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Electronic Instrumentation Quiz 1 Review Op Amps (Continued) Notes Test is TOMORROW 6:00 p.m.-8:00 p.m. DCC 308 If you’ve just started Project 1 you are a week behind! (Get to open shop this weekend) Make sure to find the studio attendance sheet if it doesn’t make it around to you Review Topics Quiz Review Tally 80 70 60 Arb. Units 50 40 30 20 10 0 Transfer Functions Resonant Frequency (Filters) Transformer Topics Circuit Analysis Pspice Transfer Functions and Phasors Apply the voltage divider equation and parallel and series combination rules to find transfer functions using complex impedance expressions Simplify the transfer function to find a function which governs behavior at low and high frequencies. Find an expression (or value) for the magnitude and phase of the simplified transfer function at the corner or resonant frequency Find Vout or Vin from the transfer function (magnitude and phase) Crib Sheet Highlighter Crib Sheet Highlighter Crib Sheet Highlighter Transfer Functions and Phasors 1) Find the transfer function for the above circuit. Write in terms of Z impedance first Transfer Functions and Phasors Z ZR R Z ZC Z 1) Transfer function for the above circuit. a) Combine Impedances: Find ZR2C1 b) Use voltage divider to find H=Vout/V1 c) Substitute component values d) Simplify j R 2 C 1 1 H j j C 1 R 1 R 2 1 1 j C Transfer Functions and Phasors 2) Assume R1=R2=1KΩ and C1=1μF, evaluate the magnitude of the transfer function at ω=0 and ω=∞. a) Magnitude of the transfer function at ω=0 For low frequencies, the lowest power dominates H j j R 2 C 1 1 j C 1 R 1 R 2 1 lim H j 0 1 1 Transfer Functions and Phasors 2) Assume R1=R2=1KΩ and C1=1μF, evaluate the magnitude of the transfer function at ω=0 and ω=∞. b) Magnitude of the transfer function at ω= ∞ For high frequencies, the highest power dominates H j j R 2 C 1 1 j C 1 R 1 R 2 1 lim H j 1 2 Transfer Functions and Phasors 3) V1 V1 V1 A e j t j 4 5e 5e j ( 0.79) See crib sheet When doing analysis ωt will eventually drop out so just θ is used Transfer Functions and Phasors 4 Given R1=1K , R2=1K , and C1=1F, what is the output phasor Vout 2 1 KHz H j H j Need both Magnitude and Phase from the transfer function. j R2 C1 1 j C1 R1 R2 1 2 R2 C1 12 C1 R1 R2 6.283 2 12.566 2 12 12 2 12 0.505 Magnitude Transfer Functions and Phasors 2 1 KHz H j PhaseH j R2 C1 Need both Magnitude and Phase from the transfer function. 1 j C1 R1 R2 1 R2 C1 C1 R1 R2 1 tan tan 1 1 1 PhaseH atan 2 1 103 1 103 1 10 6 atan 2 1 103 2 103 1 10 6 PhaseH 0.078 Vout H j V1 Vout H j V1 H A e .505 5 e j phaseH j ( 0.790.078) Phase 2.5 e j 0.71 Transfer Functions and Phasors Using Transfer Functions (on crib sheet!) Vout H j V1 Vout H j V1 A e j phaseH H .505 5 e j ( 0.790.078) 2.5 e In the form Vout(t)=ACos( t+), i.e. find A, , and V out ( t) 2.5Cos 2 1KHz t 0.71 j 0.71 Filters Understand how capacitors and inductors work at very low and high frequencies Redraw a given RL, RC or RLC circuit at low and/or high frequencies and identify low pass, high pass, band pass and band reject filters Find resonant frequency of RLC circuits Find corner frequency of RC and RL circuits Identify whether a signal of a certain frequency be passed rejected or something in between by a filter Crib Sheet Highlighter Crib Sheet Highlighter Filters RS 1k R2 4k VS 8Vac 2 C1 10uF Va Vb L1 100uH R3 20k 1 0 1) Let Rs=0Ω Redraw the circuit above for very low frequencies 2) What is Va and Vb at very low frequencies? Filters RS 1k R2 4k VS 8Vac Va Vb