0 0 T No, not really We often use the Effective Value, or RMS T V EFF = e 1/T f v O2 (t) dt = e 1/T f 2 sin 2 (t) dt 0 0 T = / S 2 This IS useful Images from Wikipedia Basic Electronics Special Lecture for TIPP 2011 4
Resistive Circuits with Steady State Sources Consider again a resistive network, but this time with a sinusoidal source v s (t) I 1 v s (t) = V p sin( t) What is the instantaneous power consumed by? p 1 (t) = v 1 (t) i 1 (t) v 1 (t) = V p sin( o t) [V p sin( o t)] / R = [V p2 / ] sin 2 1 ( o t) What is the average power consumed by R1? P AVG = 1/T f p 1 (t) dt 0 T = 1/T f [2 / ] sin 2 ( o t) dt 0 T Just as before, we can use Kirchoff s Laws: i 1 (t) v s (t)= 0 i 1 (t) = v s (t) / v O (t) = v S (t) P AVG = ½ 2 / In general, = V EFF2 /, since V EFF = / S2 P AVG = V EFF I EFF Says that Effective Values of a sinusoid produce average power equivalent to comparable DC values Basic Electronics Special Lecture for TIPP 2011 5
Resistive Circuits with Steady State Sources In general, can perform the same analysis with any resistive network Voltage Divider From before: v s (t) I 1 R 2 V o =? I 1 + I 1 R 2 V s = 0 I 1 = V s / ( + R 2 ) Then: V 1 = I 1, V 2 = I 1 R 2 V O = V S R 2 / ( + R 2 ) v s (t) = V p sin( t) Inserting the time-varying source: v o (t) = V p [R 2 / ( + R 2 )] sin( o t) P AVG = V EFF I EFF = ½ V p [R 2 / ( +R 2 )] [V p / ( +R 2 )] P AVG = ½ 2 [R 2 / ( + R 2 ) 2 ] Basic Electronics Special Lecture for TIPP 2011 6
Introduction to Steady State (AC) Analysis Sinusoidal Sources Why are sinusoids important? Reason #1: Our electrical grid works on sinusoidal power Famous shoot-out between Nikola Tesla and Thomas Edison, 1893 Worlds Fair in Chicago (Tesla won, but died a pauper ) In the USA: f = 60 Hz V EFF = 120V = 170V Images from Wikipedia Basic Electronics Special Lecture for TIPP 2011 7
Introduction to Steady State (AC) Analysis Sinusoidal Sources (Continued) Why are sinusoids important (Cont.)? Reason #2: Convenient representation in the Frequency Domain Time Domain Magnitude Frequency Domain t O - T Angle O v(t) = cos( O t+ O ) Define phasors: F() = Mag Angle If: v(t) = cos( O t+ O ) Phasor form: V() = O Has connection with Complex Number Analysis More on this later Basic Electronics Special Lecture for TIPP 2011 8
Introduction to Steady State (AC) Analysis Sinusoidal Sources (Continued) Why are sinusoids important (Cont.)? Reason #3: Any periodic waveform can be represented as an infinite sum of sinusoids, with frequencies that are multiples of the fundamental frequency O Fourier Series For any periodic waveform V(t), with fundamental frequency O v(t) t T O = (2 ) / T +h v(t) = E n e (jnot), where j = S -1 n=-h The coefficients E n are given by: T/2 E n = 1/T f V(t) e -jnot -T/2 Basic Electronics Special Lecture for TIPP 2011 9 dt n 0 are the harmonic frequencies of v(t)
Introduction to Steady State (AC) Analysis Sinusoidal Sources (Continued) Why are sinusoids important (Cont.)? Example: Fourier Series of a Square Wave Time Domain Frequency Domain 2 / T O = (2 ) / T t O O O O O O O O O O +h +h V(t) = E n e (jnot) n=-h n=-h where j = S -1 where E n = (2 / n) sin(n / 2) Notice that only the odd harmonics of v(t) are present in this particular example Basic Electronics Special Lecture for TIPP 2011 10
