Lecture 2 Analog circuits. Seeing the light..

Transcription

1 Lecture 2 Analog circuits Seeing the light.. I t

2 IR light V1 9V +V Q1 OP805 RL IR detection Vout Noise sources: Electrical (60Hz, 120Hz, 180Hz.) Other electrical IR from lights IR from cameras (autofocus) Visible light Vout ~ mv What we want: 0 5 V DC signal representing the IR amplitude. t

3 Analog circuits filtering and detection IR DC Amplify Filter detect block Peak detect

4 Analog circuits filtering and detection IR DC Amplify Filter detect block Can put these together into a single circuit But today we ll keep them separate for clarity. Peak detect

5 Analog circuits filtering and detection IR DC Amplify Filter detect block Peak detect

6 Discrete devices: BJT Bipolar Junction Transistors b ~ or more + 6 Diagrams courtesy of: University of St. Andrews, St Andrews, Fife KY16 9SS, Scotland

7 Analog circuits discrete devices: BJT Application: light detection Phototransistor: Acts like BJT except charge carriers generated by incident light add to the base current. V1 9V +V I c Q1 OP805 Vout In other words, I c ;Incident light RL

8 IR detection Build a circuit that: Uses an OP805 and a resistor to detect variations in light with a voltmeter. Determine whether increasing or decreasing the load resistance makes it more sensitive v Note: OP805 will see some room light use your hand to block it, and use the voltmeter to detect the change in signal.

9 + Selecting R L. Vout = I c * R L

10 Analog circuits filtering and detection IR DC Amplify Filter detect block Peak detect

11 Analog circuits DC block C R Capacitors: Block DC Pass high frequencies > 1/(2πRC)

12 DC block Imaging adding a DC block to your photodetector circuit: Which circuit would you build? Why? 1) 2)

13 Analog circuits filtering and detection IR DC Amplify Filter detect block Peak detect

14 Analog circuits filtering and detection What is the result of the following: Z 2 /Z 1 = 3 V 0V t Z Z 2 V out V out = 1) 3) 2) 4)

15 Analog circuits: Op-amps Eg: Inverting amplifier. + Vin Z 1 V- - Vout V - = 0 I 1 = V in/ Z 1 I 1 Z 2 V out = 0 Z 2 I 1 Eg 1: Z 2 = 100kΩ Z 1 = 10kΩ Eg 2: Z 2 = 100kΩ V out = - 10 V in Z 1 = 1 Ω V out = - 100,000 V in!! V out = - (Z 2 /Z 1 ) V in 10x gain is a reasonable value Not likely.

16 Analog circuits: Real Op-amps Vin Z 1 V- + - Vout I 1 Z 2 Eg 2: Z 2 = 100kΩ Z 1 = 1 Ω V out = - 100,000 V in!! Several problems: I 1 = 1A for V in = 1 V!! (excessive load for upstream circuitry) Gain Bandwidth product ~ 3 MHz. This would limit the bandwidth of the amplifier from DC up to 30 Hz (i.e. not a very responsive system!).

17 Analog circuits: Real Op-amps Things to consider: Input impedance Gain Bandwidth product Bias Currents Voltage limitations Vin Z 1 V- + - Vout Output current limitations I 1 Z 2 Since V - is a virtual ground, input impedance seen by V in is Z 1

18 Analog circuits: Real Op-amps Things to consider: Input impedance Gain Bandwidth product Bias Currents Voltage limitations Output current limitations Vin V- + - Vout Since Op-amp inputs source or sink very little current (depends on type), input impedance in this case is very high. This is a commonly used buffer to separate your low impedance circuit from a sensitive source that you need to measure without drawing current.

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20 Analog circuits: Real Op-amps Things to consider: Input impedance Gain Bandwidth product Bias Currents Voltage limitations Output current limitations Slew-rate is a similar limit: it is a limit on the rate of change of output voltage log K Vin Z 1 I 1 Open loop gain (K) 20 V- 100 khz + - Z 2 Vout Frequency log ω Gain-Bandwidth limit (Hz) = Gain * Max. Frequency = CONSTANT

21 TL082: Gain*Bandwidth = 3 MHz è This means that at a gain of 100, Bandwidth is 30 khz.

22 Analog circuits: Real Op-amps Things to consider: Input impedance Gain Bandwidth product Bias Currents + Vout Voltage limitations Vin Z 1 V- - Output current limitations I 1 Op-amp terminals can act as small current sources. These Bias Currents can become large error or offset voltages if the resistors in the circuit are large. Z 2 Eg: 20 na bias current * 10 MΩ = 200 mv!

