IPOLAR JUNCTION TRANSISTORS (JTs) Dr Derek Molloy, DCU
What are JTs? Two PN junctions joined together is a JT Simply known as a transistor! ipolar? Current carried by electrons and holes Will see FETs (Field Effect Transistors) JTs have a higher gain (amplification). JTs can supply more current. FETs are less complex and require less power.
Production Example of a bipolar transistor production
Transistor types So, they are like diodes: Semiconductor material Doped p and n regions Unlike a diode: 3 alternated doped regions p-n-p or n-p-n Narrow channel between 2 terminals is controlled by a voltage on a 3 rd terminal. Transistor controlled to operate as a switch or a variable resistor Key element in the design of amplifier
JT CONFIGURATIONS
JT PRINCIPLE OF OPERATION Common Emitter Configuration I C Collector I b V E + ase n p n Emitter + V CE A Current Amplifier! A small current flowing in the base-emitter circuit can control the amount of a much larger current flowing in the collector-emitter circuit
JT PRINCIPLE OF OPERATION - E diode is forward biased. Closed circuit between the collector and emitter. Current flowing between C-E, E diode is reverse biased. Open circuit, high resistance between the collector and emitter. Small current flowing between C-E, I CEO
SOME CHARACTERISTICS I C (ma) I C β=h FE I ( A) 0.7V V E Current transfer characteristic Current-voltage transfer characteristic (slightly non-linear) (highly non-linear) IF V CE is large I C α I I C depends on V E
ASICS JT I E =I C +I Numbers for illustration of measurements only Not a worked example!
CURRENT GAIN DC or large signal gain: h FE or β h FE I I C For most practical purposes, h fe and h FE are considered equal Small signal gain: h fe h fe di di C I I Voltage Gain is the ratio of output voltage, to input voltage. Current Gain is the ratio of output current, to input current. Transconductance is the ratio of output current, to input voltage. Transimpedance is the ratio of output voltage, to input current. C
COMMON EMITTER IPOLAR AMPLIFIER Input signal between base and emitter. Output signal between collector and emitter. h = 100 FE ase needs to be biased. R = 910k ase-emitter is forward biased due to R R = 4.7k R C sets the quiescent (steady-state with no input signal applied) base I Q, I CQ, V OQ, = V10V CC CEQ V E =0.7V I Q V CC V R E 10V 0. 7V 910k 10. 2A I h I 100 10. 2A 1. 02mA CQ FE Q Coupling Capacitor acts as a high-pass filter, allowing AC signal voltage on to the transistor, while blocking all DC voltage from being shorted through the AC signal source. V i R V E C E R C V CC V cc =10V R =910k R C =4.7k h FE =100 V O 0V 0.00102 4700 5. V VOQV CC ICQRC 10 206
INPUT CHARACTERISTICS I C (ma) I C I Current transfer characteristic (slightly non-linear) I ( A ) 0.7V V E Current-voltage transfer characteristic (highly non-linear) 0.7V V E DC input characteristics I I ev kt 1I AC input impedance h ie S e dve 1 di 40I E ( ) S e 40V E at room temperature I S is a constant determined by the base characteristics
OUTPUT CHARACTERISTICS
OUTPUT CHARACTERISTICS Saturation Region: V CE is not large enough, I C is Independent of I and depends on V CE. Active Region: V CE is large enough, I C is independent of V CE and depends on I (and the gain). Saturation region Active region I c (ma) 60 50 40 30 20 10 0 I (µa) V CE (V) 600 500 400 300 200 100 0 For amplification, JT operates in active region For switch or digital applications, JT swings between saturation (switch on ) and cut-off (switch off)
OPERATING REGION I c A : Maximum of I c Damage Transistor C C: Maximum Power Dissipation V CE and I c P = V CE x I c A: Saturation region Highly non-linear I D D: Maximum V CE Result in Avalanche reakdown of the transistor V CE Operating point placed in the white area A suitable operating point, I cmax /2 to avoid any red zones, close to centre Maximum swing possible with AC input.
VARIATION AROUND THE QUIESCENT POINT I c I i c i Operating Point V CE v CE V CC V I V RC V CE V out Avoid: clamping the wave
Saturation region DC DESIGN PARAMETERS OF A COMMON EMITTER AMPLIFIER The load line graphical method I c (ma) 60 50 40 30 20 10 0 Active region 1 R c h = 100 FE R = 910k R = 4.7k C V = 10V CC I (µa) A V cc 600 500 400 300 200 100 0 V CE (V) I c V cc V R c R ce V R ce c R C V R cc c V CC E V 0CE V i A: cut-off : saturation C mx b 1) Chose an operating point (I C, V CE ) and a power rail V CC. Then draw the load line and calculate the slope. R C is given by -1/slope. 2) Chose R C and V CC. Draw a load line through V CE = V CC with a slope of -1/R C. Then select an operating point somewhere along this load-line.