* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Discrete devices: FETs - UBC Engineering Physics Project Lab
Mechanical filter wikipedia , lookup
Audio crossover wikipedia , lookup
Digital electronics wikipedia , lookup
Spectrum analyzer wikipedia , lookup
Analog television wikipedia , lookup
Telecommunication wikipedia , lookup
Broadcast television systems wikipedia , lookup
Oscilloscope types wikipedia , lookup
Regenerative circuit wikipedia , lookup
Schmitt trigger wikipedia , lookup
Phase-locked loop wikipedia , lookup
Index of electronics articles wikipedia , lookup
Analog-to-digital converter wikipedia , lookup
Oscilloscope history wikipedia , lookup
Tektronix analog oscilloscopes wikipedia , lookup
Surge protector wikipedia , lookup
Analogue filter wikipedia , lookup
Zobel network wikipedia , lookup
Electronic engineering wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Power electronics wikipedia , lookup
Wien bridge oscillator wikipedia , lookup
Distributed element filter wikipedia , lookup
Valve audio amplifier technical specification wikipedia , lookup
RLC circuit wikipedia , lookup
Transistor–transistor logic wikipedia , lookup
Current mirror wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Valve RF amplifier wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Network analysis (electrical circuits) wikipedia , lookup
Flexible electronics wikipedia , lookup
Rectiverter wikipedia , lookup
Integrated circuit wikipedia , lookup
Radio transmitter design wikipedia , lookup
Discrete devices: FETs Field Effect Transistors Gate voltage either ENHANCES or DEPLETES the conduction channel. JFET = Junction FET MOSFET = Metal Oxide Semiconductor FET MOSFETS have an insulating layer at gate so draw less current. Current passing from source to drain now controlled by VOLTAGE at the gate (rather than by CURRENT into the base as in a BJT). 1 12V +V 12V +V R8 10k Discrete devices: FETs R1 10k Q3Effect Transistors Q4 Field MTP2955 MTP2955 R2 R7 There are FOUR 100 100 kinds of MOSFETs: Q8 2N3904 HB-2 Enhancement Mode: Q5 2N3904 N type HB-1 P type M1 12V +V 12V +V 12V +V Q7 2N3906 Q6 2N3906 Id R6 100 R8 10k Q1 Q2 HUF 75321MTP3055 MTP3055 + R5 10k 12V +V Vgs - - Vgs + Q3 R3 Q4 MTP2955 MTP2955 100IRF5305 R7 100 Q8 2N3904 HB-2 R4 10k Id Q5 2N39 M1 12V +V Increasing Vgs increases Id. Q7 2N3906 R6 100 Depletion : Increasing Vgs decreases Id 2 2 Q1 Q2 3 Discrete devices: FETs Enhancement: N type 12V +V P type 12V +V 12V +V 12V +V R7 100 L1 R8 10k Q3 S1 IRF5305 MTP2955 HUF 75321 MTP3055V R7 100 R8 10k L1 S1 4 Discrete devices: FETs 12V +V 5V 12V +V R7 100 L1 S1 HUF 75321 MTP3055V R8 10k gds GND 5 Discrete devices: FETs 12V +V 5V 12V +V R7 100 L1 S1 HUF 75321 MTP3055V R8 10k GND 6 Discrete devices: FETs If after closing S1 (and turning on the LED) we cut the circuit at the red line: 12V +V 1. The LED will stay lit 2. The LED will go off 12V +V R7 100 L1 S1 HUF 75321 MTP3055V R8 10k 7 Discrete devices: FETs 12V +V 5V 12V +V R7 100 L1 S1 MTP3055V R8 10k GND Gate is floating, but lamp is still lit!! 8 Lecture 2 – Analog circuits Seeing the light….. I t IR detection Noise sources: • Electrical (60Hz, 120Hz, 180Hz….) • Other electrical • IR from lights • IR from cameras (autofocus) • Visible light V1 9V +V IR light Q1 OP805 RL Vout Vout ~ mV t What we want: 0 – 5 V DC signal representing the IR amplitude. Analog circuits – filtering and detection IR DC detect block Amplify Peak detect Filter Analog circuits – filtering and detection IR DC detect block Amplify Peak detect Filter Analog circuits – discrete devices: BJT Application: light detection Phototransistor: Acts like BJT except charge carriers generated by incident light add to the base current. In other words, Ic Incident light V1 9V +V Ic + Q1 OP805 RL Vout For the circuit below, 1) Smaller R increases the DC sensitivity to light 2) Larger R increases the DC sensitivity to light + Selecting RL …. Vout = Ic * RL Analog circuits – filtering and detection Z2/Z1= 3 What is the result