LM311 comparator open collector output. LM311 comparator open collector output. LM311 comparator open collector output

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1 00k 40% LM3 comparator open collector output LM3 comparator open collector output ON OFF LM3 comparator open collector output Example: QRD4 reflectance sensor V V V6 V 330 R4 47K V V V QRD4 R3 V5 V LM3 UA LM339 J V R 47k U 74LS44 OEa Ia3 Ya3 Ia Ya Ia Ya Ia0 Ya0 OEb Ib3 Ib Ib Ib0 Yb3 Yb Yb Yb0 ON OFF TINAH Board 3 4 QRD4 simple analog connection QRD4 reflectance sensor 5 Watch for direct crosstalk: diode and transistor can come out of the housing. 6

2 Lecture Analog circuits Seeing the light.. light V 9V V ion Noise sources: Electrical (60Hz, 0Hz, 80Hz.) Other electrical from lights from cameras (autofocus) Visible light I Q OP805 RL ~ mv t t What we want: 0 5 V signal representing the amplitude. Analog circuits filtering and ion Analog circuits filtering and ion Peak Peak Analog circuits discrete devices: BJT Application: light ion ion Phototransistor: Acts like BJT except charge carriers generated by incident light add to the base current. In other words, I c Incident light V 9V V Q OP805 RL I c Build a circuit that: Uses an OP805 and a resistor to variations in light with a voltmeter. Determine whether increasing or decreasing the load resistance makes it more sensitive Note: OP805 will see some room light use your hand to it, and use the voltmeter to the change in signal.

3 Analog circuits filtering and ion What is the result of the following: Z / = 3 V out Z V out = ) 3) Selecting R L. ) 4) = I c * R L Analog circuits filtering and ion Analog circuits Capacitors: Peak Block Pass high frequencies Add a to your photoor circuit: Move your hand or light over the phototransistor and make sure the output responds only to changes in light. Add a to your photoor circuit: This will be needed for lecture to see voltage changes without a scope, but is NOT THE RIGHT USE OF AN ELECTROLYTIC CAPACITOR. 3

4 This is a much better circuit: Analog circuits filtering and ion Why? Peak Analog circuits: Opamps Ideal opamp: With feedback to limit gain: V V V V When negative feedback is applied to an opamp, V = V Vcc Vcc.if the amplifier is within its operating range. = K (V V ) K ~ 0 6 (at!) V = V out V out = K (V V out ) V out (K) = KV V out =V (K/(K)) V out = V The next few slides will discuss limitations Build a circuit that: Inverting Opamp Wire a TL08 as an inverting amplifier with 3.3 gain Use the potentiometer as input and the voltmeter to measure the output Use both 9V batteries Experiment with gain start with a signal between and V, to get an output of 9 to 9V 9 to 9V adjustable Signal (set near zero) Test output with voltmeter Analog circuits: Opamps Eg: Inverting amplifier. I Eg : Z = 00kW = 0kW Eg : Z = 00kW V Z V out = 0 V in V = 0 = W V out = 00,000 V in!! Not likely. I = V in/ V out = 0 Z I V out = (Z / ) V in 0x gain is a reasonable value Analog circuits: Real Opamps I Eg : Z = 00kW V Z = W V out = 00,000 V in!! Several problems: I = A for V in = V!! (excessive load for upstream circuitry) ~ 3 MHz. This would limit the bandwidth of the amplifier from up to 30 Hz (i.e. not a very responsive system!). 4

5 Analog circuits: Real Opamps Analog circuits: Real Opamps V V Output current limitations I Z Output current limitations Since V is a virtual ground, input impedance seen by V in is Since Opamp 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. Analog circuits: Real Opamps I Open loop gain (K) log K V Z Output current limitations Slewrate is a similar limit: it is a limit on the rate of change of output voltage 0 00 khz log w GainBandwidth limit (Hz) = Gain * Max. = CONSTANT Analog circuits: Real Opamps V TL08: Gain*Bandwidth = 3 MHz This means that at a gain of 00, Bandwidth is 30 khz. I Output current limitations Opamp 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. Eg: 0 na bias current * 0 MW = 00 mv! Z 5

6 Analog circuits: Real Opamps Output current limitations V V Opamp 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 Opamp to behave unexpectedly. Railtorail Opamps are an expensive solution to this limitation. Vcc Vcc Analog circuits: Real Opamps Output current / voltage limitations V V Opamp 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 Opamps exist that can provide much higher current levels. Output voltage range is also limited within a few volts of the power rails. Vcc Vcc Analog circuits: Real Opamps Summary: Keep resistors in K 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 opamp (power opamps exist for this) Keep input voltages away from the opamp voltage rails (unless using railtorail opamps) 6

