* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Transistors - SFA Physics and Astronomy
Audio power wikipedia , lookup
Mercury-arc valve wikipedia , lookup
Negative feedback wikipedia , lookup
Stepper motor wikipedia , lookup
Power inverter wikipedia , lookup
Pulse-width modulation wikipedia , lookup
Thermal runaway wikipedia , lookup
History of electric power transmission wikipedia , lookup
Three-phase electric power wikipedia , lookup
Electrical substation wikipedia , lookup
Variable-frequency drive wikipedia , lookup
Electrical ballast wikipedia , lookup
Stray voltage wikipedia , lookup
Voltage optimisation wikipedia , lookup
Power electronics wikipedia , lookup
Mains electricity wikipedia , lookup
Voltage regulator wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Power MOSFET wikipedia , lookup
Alternating current wikipedia , lookup
Buck converter wikipedia , lookup
Current source wikipedia , lookup
Two-port network wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Schmitt trigger wikipedia , lookup
Semiconductor device wikipedia , lookup
Wilson current mirror wikipedia , lookup
History of the transistor wikipedia , lookup
Transistors Transfer Resistor Chapter 9 Bipolar Transistors Collector Base Emitter Two PN junctions joined together Two types available – NPN and PNP The regions (from top to bottom) are called the collector (C), the base (B), and the emitter (E) Operation Begin by reverse biasing the CB junction Here we are showing an NPN transistor as an example Now we apply a small forward bias on the emitter-base junction Electrons are pushed into the base, which then quickly flow to the collector The result is a large emitter-collector electron current (conventional current is C-E) which is maintained by a small E-B voltage Some of the electrons pushed into the base by the forward bias E-B voltage end up depleting holes in that junction This would eventually destroy the junction if we didn’t replenish the holes The electrons that might do this are drawn off as a base current Currents Conventional View Origin of the names the Emitter 'emits' the electrons which pass through the device the Collector 'collects' them again once they've passed through the Base ...and the Base?... Original Manufacture Base Thickness The thickness of the unmodified Base region has to be just right. Too thin, and the Base would essentially vanish. The Emitter and Collector would then form a continuous piece of semiconductor, so current would flow between them whatever the base potential. Too thick, and electrons entering the Base from the Emitter wouldn't notice the Collector as it would be too far away. So then, the current would all be between the Emitter and the Base, and there'd be no Emitter-Collector current. Amplification Properties The C-B voltage junction operates near breakdown. This ensures that a small E-B voltage causes avalanche Large current through the device Common Base NPN Common Emitter NPN Common Collector NPN How does IC vary with VCE for various IB? Note that both dc sources are variable Set VBB to establish a certain IB Collector Characteristic Curve If VCC = 0, then IC = 0 and VCE = 0 As VCC ↑ both VCE and IC ↑ When VCE 0.7 V, base-collector becomes reverse-biased and IC reaches full value (IC = bIB) IC ~ constant as VCE ↑. There is a slight increase of IC due to the widening of the depletion zone (BC) giving fewer holes for recombinations with e¯ in base. Since IC = bIB, different base currents produce different IC plateaus. NPN Characteristic Curves PNP Characteristic Curves Load Line Slope of the load line is 1/RL For a constant load, stepping IB gives different currents (IC) predicted by where the load line crosses the characteristic curve. IC = bIBworks so long as the load line intersects on the plateau region of the curve. Saturation and Cut-off Cut-off Note that the load line intersects the 75 mA curve below the plateau region. This is saturation and IC = bIB doesn’t work in this region. Example We adjust the base current to 200 mA and note that this transistor has a b = 100 Then IC = bIB = 100(200 X 10-6A) = 20 mA Notice that we can use Kirchhoff’s voltage law around the right side of the circuit VCE = VCC – ICRC = 10 V – (20 mA)(220 W) = 10 V – 4.4 V = 5.6 V Example Now adjust IB to 300 mA Now we get IC = 30 mA And VCE = 10 V – (30 mA)(220 W) = 3.4 V Finally, adjust IB = 400 mA IB = 40 mA and VCE = 1.2 V Plot the load line VCE IC 5.6 V 20 mA 3.4 V 30 mA 1.2 V 40 mA Gain as a function of IC As temperature increases, the gain increases for all current values. Operating Limits There will be a limit on the dissipated power PD(max) = VCEIC VCE and IC were the parameters plotted on the characteristic curve. If there is a voltage limit (VCE(max)), then you can compute the IC that results If there is a current limit (IC(max)), then you can compute the VCE that results Example Assume PD(max) = 0.5 W VCE(max) = 20 V IC(max) = 50 mA PD(max) VCE 0.5 W IC 5V 100 mA 10 50 15 33 20 25 Operating Range Operating Range Voltage Amplifiers Common Base PNP Now we have added an ac source The biasing of the junctions are: BE is forward biased by VBB - thus a small