R3 20k 0 2) Va=0V, Vb=0V Filters RS 1k R2 4k VS 8Vac 2 C1 10uF Va Vb L1 100uH R3 20k 1 0 1) Let Rs=0Ω Redraw the circuit above for very high frequencies 2) What is Va at very high frequencies? Filters RS 1k R2 4k VS 8Vac Va Vb R3 20k 0 2) Va=8V If Va is considered the output, what type of filter is this? In Class Problem: Filters: Resonant Frequency 1) 2) 3) 4) Redraw this circuit at low frequencies Redraw this circuit at high frequencies What type of filter is this? If L1=2mH, L2=2mH, C1=0.5uF and R13K, what is the resonant frequency in Hertz? In Class Answer: Filters: Resonant Frequency Low Frequencies Inductors: short Capacitor: open Vout=0V High Frequencies Inductors: open Capacitor: short Vout=V1 High Pass Filter In Class Answer: Filters: Resonant Frequency Resonant frequency L1 2mH L2 2mH Ltotal L1 L2 1 0 0 f0 Ltotal C1 2.236 0 2 104 rad s f0 3.559 103 Hz C1 0.5 F Transformers and Inductors How to apply transformer equations Basic characteristics of transformers Calculate unknown inductance given the capacitance or visa versa Calculate resonant frequency given inductance or capacitance or visa versa Estimate inductance of a coil given some dimensions Crib Sheet Highlighter Crib Sheet Highlighter Transformers (Homework 3, #9) You are given the transformer pictured below. What is the voltage across the load resistor if “a” is 9, R2 is 5K ohms, R1 is 3K ohms, and V1 is 62 Volts? Note that R1 is large so you CANNOT assume that it has negligible resistance. Give your answer in volts. Transformers (Homework 3, #9) You are given the transformer pictured below. What is the voltage across the load resistor if “a” is 9, R2 is 5K ohms, R1 is 3K ohms, and V1 is 62 Volts? Note that R1 is large so you CANNOT assume that it has negligible resistance. Give your answer in volts. N L VL LL I S a N S VS LS I L RL Z in 2 a Transformers (Homework 3, #9) 1. Find Zin 2. Use the voltage divider to find Vs (due to nonnegligible voltage drop across R1 3. Use ratio relationship below N L VL LL I S a N S VS LS I L RL Z in 2 a Transformers (Homework 3, #9) 5k Zin 1.25 92 Zin Vs Vs 61.728 62 Z 3 K in Zin a VL Vs VL Vs 9 VL 11.25 Circuit Analysis Handle combinations of parallel and/or series resistors Give resistance expressions in equation form, rather than as a number. Find voltages or currents through any resistor Find the total resistance or current Know the voltage divider equation Find the voltage across a resistor in a voltage divider configuration. Crib Sheet Highlighter Crib Sheet Highlighter Circuit Analysis R1 R3 V1 R2 R4 0 1) Find the total resistance of the circuit, seen from the voltage source Write equation first then put in the numbers! Given V1=5V, R1=2000Ω, R2=3000Ω, R3=200Ω, R4=800Ω Circuit Analysis R1 R3 V1 R2 R4 0 2) Find the voltage across R1 3) Find the current through R3 Given V1=5V, R1=2000Ω, R2=3000Ω, R3=200Ω, R4=800Ω PSpice Instrumentation and Components Know which trace corresponds to which voltage point on a simple circuit Describe specific steps you’d follow to obtain a certain output for AC sweep, DC sweep or Transient Analysis Understand how to set parameters for function generator Understand how to use the oscilloscope Op-Amp Circuits: Quick Review Op-Amps are most commonly used to ________ a signal. Inputs to the op-amp are called the _______ and _______ inputs. Unpredictable high gain that is multiplied by the input signal is called ____-____ ____ or ______ ______. Extreme gain causes __________. What day, time, and location is Quiz 1? Op-Amp Circuits use Negative Feedback A