7 Analog circuits filtering and ion To understand filters you should first understand the difference between the TIME DOMAIN and FREQUENCY DOMAIN Peak Which is the correct match between the following timedomain (left) signals and their Fourier Transforms (right)? a) b) t i) ii) f ) ai, biv, cii, diii ) aii, biv, ciii, di 3) aiii, biv, ci, dii Demo: generator and Spectrum analyzer c) iii) 4) aiv, bii, ciii, di Generator signal Spectrum Analyzer d) iv) ground Transfer Function = / = w) So: V out (w) = w)*v in (w) This is all in terms of w since, in general, impedances are functions of w. Z cap = /j wc Z ind = j wl Z res = R Similar to voltage divider: except w dependent. V out (w) = [V in (w)/( Z )]*Z So: w) = Z /( Z ) For resistors, this is just the well known voltage divider: R /(R R ) Z w) Z Z Generator V in V out Z Spectrum Analyzer Generator V in V out Z Spectrum Analyzer ground ground 7

8 Now plug in a resistor and a capacitor: / jwc Z = /j wc H ( w) R / jwc = R w) jwrc For low frequencies (small w), H = For high frequencies (large w), H 0 This is a LOW PASS FILTER jwrc At w = /RC, H begins to decrease in amplitude. Generator 3300 k C 50 nf 00 nf Spectrum Analyzer Generator 3300 k C 50 nf 00 nf Spectrum Analyzer Z f 0 = /(prc) = 30 Hz Z Low Pass Low Pass Build a circuit that: Uses a capacitor, a switch, an LED and resistors, to implement a low pass filter. When the button is pressed and held, the LED should light slowly. Calculate the time constant see if it matches what you see. Analog circuits: Transfer Functions Bode plots: a graphical representation of frequency response on logarithmic axes. (0 is used instead of 0 so the Vertical axis: result will represent power ~ V ) log 0 (H) dbv = 0log 0 ( V ) Horizontal axis: log 0 (f) 3 db = ½ as much power as 0 db V out is / of V in at 3dB Log of frequency is used to ensure linear plots from /f or /f n functions Bode Plot: H ( w) jwrc Pole: /(jw/w 0 ) 0 db/decade in amplitude after w 0, 90 phase Zero: (jw/w 0 ) 0 db/decade in amplitude after w 0, 90 phase 3dB, 30 Hz 0db/decade / pole 45 deg, 30 Hz 90 deg / pole 8

9 Analog circuits: Transfer Functions Bode plots: a graphical representation of frequency response on logarithmic axes. jwc w) ( jwr C )( jwr C ) Zero at w=0 Bode Plot: Analog circuits: Simple Pole H ( w) jwrc Pole at w=/(r C ) Pole at w=/(r C ) Pole: /(jw/w 0 ) 0 db/decade in amplitude after w 0, 90 phase Zero: (jw/w 0 ) 0 db/decade in amplitude after w 0, 90 phase 3dB, /RC 0db/decade 45 deg, /RC 90 deg Analog circuits: Simple Zero w ) jwrc Bode Plot: C How does this circuit affect the following waveform: ) 3) 3dB, /RC 0db/decade 45 deg, /RC 90 deg ) 4) High Pass Build and explain what this circuit does: Active Band Pass: Analog circuits: Active s C Combines a high and a low pass filter to create a pass band. R C U TL08 Z Zero at w=0 jwc w) ( jwr C )( jwr C ) Pole at w=/(r C ) Pole at w=/(r C ) 9

10 Analog circuits: Active s 0log( H ) Idealized Bode Plot: 0log(R /R ) db () () Analog circuits filtering and ion Band Pass 0 db () () () (3) 0db/decade () w /(R C ) /(R C ) Zero at w=0 jwc w) ( jwr C )( jwr C ) () (3) Pole at w=/(r C ) Pole at w=/(r C ) Low pass R C C U TL08 R C C U TL08 Use multiple stages to get steeper filter rolloffs H tot (w) = H (w) * H (w) *H 3 (w) Remember 0dB/dec for each POLE R C C U TL08 More advanced filters: Biquad Discrete devices: diodes D N5400 V 9V L V in = 9V, V out is connected to scope only. D DIODE DISCUSSION Q: What is wrong with this circuit, when implemented with a TL08 OPAMP? a) V out = 9 V if = K b) V out = ~8.3 V regardless of c) V out = ~8.3 V if = K d) V out = 9 V if is removed e) V out = ~8.3 V if is removed k Vcc V U UA74 D N94 Vcc V U UA74 The following are true:. a (Try it with your own meter) Vee V C Vee V. b,c,e D N94 3. c,d 4. a,e 5. c 59 0

11 Debugging Circuits Learn to systematically check your circuits: Power rails: Check that is really ; 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 if ok plug into stage and see if stage output is degraded. If ok check output of stage etc Keep wiring TIDY! Lab 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 signal!