resistance BC is reverse biased by VCC – and a large resistance Since IB is small, IC IE Equivalent ac Circuit rE = internal ac emitter resistance IE = Vin/rE (Ohm’s Law) Vout = ICRC IERC Vout AV voltage gain Vin I E RC RC AV I E rE rE Recall the name – transfer resistor Current Gains Common Base a = IC/IE < 1 Common Emitter b = IC/IB From Kirchhoff' s Current Law I E IC I B IE IB 1 IC IC 1 a 1 1 b 1 1 b a b b a ab b a( 1 b ) b a 1 b Example If b = 50, then a = 50/51 = 0.98 Recall a < 1 Rearranging, b = a + ab b(1-a) = a b = a/(1-a) Transistors as Switches The operating points We can control the base current using VBB (we don’t actually use a physical switch). The circuit then acts as a high speed switch. Details In Cut-off All currents are zero and VCE = VCC In Saturation IB big enough to produce IC(sat) bIB Using Kirchhoff’s Voltage Law through the ground loop VCC = VCE(sat) + IC(sat)RC but VCE(sat) is very small (few tenths), so IC(sat) VCC/RC Example a) What is VCE when Vin = 0 V? Ans. VCE = VCC = 10 V b) What minimum value of IB is required to saturate the transistor if b = 200? Take VCE(sat) = 0 V IC(sat) VCC/RC = 10 V/1000 W = 10 mA Then, IB = IC(sat)/b = 10 mA/200 = 0.05mA Example LED If a square wave is input for VBB, then the LED will be on when the input is high, and off when the input is low. Transistors with ac Input Assume that b is such that IC varies between 20 and 40 mA. The transistor is constantly changing curves along the load line. Pt. A corresponds to the positive peak. Pt. B corresponds to the negative peak. This graph shows ideal operation. Distortion The location of the point Q (size of the dc source on input) may cause an operating point to lie outside of the active range. Driven to saturation Driven into Cutoff Base Biasing It is usually not necessary to provide two sources for biasing the transistor. The red arrows follow the base-emitter part of the circuit, which contains the resistor RB. The voltage drop across RB is VCC – VBE (Kirchhoff’s Voltage Law). The base current is then… VCC VBE IB RC and IC = bIB Base Biasing Use Kirchhoff’s Voltage Law on the black arrowed loop of the circuit VCC = ICRC + VCE So, VCE = VCC – ICRC VCE = VCC – bIBRC Disadvantge b occurs in the equation for both VCE and IC But b varies – thus so do VCE and IC This shifts the Q-point (b-dpendent) Example Let RC = 560 W RB = 100 kW VCC = +12 V @ 25 °C b = 100 @ 75 °C b = 150 @ 25 C VCC VBE 12 V - 0.7 V IB 113mA RB 100,000 W IC b I B (100)(113 mA) 11.3 mA VCE VCC b I B R C @ 75 °C IB is the same IC = 16.95 mA VCE = 2.51 V 12 V - (100)(113 mA)(560 W ) IC increases by 50% 5.67 V VCE decreases by 56% Transistor Amplifiers Amplification The process of increasing the strength of a signal. The result of controlling a relatively large quantity of current (output) with a small quantity of current (input). Amplifier Device use to increase the current, voltage, or power of the input signal without appreciably altering the essential quality. Class A Entire input waveform is faithfully reproduced. Transistor spends its entire time in the active mode Never reaches either cutoff or saturation. Drive the transistor exactly halfway between cutoff and saturation. Transistor is always on – always dissipating power – can be quite inefficient Class A Class B No DC bias voltage The transistor spends half its time in active mode and the other half in cutoff Push-pull Pair Transistor Q1 "pushes" (drives the output voltage in a positive direction with respect to ground), while transistor Q2 "pulls" the output voltage (in a negative direction, toward 0 volts with respect to ground). Individually, each of these transistors is operating in class B mode, active only for one-half of the input waveform cycle. Together, however, they function as a team to produce an output waveform identical in shape to the input waveform. Class AB Between Class A (100% operation) and Class B (50% operation). Class C IC flows for less than half then cycle. Usually get more gain in Class B and C, but more distortion Common Emitter Transistor Amplifier Notice that VBB forward biases the emitter-base junction and dc current flows through the circuit at all times The class of the amplifier is determined by VBB with respect to the input signal. Signal that adds to VBB causes transistor current to increase Signal that subtracts from VBB causes transistor current to decrease Details At positive peak of input, VBB is adding to the input Resistance in the transistor is reduced Current in the circuit increases Larger current means more voltage drop across RC (VRC = IRC) Larger voltage drop across RC leaves less voltage to be dropped across the transistor We take the output VCE – as input increases, VCE decreases. More details As the input goes to the negative peak Transistor resistance increases Less current flows Less voltage is dropped across RC More voltage can be dropped across C-E The result is a phase reversal Feature of the common emitter amplifier The closer VBB is to VCC, the larger the transistor current. PNP Common Emitter Amplifier NPN Common Base Transistor Amplifier Signal that adds to VBB causes transistor current to