Introduction to Steady State (AC) Analysis More generally, use the Fourier Transform For a periodic function v(t) with period T, the Fourier Transform is defined as: h F [v(t)] = V(j) = f v(t) e jt Some important Transforms v(t) = cos( O t) -h dt where j = S -1 V(j) t O O - v(t) = sin( O t) V(j) t O O - Generally, we will ignore the negative frequency space Basic Electronics Special Lecture for TIPP 2011 11
Introduction to Steady State (AC) Analysis Fourier Transform Properties For a periodic function v(t) with period T, the Fourier Transform is defined as: h F [v(t)] = V(j) = f v(t) e jt The Inverse Fourier Transform is given as: F -1 h [V(j)] = v(t) = 1/(2 f V(j) e jt An important Transform property that we will use: F [d/dt v(t)] = j V(j) -h -h dt where j = S-1 d Fourier Transforms generally generate Complex Numbers Basic Electronics Special Lecture for TIPP 2011 12
Introduction to Steady State (AC) Analysis A Review of Complex Numbers For F(j) = a REAL + j b IMAG Think of Complex Plane IMAG F(j) b F(j) = w a REAL2 + b IMAG 2 a REAL = tan -1 [b IMAJ / a REAL ] These are related to Phasors Phasor form: F(j) = F(jw) F(j) = Mag Angle Basic Electronics Special Lecture for TIPP 2011 13
Frequency Domain Analysis Basic Principles We will generally use sinusoidal excitation to evaluate the performance of circuits, and sweep the frequency over a range of interest Time Domain Magnitude Frequency Domain Log t O - T Angle O Log O f O = 2 / T In general, when energy storage elements are involved, we will calculate circuit response in the frequency domain Use Complex Analysis REAL & IMAG Phasor forms Use Fourier Transforms Basic Electronics Special Lecture for TIPP 2011 14
Frequency Domain Analysis Capacitors Revisited i(t) = C dv(t) / dt F [i(t)] = F [ C dv(t) / dt ] I(j) = j C V(j) 20 Log Z C Z C Log Z C (j) = V(j) / I(j) = 1 / (j C) Inductors Revisited v(t) = L di(t) / dt F [v(t)] = F [ L di(t) / dt ] V(j) = j L I(j) Z L (j) = V(j) / I(j) = j L -90 20 Log Z L Z L +90 Log Log Log Basic Electronics Special Lecture for TIPP 2011 15
Frequency Domain Analysis Example 1 Simple RC Circuit v i v C O 1 v i (t) = V p cos( t) v O (t) =? Complex Numbers F(j) = a + j b IMAG b REAL a First, express v i (t) in complex form v i (t) = cos( O t) IMAG Mag = w a 2 + b 2 Angle = tan -1 [b / a] Phasor Notation: V i (j) = REAL F(j) = Mag Angle Phasor form: V i (j) = Basic Electronics Special Lecture for TIPP 2011 16
Example 1 (Cont.) v i Frequency Domain Analysis I 1 v i (t) = V p cos( t) Next, express as an impedance in complex form Z C1 (j) = 1 / ( j ) Find V O (j) using Kirchoffs Voltage Law I v 1 + I 1 Z C1 (j) V i (j) = 0 C O 1 Solution Voltage Divider: V O (j) = V O (j) = V i (j) Z C1 (j) + Z C1 (j) V i (j) [ 1 / ( j ) ] + [ 1 / (j ) ] Rearrange, using complex algebra: V i (j) V O (j) = j R1 + 1 Basic Electronics Special Lecture for TIPP 2011 17
Example 1 (Cont.) Frequency Domain Analysis Next, express in Phasor Form V i (j) V O (j) = j + 1 v i v C O 1 v i (t) = V p cos( t) = 0 e ( ) 2 + 1 tan -1 [ ] = tan -1 [ ] e ( ) 2 + 1 Solution: V O (j) = v O (t) = e ( ) 2 + 1 e ( ) 2 + 1 tan -1 [ ] cos ( o t tan -1 [ ] Basic Electronics Special Lecture for TIPP 2011 18
Example 1 (Cont.) v O (t) = Frequency Domain Analysis Try some numbers e ( ) 2 + 1 a) Let = 1 K, = 1 F, = 1V, o = 10 rad/sec cos ( o t tan -1 [ ( ] v i v i (t) = V p cos( t) v O Solution: v O (t) y 1 cos ( o t 0 o ) Almost no change v o (t) y v i (t) b) Let = 1 K, = 1 F, = 1V, o = 1000 rad/sec X100 Solution: v O (t) = 0.707 cos ( o t 45 o ) Moderate change v o (t) decreasing c) Let = 1 K, = 1 F, = 1V, o = 100,000 rad/sec X 10,000 Solution: v O (t) = 0.099 cos ( o t 84.2 o ) Big change v o (t) 0 We could plug in numbers all day long, but let s find a better way Basic Electronics Special Lecture for TIPP 2011 19
