المملكة العربية السعودية وزارة التعليم العالي - جامعة أم القرى كلية الهندسة و العمارة الإسلامية قسم الهندسة الكهربائية ELECTRONIC DEVICES K INGDOM OF S AUDI A RABIA Ministry of Higher Education Umm Al-Qura University College of Engineering and Islamic Architecture Electrical Engineering Department Lecture 7 By: Dr Tarek Abdolkader

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader What is a Bipolar junction transistor? Construction and symbol Active mode of operation Cut-off and saturation modes BJT characteristics BJT specifications BJT as an amplifier BJT as a switch 2

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader A transistor is a semiconductor device used to amplify and switch electronic signals. The name transistor is a combination of the words transfer and resistor - a transferred resistor This is because it can amplify electrical signals by transferring a current I from a low to a high- resistance circuit. 3 Bipolar junction transistor (BJT) is one of the most famous types of transistors. Bipolar: current is carried by two types of carriers (electrons and holes). Junction: current passes through pn-junctions BJT is a three terminal device.

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader4 Current increases with v i but v o has the same amplitude as v i (no amplification) Amplification of signals

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader There are two main types of BJT: 1. npn transistor2. pnp transistor 5

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader6 The arrow on the emitter is in the direction from p to n.

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader7 In the normal operation of the transistor (when it is used as an amplifier), The Base-Emitter junction ( BE) is forward biased and the Base-Collector junction (BC) is reverse biased.

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader8 p  EfEf EfEf  n p p   np at equilibrium BE is forwarded and BC is reversed n p n   n p n  

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader9 The emitter is made heavily doped to emit large number of carriers into base. The base is made very thin so the least number of carriers injected from the emitter is lost by recombination before reaching the collector. The collector collects the minority carriers from the emitter through the base. The collector usually has the lightest doping concentrations of the three regions. The main source of current collected at the collector is the current sent from the emitter. The collector own current is very small (reverse-biased junction). The collector current is controlled by the voltage in a different circuit ( V BE ). Small changes in V BE leads to large changes in collector current ( I C ). The main difference between transistor and p-n junction is that the current is collected in a different circuit with different resistance (transfer resistor).

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader10 p n p 3.Some of the holes injected from the emitter are recombined with some of the electrons diffusing from the base (small portion because the base is thin). 4.Most of the hole current injected from the emitter is collected by the collector. 5.Minority holes are drifted by the reverse-bias field of the BC junction. 6.Minority electrons are drifted by the reverse-bias field of the BC junction. 1.Hole diffusion in the forward- biased BE junction. 2.Electron diffusion in the forward-biased BE junction (it is less than hole diffusion because base is lightly doped while emitter is heavily doped). Transistor currents:

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader11 Transistor equations in active mode: The common-emitter dc current gain β DC : β DC >> 1 The common-base dc current gain α DC : α DC < 1 From KCL: I E = I C + I B I C = β DC I B I E = (1 + β DC ) I B Remember that for forward-biased Si pn junction, V BE = 0.7 V All above equations are valid only in the active mode. V BE = V BC + V CE

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader Determine all currents and voltages in the circuit shown. β DC =

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader13 Cut-off mode: BE is reverse-biased and BC is also reverse-biased No current will be in both circuits except the reverse saturation currents (can be neglected). I B = 0, I C = 0, I E = 0 n p n   n p n  

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader14 Saturation mode: BE is forward-biased and BC is also forward-biased The field in the BC junction is no longer sufficient to collect all the current injected by the emitter. In this case, increasing the base current will not result in an increase in the collector current. We say that the current is saturated. In this case I C < β DC I B. Saturation starts before the BC junction is completely forward- biased ( V BC 0.5 ~ 0.7 V, V CE(sat) 0 ~ 0.2 V). n p n   n p n  

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader15 V BE V BC active saturation Cut-off inverse V BE V BC active saturation Cut-off inverse npn pnp As long as BC junction is reverse biased ( V CE > V CE(sat) ), the collector is able to withdraw α DC I E ( β DC I B ) and the current will depend only on I B and not on V CE. Active region I C = β DC I B saturation

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader Sketch an ideal family of collector curves for the circuit shown for I B = 5  A to 25  A in 5  A increments. β DC = V

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader17 Determine I B from the input circuit. (Use V BE = 0.7 V for forward-biased BE junction). Draw the load line and determine the operating point. Load line analysis: V CC = I C R C + V CE Note that the onset of saturation for all base currents is nearly the same. This value is called V CE(sat) ~ 0 → 0.2 V I C = β DC I B

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader Determine the mode of operation of the transistor of the circuit shown. Assume V CE(sat) = 0.2 V, β DC = Draw the collector characteristics and the load line. Then determine the Q-point. 2.Assume the transistor is in the active mode ( I C = β DC I B ), then find V CE. If V CE > V CE(sat), then the transistor is really in active mode. Otherwise, it is saturation. 3.Assume the transistor is in the saturation mode ( V CE = V CE(sat) ), then find I C. If I C < β DC I B, then the transistor is really in saturation mode. Otherwise, it is active. There are three methods to solve this problem:

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader19 BJT data sheet:

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader20 Amplification is the process of linearly increasing the amplitude of an ac signal. The transistor in the active mode amplifies the base current linearly ( I c = β I b ) Large changes in collector current leads to large changes in the collector voltage V c. Voltage gain is the ratio V c / V in. V in IbIb IcIc VcVc

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader21 Input voltage is zero (low). Transistor is in cut-off mode. Base current is zero. Collector current is zero. CE junction is open-circuited. V CE = V CC (high) Input voltage is + V BB (high). Transistor is in saturation mode. Base current is high. Collector current is I C(sat) (high). CE junction is nearly short- circuited. V CE = V CE(sat) ~ 0.2 V (low) Transistor switch is called an inverter because low input results in high output and vice versa.

10/1/1434Electronic devices (802311) Lecture 7 Dr Tarek Abdolkader For the circuit shown, (a) Find V CE for V IN = 0, (b) Find the maximum value of R B to drive the transistor into saturation with V IN = 5 V, β DC = 200. Neglect V CE(sat). 22