increase Signal that subtracts from VBB causes transistor current to decrease • At positive peak of input, VBB is adding to the input • Resistance in the transistor is reduced • Current in the circuit increases • Larger current means more voltage drop across RC (VRC = IRC) • Collector current increases • No phase reversal PNP Common Base Amplifier NPN Common Collector Transistor Amplifier Also called an Emitter Follower circuit – output on emitter is almost a replica of the input Input is across the C-B junction – this is reversed biased and the impedance is high Output is across the B-E junction – this is forward biased and the impedance is low. Current gain is high but voltage gain is low. PNP Common Collector Transistor Amplifier Gain Factors IC a IE Usually given for common base amplifier IC b IB Usually given for common emitter amplifier IE IB Usually given for common collector amplifier Gamma Recall from Kirchhoff’s Current Law a IB + I C = IE And since b 1-a IB IC IE 1 IB IB 1 b Ex. For b = 100 LCD a 1 1-a 1-a a 1 1-a 1-a a = b/(1+b) = 0.99 = 1 + b = 101 Bringing it Together Type Common Base Common Emitter Common Collector Relation between input/output phase 0° 180° 0° Voltage Gain High Medium Low Current Gain Low (a) Medium (b) High () Power Gain Low High Medium Input Z Low Medium High Output Z High Medium Low Hybrid Parameters Condition hi Input resistance Output shorted hr Voltage feedback ratio Input open hf Forward current gain Output shorted ho Output conductance Input open Second subscript indicates common base (b), common emitter (e), or common collector (c) Hybrid Parameters =b = Slope of curve Hybrid Parameters hie = VB/IB Ohm’s Law hie =input impedance hre = VB/VC Hybrid Parameters hfe = IC/IB Equivalent of b hoe = IC/VC Various Forms Common Emitter (e) Common Base (b) Common Collector (c) hi (ohms) VB/IB VE/IB VB/IB hr (unitless) VB/VC VE/VC VB/VE hf (unitless) IC/IB IC/IE IE/IB ho (watts) ICVC ICVC IEVE Pin-outs No standard – look at the spec sheet or the case Loudness When the energy (intensity) of the sound increases by a factor of 10, the loudness increases by 1 bel Named for A. G. Bell One bel is a large unit and we use 1/10th bel, or decibels When the energy (intensity) of the sound increases by a factor of 10, the loudness increases by 10 dB Decibel Scale For intensities L = 10 log(I/Io) For energies L = 10 log(E/Eo) For amplitudes L = 20 log(A/Ao) Threshold of Hearing The Io or Eo or Ao refers to the intensity, energy, or amplitude of the sound wave for the threshold of hearing Io = 10-12 W/m2 Loudness levels always compared to threshold Relative measure Common Loud Sounds 160 Jet engine - close up 150 Snare drums played hard at 6 inches away Trumpet peaks at 5 inches away 140 Rock singer screaming in microphone (lips on mic) 130 Pneumatic (jack) hammer Planes on airport runway Cymbal crash 120 Threshold of pain - Piccolo strongly played Fender guitar amplifier, full volume at 10 inches away Power tools 110 Subway (not the sandwich shop) 100 Flute in players right ear - Violin in players left ear Common Quieter Sounds 90 Heavy truck traffic Chamber music 80 Typical home stereo listening level Acoustic guitar, played with finger at 1 foot away Average factory 70 Busy street Small orchestra 60 Average office noise 50 Quiet conversation 40 Quiet office 30 Quiet living room 20 Conversational speech at 1 foot away 10 Quiet recording studio 0 Threshold of hearing for healthy youths The Math l1 = 10 log(I1/Io) l2 = 10 log(I2/Io) l2 – l1 = Dl = 10(log I2 – log Io – log I1 + log Io) = 10(log I2 – log I1) l2 – l1 = Dl = 10 log(I2/I1) Threshold of Hearing when I = Io l = 0 dB Threshold of Pain when I 1012 Io l = 120 dB Example A loudspeaker produces loudness rated at 90 dB (l1) at a distance of 4 ft (d1). How far can the sound travel (d2) and still give a loudness at the listener’s ear of 40 dB (l2 - conversation at 3 ft.)? Sound follows the inverse square law I1/I2 = d22/d12 Dl = 50 dB = 10 log(I2/I1) log(I2/I1) = 5 which means I2/I1 = 105 If d1 = 4 ft, then d22 = (I1/I2) d12 = 105 (4 ft)2 d2 = 1260 ft (about ¼ mile) Common Emitter Current Gain 0 dB -3 dB hfe Frequency For the -3 dB point Dl = 3 dB = 10 log (I1/I2) I1/I2 = 2 = P1/P2 so 3 dB below initial level mean half the power Why do Frequency limits occur? It takes a certain time for e- to travel from emitter to collector (transit time) If frequency is too high, applied current varies too rapidly Electrons may be unable to dislodge rapidly enough to move from E to C before current surges in the other direction. Making the base thinner reduces transit time and improves frequency response Interelement Capacitance As reverse bias increases on the C-B junction, the depletion zone increases and C decreases (C = eA/d and d increasing). As emitter current increases, C increases (d decreasing). If capacitance changes, so does capacitive reactance X 1 C 2 f C Increasing C decreases XC Feedback Small base current provides a path back to input If the feedback voltage aids the input voltage, then it is positive (regenerative) feedback If the feedback is too large, the amplifier will oscillate Superheterodyne Receiver