of the following: Vout + Z1 Z2 Vout = 1) 3) 2) 4) Analog circuits – filtering and detection IR DC detect block Amplify Peak detect Filter Analog circuits – DC block Capacitors: • Open circuits for DC • Short circuit at high frequencies Analog circuits – filtering and detection IR DC detect block Amplify Peak detect Filter Analog circuits: Op-amps +Vcc V+ + Ideal op-amp: V- Vout - K ~ 106 (at DC!) -Vcc With feedback to limit gain: Vout = K (V+ - V-) V+ + V- - Vout V- = Vout Vout = K (V+ - Vout) When negative feedback is applied to an op-amp, V+ = V- ….if the amplifier is within its operating range. Vout(1+K) = KV+ Vout=V+(K/(1+K)) Vout= V+ The next few slides will show some bad design choices Analog circuits: Op-amps Eg: Inverting amplifier. Vin + V- Z1 - V- = 0 I1 = Vin/Z1 Z2 I1 Vout Vout = 0 – Z2I1 Vout= - (Z2/Z1) Vin Eg 1: Z2 = 100kW Z1 = 10kW Vout= - 10 Vin Eg 2: Z2 = 100kW Z1 = 1 W Vout= - 100,000 Vin !! Not likely…. 10x gain is a “reasonable” value Analog circuits: Real Op-amps + Vin Z1 I1 V- Vout Z2 Eg 2: Z2 = 100kW Z1 = 1 W Vout= - 100,000 Vin !! Several problems: • I1 = 1A for Vin = 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!). Analog circuits: Real Op-amps Things to consider: • Input impedance • Gain Bandwidth product • Bias Currents + • Voltage limitations Vin • Output current limitations Z1 I1 V- Z2 Since V- is a virtual ground, input impedance seen by Vin is Z1 Vout Analog circuits: Real Op-amps Things to consider: • Input impedance • Gain Bandwidth product • Bias Currents • Voltage limitations Vin + V- Vout - • Output current limitations 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. Analog circuits: Real Op-amps + Things to consider: Vin • Input impedance • Voltage limitations Open loop gain (K) log K • Output current limitations Slew-rate is a similar limit: it is a limit on the rate of change of output voltage - Z2 I1 • Gain Bandwidth product • Bias Currents V- Z1 Vout 20 Frequency 100 kHz log w Gain-Bandwidth limit (Hz) = Gain * Max. Frequency = CONSTANT TL082: Gain*Bandwidth = 3 MHz This means that at a gain of 100, Bandwidth is 30 kHz. Analog circuits: Real Op-amps Things to consider: • Input impedance • Gain Bandwidth product • Bias Currents • Voltage limitations + Vin Z1 V- Vout Z I1 2 • Output current limitations 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. Eg: 20 nA bias current * 10 MW = 200 mV! Analog circuits: Real Op-amps +Vcc Things to consider: • Input impedance • Gain Bandwidth product V+ + V- Vout -Vcc • Bias Currents • Voltage limitations • Output current limitations 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. Analog circuits: Real Op-amps +Vcc V+ Things to consider: • Input impedance + V- • Gain Bandwidth product Vout -Vcc • Bias Currents • Voltage limitations • Output current / voltage limitations 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. Q: We wish to amplify a 100 kHz, 10mV p-p sine wave, as much as possible. Which is the best circuit? 100k V2 -5m/5mV 1) 100k 9V 2k 9V + 2k + Vout 100kHz TL082 120k V2 -5m/5mV 2) 120k 9V 10k 10k 100kHz 10k 120k 10k TL082 -9V TL082 120k 9V + Vout -9V -9V 9V 10k + TL082 120k 3) 9V + TL082 -9V V2 -5m/5mV 100k 9V + 100kHz TL082 9V + 10k TL082 -9V + Vout TL082 -9V Analog circuits: Real Op-amps Application: amplifier stages R3 30k U1 TL082 Vin Vout 1k R2 R6 1k 10k R1 Total gain: 30*30*11 = 9900 Input impedance = high Max Bandwidth = 100 kHz R5 30k TL082 U2 1k R4 TL082 U3 Analog circuits – filtering and detection IR DC detect block Amplify Peak detect Filter Analog circuits: Filters To understand filters you should first understand the difference between the TIME DOMAIN and FREQUENCY DOMAIN Which is the correct match between the following time-domain (left) signals and their Fourier Transforms (right) ? a) t i) f 1) a-i, b-iv, c-ii, d-iii 2) a-ii, b-iv, c-iii, d-i b) ii) 3) a-iii, b-iv, c-i, d-ii 4) a-iv, b-ii, c-iii, d-i c) iii) d) iv) Analog circuits: Filters Demo: Frequency generator and Spectrum analyzer Frequency