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24 Analog circuits: Real Op-amps Things to consider: Input impedance Gain Bandwidth product Bias Currents Voltage limitations Vin Z 1 Z 3 V- + - Vout Output current limitations I 1 Z 2 The circuit above corrects for bias current induced error and is now only subject to offset current. Z3 = Z1 Z2

25 Analog circuits: Real Op-amps Things to consider: Input impedance Gain Bandwidth product Bias Currents Voltage limitations Output current limitations V+ V- Op-amp input voltages (V +, V - ) must be at least a few volts away from the power rails (+Vcc, -Vcc). Applying input voltages equal or near the power rails will cause the Op-amp to behave unexpectedly. Rail-to-rail Op-amps are an expensive solution to this limitation Vcc -Vcc Vout

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27 Analog circuits: Real Op-amps Things to consider: Input impedance Gain Bandwidth product Bias Currents Voltage limitations Output current / voltage limitations V+ V- Op-amp output terminals can only provide a few ma of current. Motors, lamps and similar high current devices cannot typically be driven by a normal OPamp. High power Op-amps exist that can provide much higher current levels. Output voltage range is also limited within a few volts of the power rails Vcc -Vcc Vout

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29 Analog circuits: Real Op-amps Summary: Keep resistors in 1K to 500K range unless you really know what you re doing. Don t ask a single amplifier to provide huge gains (>30?) Don t drive motors, lamps, or other heavy loads with a normal op-amp (power op-amps exist for this, or use a transistor) Keep input voltages away from the op-amp voltage rails (unless using rail-to-rail opamps)

30 Analog circuits filtering and detection IR DC Amplify Filter detect block Peak detect

31 Analog circuits: Filters To understand filters you should first understand the difference between the TIME DOMAIN and FREQUENCY DOMAIN

32 Which is the correct match between the following time-domain (left) signals and their Fourier Transforms (right)? a) i) 1) a-i, b-iv, c-ii, d-iii t f 2) a-ii, b-iv, c-iii, d-i b) ii) 3) a-iii, b-iv, c-i, d-ii c) iii) 4) a-iv, b-ii, c-iii, d-i d) iv)

33 Analog circuits: Filters Transfer Function = Vout/Vin = H(ω) So: V out (ω) = H(ω)*V in (ω) This is all in terms of ω since, in general, impedances are functions of ω. Z cap = 1/j ωc Z ind = j ωl Z res = R Similar to voltage divider: except ω dependent. Frequency Generator V in Z 1 V out Z 2 Spectrum Analyzer ground

34 V out (ω) = [V in (ω)/(ζ 1 +Ζ 2 )] Ζ 2 So: H(ω) = Z 2 /(Z 1 +Z 2 ) Analog circuits: Filters For resistors, this is just the well known voltage divider: R 2 /(R 1 +R 2 ) H ( ω) = Z 1 Z + 2 Z 2 Frequency Generator V in Z 1 V out Z 2 Spectrum Analyzer ground

35 Analog circuits: Filters Now plug in a resistor and a capacitor: Z 2 = 1/j ωc Z 1 = R H ( ω) = 1/ jωc R + 1/ jωc = 1+ 1 jωrc Z 1 Frequency Generator Vin R k Vout C nf nf Spectrum Analyzer Z 2

36 H( ω) 1 = 1 + jωrc Analog circuits: Filters For low frequencies (small ω), H = 1 For high frequencies (large ω), Η = 0 This is a LOW PASS FILTER At ω = 1/RC, H begins to decrease in amplitude. Z 1 Frequency Generator Vin R k Vout C pf nf Spectrum Analyzer f 0 = 1/(2πRC) = 30 khz Z 2

37 Analog circuits: Filters Vin C1 Vout R1 How does this circuit affect the following waveform: 1) 3) 2) 4)

38 Analog circuits: Transfer Functions Bode plots: a graphical representation of frequency response on logarithmic axes. (20 is used instead of 10 so the Vertical axis: result will represent power ~ V 2) 20log 10 (H) Horizontal axis: log 10 (f) -3 db = ½ as much power as 0 db V out is 1/ 2 of V in at -3dB Log of frequency is used to ensure linear plots from 1/f or 1/f n functions Pole: 1/(1+jω/ω 0 ) -20 db/decade in amplitude after ω 0, -90 phase Zero: (1+jω/ω 0 ) +20 db/decade in amplitude after ω 0, +90 phase