balancing act between gain and negative feedback for a stable circuit How do you “design” negative feedback in the circuit? Op-Amp Analysis We assume we have an ideal op-amp: • • • • infinite input impedance (no current at inputs) zero output impedance (no internal voltage losses) infinite intrinsic gain instantaneous time response The Inverting Amplifier Vout Rf Rin Vin A Rf Rin Is this the same as intrinsic gain? Inverting Amplifier Analysis Step 0: Understand the Golden Rules! Rule 1: VA = VB (feedback network brings the input differential to zero) Rule 2: IA = IB = 0 (inputs draw no current) Inverting Amplifier Analysis inverting input (-): non-inverting input (+): Step 1: Re-draw the circuit Remove the op-amp from the circuit and draw two circuits (one for the + and one for the – input terminals of the op amp). Inverting Amplifier Analysis Step 2: Write equations for the two circuits inverting input (-): non-inverting input (+): inverting input (-): i non-inverting input (+): V V in V B V B V out R R in Rf VA=0 Inverting Amplifier Analysis Step 3: Simplify using Golden Rules and solve for Vout/Vin V in 0 0 V out VA=VB=0 therefore R in Rf Golden Rule! V in V out R in Rf V out R f V in R in What is this saying about how you can design your gain? PSpice Inverting Amplifier Use the uA741 op amp to model your circuits Can’t find it? It is in the EVAL library Add library “Eval” Inverting Amplifer R3 Input amplitude: 200mV Rf=10k Ω Rin=1k Ω 10k V2 U1 7 3 V+ OS2 V + 0 R1 2 V3 1k OUT - 4 uA741 VOFF = 0 VAMPL = 200mV FREQ = 1kHz OS1 V- 5 V 9V 6 What should the simulated output look like? 1 V1 -9V R2 1k 0 The Non-Inverting Amplifier Rf Vout 1 R g Rf A 1 Rg Vin Non-inverting Amplifier Analysis inverting input (-): non-inverting input (+): Step 1: Re-draw the circuit Remove the op-amp from the circuit and draw two circuits (one for the + and one for the – input terminals of the op amp). Non-inverting Amplifier Analysis Step 2: Write equations for the two circuits inverting input (-): non-inverting input (+): Voltage Divider inverting input (-): non-inverting input (+): VA=Vin VB Rg Rf Rg V out Non-inverting Amplifier Analysis Step 3: Simplify using Golden Rules and solve for Vout/Vin VA=VB=Vin Golden Rule! therefore V out V in V in Rg Rf Rg 1 Rf Rg V out PSpice Non-inverting Amplifier Non-Inverting Amplifer 0 0 V5 9V V 7 3 V+ OS2 + OUT 2 - OS1 V- 4 U2 uA741 5 6 V V6 VOFF = 0 VAMP L = 200mV FREQ = 1kHz Using the uA741 op amp Input amplitude: 200mV Rf=1k Ω Rg=1k Ω 1 V4 -9V R40 1k R5 1k 0 What should the simulated output look like? The Voltage Follower Vout 1 Vin Unity gain amplifier analysis : 1] VA Vout VA VB 2] VB Vin therefore, Vout Vin Why is it useful? In this voltage divider, we get a different output depending upon the load we put on the circuit. Why? We can use a voltage follower to convert this real voltage source into an ideal voltage source. The power now comes from the +/- 15 volts to the op amp and the load will not affect the output. Integrators and Differentiators General Op-Amp Analysis Differentiators Integrators Comparison General Analysis Example(1) into inverting input Assume we have the circuit above, where Zf and Zin represent any combination of resistors, capacitors and inductors. General Analysis Example(2) We remove the op amp from the circuit and write an equation for each input voltage. Note that the current through Zin and Zf is the same, because equation 1] is a series circuit. General Analysis Example(3) I Since I=V/Z, we can write the following: Vin VA VA Vout I Z in