Frequency Domain Analysis A useful tool to aid in analyzing circuits in the frequency domain is the concept of a Transfer Function Suppose have a network that contains R s, L s, C s and even transistors Suppose have a single input F(j) (called a Forcing Function) Could be either a voltage or a current Suppose have a single output variable of interest E(j) that responds in some way to the input Could also be either a voltage or a current Denote the network as H(j) F(j) H(j) E(j) Then H(j) = E(j) F(j) = Output of Interest Forcing Function We will use this concept extensively in Frequency Domain Analysis Basic Electronics Special Lecture for TIPP 2011 20
Back to Example 1: Frequency Domain Analysis Back to Frequency Domain v i v O v O (t) = e ( ) 2 + 1 cos ( o t tan -1 [ ( ] = 1 K, = 1 F V O (j) = V i (j) tan -1 [ ( ] Phasor Form e ( ) 2 + 1 To find general solution, plot V o (j) / V i (j) Transfer Function V O (j) 1 = V e ( ) 2 i (j) + 1 tan -1 [ ( ] = Mag Angle H(j) = Output (j) Input (j) Now, plot V o / V i vs., and V o / V i vs. (Actually, plot 20 Log V o / V i vs. Log, and V o / V i vs. Log ) Called a Bode Plot Basic Electronics Special Lecture for TIPP 2011 21
Example 1 (Cont.) Frequency Domain Analysis Plotting Transfer Functions v O V O (j) 1 H(j) = = V e ( ) 2 i (j) + 1 tan -1 [ ( ] 20 Log H(j) db v i = 1 K, = 1 F H(j) = 1 e ( ) 2 + 1 Log H(j) = tan -1 [ ( ] H(j) o Log Basic Electronics Special Lecture for TIPP 2011 22
Example 1 (Cont.) Frequency Domain Analysis Examining Bode Plots for features v O V O (j) 1 H(j) = = V e ( ) 2 i (j) + 1 tan -1 [ ( ] v i = 1 K, = 1 F 20 Log H(j) db H(j) o Log Log Response starts out flat At critical frequency, see inflection point What is that frequency? Response falls at -20 db/decade Response starts out flat At critical frequency, see half-way point What is that frequency? Response ends flat, with -90 o phase shift = 1 / = 1 / Called the pole frequency Again, the pole frequency Actual: 3 db Actual: 45 o Basic Electronics Special Lecture for TIPP 2011 23
Example 1 (Cont.) Frequency Domain Analysis General Guidelines for Bode Plots (Note: The following is true ONLY for single pole circuits (1 energy storage element) 20 Log H(j) db H(j) o v i = 1 K, = 1 F v O Log Log Straight line starts out flat at 0 db Inflection point at: = 1 / Actual curve is -3 db down from intersection of the two straight lines Straight line falls at -20 db/decade Straight line starts out flat at 0 o Inflection points at: = 0.1 / = 10 / Curve = -45 o at: = 1 / Straight line ends flat, with -90 o phase shift Basic Electronics Special Lecture for TIPP 2011 24
Example 1 (Cont.) Frequency Domain Analysis This configuration is known as a Low-Pass Filter Low frequencies are passed with 0 db attenuation ( Gain = 1) High frequencies are attenuated Filter frequency = pole frequency v i = 1 K, = 1 F v O o = 1 / 20 Log H(j) db H(j) o Log Log You can create Bode Plots almost by inspection!!! Basic Electronics Special Lecture for TIPP 2011 25