Generator signal ground Spectrum Analyzer Analog circuits: Filters “Transfer Function” = Vout/Vin = H(w) Zcap = 1/j wC So: Vout (w) = H(w)*Vin(w) Zind = j wL This is all in terms of w since, in general, impedances are functions of w. Zres = R Similar to voltage divider: except w dependent. Frequency Generator Vin Z1 Vout Z2 ground Spectrum Analyzer Analog circuits: Filters Z2 H (w ) Z1 + Z 2 Vout (w) = [Vin(w)/(Z1+Z2)]*Z2 So: H(w) = Z2/(Z1+Z2) For resistors, this is just the well known voltage divider: R2/(R1+R2) Frequency Generator Vin Z1 Vout Z2 ground Spectrum Analyzer Analog circuits: Filters Now plug in a resistor and a capacitor: 1 / j wC H (w ) R + 1 / jwC Z2 = 1/j wC Z1= R 1 1 + jwRC Z1 Frequency Generator Vin R1 100kk 33 Vout C1 50 nF 100 nF Z2 Spectrum Analyzer Analog circuits: Filters 1 H (w ) 1 + jwRC For low frequencies (small w), H = 1 For high frequencies (large w), H 0 This is a LOW PASS FILTER At w = j/RC, (the pole of this function) H is infinite. At the real value w = 1/RC, H begins to decrease in amplitude. Z1 Frequency Generator Vin R1 100kk 33 Vout C1 50 nF 100 nF Z2 f0 = 1/(2pRC) = 30 Hz Spectrum Analyzer 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 ~ V2) dBV = 20log10(|V|) Horizontal axis: log10(f) -3 dB = ½ as much power as 0 dB Vout is 1/√2 of Vin at -3dB Log of frequency is used to ensure linear plots from 1/f or 1/fn functions Pole: 1/(1+jw/w0) -20 db/decade in amplitude after w0, -90 phase Zero: (1+jw/w0) +20 db/decade in amplitude after w0, +90 phase Analog circuits: Filters Bode Plot: 1 H (w ) 1 + jwRC -3dB, 30 Hz -20db/decade / pole - 45 deg, 30 Hz -90 deg / pole Analog circuits: Transfer Functions Bode plots: a graphical representation of frequency response on logarithmic axes. Zero at w=0 H (w ) - Pole at w=1/(R2C2) jwR1C2 (1 + jwR2C2 )(1 + jwR1C1 ) Pole at w=1/(R1C1) Pole: 1/(1+jw/w0) -20 db/decade in amplitude after w0, -90 phase Zero: (1+jw/w0) +20 db/decade in amplitude after w0, +90 phase Analog circuits: Simple Pole Bode Plot: 1 H (w ) 1 + jwRC -3dB, 1/RC -20db/decade - 45 deg, 1/RC -90 deg Analog circuits: Simple Zero H (w ) 1 + jwRC Bode Plot: +3dB, 1/RC +20db/decade +45 deg, 1/RC +90 deg Analog circuits: Filters 1 H (w ) 1 + jwRC Ch0 Vout Ch1 Vin R1 100kk 33 Vout C1 50 nF 100 nF Vin f0 = 1/(2pRC) = 30 Hz Gnd Analog circuits: Filters Vin C1 Vout R1 How does this circuit affect the following waveform: 1) 3) 2) 4) Analog circuits: Active Filters R1 Active Band Pass: Z1 C1 C2 R2 Z2 1 Z 2 R2 + jwC2 U1 TL082 1 Z1 ( + jwC1 ) -1 R1 H(w) = - Z1/Z2 Analog circuits: Active Filters Active Band Pass: R1 Z1 C1 C2 R2 U1 TL082 Z2 Zero at w=0 jwR1C2 H (w ) (1 + jwR2C2 )(1 + jwR1C1 ) Pole at w=1/(R2C2) Pole at w=1/(R1C1) Analog circuits: Active Filters 20log(|H|) Idealized Bode Plot: 20log(R1/R2) dB (1) (2) (1) (2) (3) (1) 0 dB 1/(R2C2) 1/(R1C1) 20db/decade (1) Zero at w=0 w jwR1C2 H (w ) (1 + jwR2C2 )(1 + jwR1C1 ) (2) Pole at w=1/(R2C2) (3) Pole at w=1/(R1C1) Analog circuits: Active Filters phase(H) Idealized Bode Plot: (1) -90 (1) (2) -180 (1) (2) (3) -270 w 1/(R2C2) 1/(R1C1) Zero at w=0 (1) jwR1C2 H (w ) (1 + jwR2C2 )(1 + jwR1C1 ) (2) Pole at w=1/(R2C2) (3) Pole at w=1/(R1C1) Analog circuits: Active Filters Which best describes the amplitude response (Bode plot) of the following transfer function (R1C1 > R2C2): 1) (1 + jwR1C1 ) H (w ) (1 + jwR2C2 ) 2) 0db 0db 1/ R1C1 1/ R2C2 3) 1/ R1C1 1/ R2C2 4) 0db 1/ R1C1 1/ R2C2 0db 1/ R1C1 1/ R2C2 Analog circuits – filtering and detection Band Pass IR DC detect block Amplify Filter R1 R1 C1 R2 Low pass C2 U1 TL082 R1 C1 R2 C2 U1 TL082 C1 R2 C2 U1 TL082 Use multiple stages to get steeper filter roll-offs… Htot(w) = H1(w) * H2 (w) *H3(w) Remember –20dB/dec for each POLE 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. • 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! DISCUSSION Q: What is wrong with this circuit, when implemented with a TL082 OP-AMP? Vcc +12V Vin + Vcc +12V U1 UA741 D2 1N914 + C1 Vee -12V D1 1N914 R1 Vee -12V U2 UA741 Vout