Zf But VA = VB = 0, therefore: Vin Z in Vout Zf Zf Vout Vin Z in General Analysis Conclusion For any op amp circuit where the positive input is grounded, as pictured above, the equation for the behavior is given by: Zf Vout Vin Z in Ideal Differentiator Phase shift j/2 - ± Net-/2 analysis : Zf Rf Vout j R f Cin 1 Vin Z in j Cin Amplitude changes by a factor of RfCin Analysis in time domain I dVCin I Cin Cin VRf I Rf R f I Cin I Rf I dt d (Vin VA ) VA Vout I Cin VA VB 0 dt Rf therefore, Vout dVin R f Cin dt Problem with ideal differentiator Ideal Real Circuits will always have some kind of input resistance, even if it is just the 50 ohms or less from the function generator. Analysis of real differentiator I 1 Z in Rin j Cin Zf Rf j R f Cin Vout 1 Vin Z in j RinCin 1 Rin j Cin Low Frequencies Vout j R f Cin Vin ideal differentiator High Frequencies Rf Vout Vin Rin inverting amplifier Comparison of ideal and non-ideal Both differentiate in sloped region. Both curves are idealized, real output is less well behaved. A real differentiator works at frequencies below c=1/RinCin Ideal Integrator Phase shift 1/j-/2 - ± Net/2 Amplitude changes by a factor of 1/RinCf analysis : Zf Vout Vin Z in 1 j C f Rin 1 j RinC f Analysis in time domain I VRin I Rin Rin I Cf C f dVCf I Cf I Rin I dt Vin VA d (VA Vout ) I Cf VA VB 0 Rin dt dVout 1 1 Vin Vout Vin dt ( VDC ) dt RinC f RinC f Analysis in time domain I VRin I Rin Rin I Cf C f dVCf I Cf I Rin I dt Vin VA d (VA Vout ) I Cf VA VB 0 Rin dt dVout 1 1 Vin Vout Vin dt ( VDC ) dt RinC f RinC f Problem with ideal integrator (2) With DC offset. Saturates immediately. What is the integration of a constant? Problem with ideal integrator (2) With DC offset. Saturates immediately. What is the integration of a constant? Miller (non-ideal) Integrator If we add a resistor to the feedback path, we get a device that behaves better, but does not integrate at all frequencies. Behavior of Miller integrator Low Frequencies High Frequencies Zf Rf Vout Vin Z in Rin Zf Vout 1 Vin Z in jRinCf inverting amplifier ideal integrator The influence of the capacitor dominates at higher frequencies. Therefore, it acts as an integrator at higher frequencies, where it also tends to attenuate (make less) the signal. Analysis of Miller integrator I Rf 1 Rf j C f Rf Zf 1 j R f C f 1 Rf j C f Zf j R f C f 1 Rf Vout Vin Z in Rin j Rin R f C f Rin Low Frequencies Rf Vout Vin Rin inverting amplifier High Frequencies Vout 1 Vin j RinC f ideal integrator Comparison of ideal and non-ideal Both integrate in sloped region. Both curves are idealized, real output is less well behaved. A real integrator works at frequencies above c=1/RfCf Problem solved with Miller integrator With DC offset. Still integrates fine. Why use a Miller integrator? Would the ideal integrator work on a signal with no DC offset? Is there such a thing as a perfect signal in real life? • noise will always be present • ideal integrator will integrate the noise Therefore, we use the Miller integrator for real circuits. Miller integrators work as integrators at > c where c=1/RfCf Comparison original signal Differentiaion v(t)=Asin(t) Integration v(t)=Asin(t) mathematically dv(t)/dt = Acos(t) v(t)dt = -(A/cos(t) mathematical phase shift mathematical amplitude change H(j electronic phase shift electronic amplitude change +90 (sine to cosine) -90 (sine to –cosine) 1/ H(jjRC -90 (-j) H(jjRC = j/RC +90 (+j) RC RC The op amp circuit will invert the signal and multiply the mathematical amplitude by RC (differentiator) or 1/RC (integrator)