Frequency Domain Analysis Example 2 Simple RL Circuit v i I 1 L 1 v O Insert expression for impedance of an inductor: Z L1 (j) = j L 1 Find V O (j) using Kirchoffs Voltage Law Voltage Divider I 1 + I 1 Z L1 (j) V i (j) = 0 V O (j) = v i (t) = V p cos( t) v O (t) =? V i (j) Z L1 (j) + Z L1 (j) V O (j) = V i (j) j L 1 + j L 1 Transfer Function: V O (j) L 1 / R1 = V i (j) e ( L 1 / ) 2 + (1) 2 tan -1 [ ( L 1 / ] Denominator is ~ the same as before Have a pole at = / L 1 = V i (j) j L 1 / 1 + j L 1 / Now have frequency content in the numerator Basic Electronics Special Lecture for TIPP 2011 26
Example 2 (Cont.) Frequency Domain Analysis Try some numbers: V O (j) L 1 / R1 = Let = 100, L 1 = 100 mh V i (j) e ( L 1 / ) 2 + (1) 2 tan -1 [ ( L 1 / ] v i L 1 v O v i (t) = V p cos( t) o = / L 1 = 1000 rad/sec L 1 / R1 H(j) = e ( L 1 / ) 2 + (1) 2 20 Log H(j) db Log H(j) = tan -1 [ ( L 1 / ] H(j) o Log Basic Electronics Special Lecture for TIPP 2011 27
Example 2 (Cont.) Frequency Domain Analysis Examining Bode Plots for features v i L 1 v O V O (j) L 1 / R1 = V i (j) e ( L 1 / ) 2 + (1) 2 tan -1 [ ( L 1 / ] 20 Log H(j) db H(j) o Log Log Response Response rises at ends flat +20 db/decade At critical frequency, see inflection point What is that frequency? = / L 1 Response starts out flat With +90 o Phase shift At critical frequency, see half-way point Response ends flat What is that frequency? = 1 / Again, the pole frequency It s a pole frequency Actual: 45 Actual: 3 db o Basic Electronics Special Lecture for TIPP 2011 28
Example 2 (Cont.) Frequency Domain Analysis This configuration is known as a High-Pass Filter High frequencies are passed with 0 db attenuation ( Gain = 1) Low frequencies are attenuated Filter frequency = pole frequency Also has a Zero at 0 frequency o = / L 1 v i L 1 v i (t) = V p cos( t) v O 20 Log H(j) db H(j) o Log Log Basic Electronics Special Lecture for TIPP 2011 29
Frequency Domain Analysis General Features of Transfer Functions & Bode Plots For a general transfer function H(j), express as: H(j) = V O (j) V i (j) = A + jb F + jg Complex Numbers = A F 1 + j B/A 1 + j G/F 1 + j B/A Form = A F e (B / A) 2 + (1) 2 tan -1 [ B / A ] e (G / F) 2 + (1) 2 tan -1 [ G / F ] Phasors Magnitude & Phase The denominators give the Poles of the Transfer Function Magnitude changes by 20 db/decade Phase lag phase changes by 90 o The numerators give the Zeros of the Transfer Function Magnitude changes by 20 db/decade Phase Lead Phase changes by 90 o Where G/F = 1 Where B/A = 1 20 Log H(j) db H(j) o Pole Zero Log Pole Zero Log Basic Electronics Special Lecture for TIPP 2011 30
General Features of Transfer Functions & Bode Plots (Cont.) Bandwidth Frequency Domain Analysis Generally concerned with points in frequency where the response begins to fall off Look for 3db points Consider a typical amplifier 20 Log H(j) db Log For this example, the Bandwidth would be stated as: 100 rad/sec to 10K rad/sec (16 Hz to 1600 Hz) Pole Pole The pass band is defined as the range of frequencies where the response is flat Exercise: Suppose that stereo has a frequency response of 20 Hz to 20 KHz, and a maximum gain of 30 db. Can you draw the frequency plot? Basic Electronics Special Lecture for TIPP 2011 31
Analog Filters Filters Many types Most use active components (i.e. op amps), and have Gain A few examples: 20 Log H(j) db Pass band 20 Log H(j) db 3-pole Low Pass Filter Log Log 20 Log H(j) db Notch 20 Log H(j) db Integrator Log Log Basic Electronics Special Lecture for TIPP 2011 32