39 Bode Plot: Analog circuits: Simple Pole H ( ω) 1 = 1 + jωrc -3dB, 1/RC -20db/decade - 45 deg, 1/RC -90 deg

40 Analog circuits: Simple Zero Bode Plot: H ( ω ) =1+ jωrc +3dB, 1/RC +20db/decade +45 deg, 1/RC +90 deg

41 Active Band Pass: Analog circuits: Active Filters R1 C1 Z 1 Combines a high and a low pass filter to create a pass band. R2 C2 U1 TL082 H = - (Z 1 /Z 2 ) Z 2 Zero at ω=0 H ( ω) = (1 + jωr1c 2 jωr C )( jωr C 1 1 ) Pole at ω=1/(r 2 C 2 ) Pole at ω=1/(r 1 C 1 )

42 Analog circuits: Transfer Functions Bode plots: a graphical representation of frequency response on logarithmic axes. Zero at ω=0 H ( ω) = (1 + jωr1c 2 jωr C )( jωr C 1 1 ) Pole at ω=1/(r 2 C 2 ) Pole at ω=1/(r 1 C 1 ) Pole: 1/(1+jω/ω 0 ) -20 db/decade in amplitude after ω 0, -90 phase Zero: (1+jω/ω 0 ) +20 db/decade in amplitude after ω 0, +90 phase

43 Analog circuits: Active Filters 20log( H ) Idealized Bode Plot: 20log(R 1 /R 2 ) db (1) (2) (1) (1) (2) (3) 20db/decade 0 db (1) 1/(R 2 C 2 ) 1/(R 1 C 1 ) Zero at ω=0 H ( ω) = (1 + jωr1c 2 jωr C )( jωr C 1 1 ) ω (2) (3) Pole at ω=1/(r 2 C 2 ) Pole at ω=1/(r 1 C 1 )

44 Analog circuits filtering and detection Band Pass IR DC Amplify Filter detect block R1 R1 R1 C1 C1 C1 High pass R2 C2 U1 TL082 R2 C2 U1 TL082 R2 C2 U1 TL082 Use multiple stages to get steeper filter roll-offs H tot (ω) = H 1 (ω) * H 2 (ω) *H 3 (ω) Remember 20dB/dec for each POLE

45 More advanced filters: Biquad 45

46 Discrete devices: diodes ΔV = 0.7 V Treat as conductor I ;V 2 I = 0 A Treat as open circuit Typical ~ 0.7 V 46 Diagrams courtesy of: University of St. Andrews, St Andrews, Fife KY16 9SS, Scotland

47 V in = 5V DC, V out is connected to scope only. Vin D1 DIODE a) V R1 out = 5 V if R1 = 1K 1k b) V out = 4.3 V regardless of R1 c) V out = 4.3 V if R1 = 1K d) V out = 5 V if R1 is disconnected (infinite resistance) e) V out = 4.3 V if R1 is disconnected (infinite resistance) Vout The following are true: 1. a 2. b,c,e (Try it with your own meter) 3. c,d 4. a,e 5. c 47

48 SUPER-DIODE rectifier circuit This circuit acts like a perfect diode, without the 0.7V deadband prior to turn-on. What bad thing happens if R1 is too big?

49 Zener Diodes Use 5V Zener Diodes to protect your TINAH Board Zener diodes (5V1) To TINAH Zener diodes conduct under reverse bias when a specific voltage is exceeded in our case 5.1V

50 Debugging Circuits Learn to systematically check your circuits: Power rails: Check that 15V is really 15V; if not, localize the component that is shorting the power rail. Check power at each chip. Physical check: Check pinouts, missing/loose wires, etc. Isolate stages where possible Check output of stage 1 if ok plug into stage 2 and see if stage 1 output is degraded. If ok, check output of stage 2 etc Keep wiring TIDY!

51 Lab 2 Tips Capacitors electrolytic capacitors have polarity, may explode if inserted backwards Gain make sure that gain does not saturate the signal, this will generate unwanted noise after filtering. + Sharp corners have highfrequency noise Saturation looks like DC signal! -