Aperiodic Signal Sources Aperiodic Sources Impulse t O Step t O Exponential t O v(t) = V p (t t O ) v(t) = V p u(t t O ) t t v(t) = V p e a(tto) u(tt o ) t For aperiodic signals, Laplace Transform h L [v(t)] = V(s) = f v(t) e -st -h L -1 C+jh [V(s)] = v(t) = f V(s) e st Cjh Two important properties: L [d/dt v(t)] = s V(s) v(0 ) t L [fv() d ] = V(s) / s 0 where s = a + j dt ds Basic Electronics Special Lecture for TIPP 2011 33
Aperiodic Signal Sources Aperiodic Sources Impulse Time Domain v(t) = V p (t) Frequency Domain L [v(t)] = V p 0 t Log s Step v(t) = V p u(t) L [v(t)] = V p / s 0 t Log s Exponential v(t) = V p e at u(t) L [v(t)] = V p / (s + a) 0 t / a Log s Basic Electronics Special Lecture for TIPP 2011 34 a
Aperiodic Signal Sources What about Impedances Z C and Z L? Replace j s Capacitors Revisited C i(t) = C dv(t) / dt L [i(t)] = L [ C dv(t) / dt ] I(s) = s C V(s) Z C (s) = V(s) / I(s) = 1 / (s C) Inductors Revisited L v(t) = L di(t) / dt L [v(t)] = L [ L di(t) / dt ] V(s) = s L I(s) Z L (s) = V(s) / I(s) = s L Basic Electronics Special Lecture for TIPP 2011 35
Aperiodic Signal Sources Recall RC circuit V i i 1 t = 0 V R1 V C1 Replace DC source and switch with Aperiodic source with step function V o v i (t) i 1 v R1 v C1 v o (t) Replace j with s Laplace Transform L [ u(t)] = V p / s Z C1 (s) = 1 / (s ) V O (s) = V i (s) Z C1 (s) + Z C1 (s) = s ( s + 1) = s [ 1 / ] [ s + 1] = s s + L -1 [V O (s)] = [ 1 e t/r1c1 ] u(t) Partial fraction expansion Easier, yes?... Basic Electronics Special Lecture for TIPP 2011 36
Linearity Two Basic Circuit Principles If a system H is linear, and has response E o to forcing function input F i, such that: E o (j) = H(j) F i (j) then if the forcing function is multiplied by a constant factor K number), the output responds as: H(j) [ K F i (j)] = K E o (j) (a real Networks that contain resistors, capacitors, and inductors are linear networks Networks that contain semiconductor devices may or may not be linear F(j) K F(j) Depends on how the semiconductors are biased or being used More on this in the next session Basic Electronics Special Lecture for TIPP 2011 37 H(j) H(j) E(j) K E(j)
Superposition Two Basic Circuit Principles If any linear network contains several independent sources (voltage sources or current sources), the quantity of interest (voltage across a component or current through a component) may be calculated by analyzing the circuit with one source at a time, with the other sources made dead : Voltage sources are replaced by short circuits ( 0 impedance) Current sources are replaced by open circuits ( infinite impedance) The complete response then is obtained by adding together the individual responses Basic Electronics Special Lecture for TIPP 2011 38
Two Basic Circuit Principles Superposition (Continued) Example: V i (j) V S (DC) H(j) V o (j) I S (DC) V i (j) 1 2 3 V S (DC) H(j) V o1 (j) H(j) V o2 (j) H(j) V o3 (j) Then: V o (j) = V o1 (j) + V o2 (j) + V o3 (j) I S (DC) Will use this idea for analyzing amplifier circuits with DC & AC sources Basic Electronics Special Lecture for TIPP 2011 39
Time-Varying Circuits Concluding Remarks Background presented here is the basis for all of modern communications How can you have 500 cable channels and mixed internet on a single coaxial cable?... Answer: Because superposition works It is also the primary method by which analog circuits are designed and analyzed Basic Electronics Special Lecture for TIPP 2011 40