ET215 Devices I Unit 4A

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1 ITT Technical Institute ET215 Devices I Unit 4A Chapter 3, Section This unit is divided into two parts; Unit 4A and Unit 4B

2 Chapter 3 Section 3.1 Structure of Bipolar Junction Transistors The basic structure of the bipolar junction transistor (BJT) determines its operating characteristics. We will see how the semi-conductive materials form a BJT, and learn the standard transistor symbols.

3 The bipolar junction transistor (BJT) is constructed with three doped semiconductor regions called emitter, base, and collector. The pn junction joining the base region and the emitter region is called the base-emitter junction. The pn junction joining the base region with the collector region is called the base-collector junction. The term bipolar refers to the use of both holes and electrons as carriers in the transistor structure.

4 Transistor Operation The two pn junctions must be supplied with external dc bias voltages to set the proper operating conditions. In figure, both the base-emitter (BE) junction is forward-biased and the base-collector (BC) junction is reverse-biased. This is called forward-reverse bias.

5 Transistor Operation The forward bias from base to emitter narrows the BE depletion region, and the reverse bias from base to collector widens the BC depletion region. Most of the electrons flowing from the emitter into the narrow lightly doped base region do not recombine but diffuse into the BC depletion region. the bottom line A small base current can control a larger collector current.

6 Transistor Currents Kirchhoff s current law (KCL) says the total current entering a junction must be equal to the total current leaving that junction. I E = I C + I B The base current, I B, is very small compared to I E or I C, which leads to the approximation I E I C, a useful assumption for analyzing transistor circuits.

7 Transistor Currents I E = I C + I B Note: emitter arrow points towards lesser voltage.

8 DC Beta (β DC ) The dc beta (β DC ), the current gain of a transistor, is the ration of the dc collector current to the dc base current. β DC = I C / I B The values of β DC vary widely and depend on the type of transistor. They are typically from 20 (power transistors) to 200 (small-signal transistors).

9 Transistor Voltages The three dc voltages for the biased transistor in figure are the emitter voltage (V E ), the collector voltage (V C ) and the base voltage (V B ). The collector power supply voltage, V CC, is shown with repeated subscript letters. Because the emitter is grounded, the collector voltage is equal to the dc supply voltage, V CC, less the drop across R C. V C = V CC - I C R C and V B = V E + V BE = V E V

10 EXAMPLE 3-1: Determine I B, I C, V B, and V C in figure where β DC is 50. I B = 0.23 ma I C = 11.5 ma V B = 0.7V Solution: Since V E is ground, V B = 0.7V. The drop across R B is V BB V B, so I B is calculated as: I B = (V BB V B ) / R B = (3V 0.7V ) / 10kΩ = 0.23 ma Now: I C = β DC I B = (50)(0.23mA) = 11.5 ma

11 EXAMPLE 3-1: Determine I B, I C, V B, and V C in figure where β DC is 50. I B = 0.23 ma I C = 11.5 ma V B = 0.7V V C = 8.5 ma Solution: I E = I C + I B = 11.5 ma ma = 11.7 ma V C = V CC - I C R C = 20 V (11.5 ma)(1kω) = 8.5 V

12 Characteristic Curve for a BJT Note that the IV curve is identical to that of an ordinary diode. You can look for the 0.7V across the base-emitter junction (as in a forward-biased diode) to determine if the transistor is conducting.

13 Collector Characteristic Recall that the collector current is proportional to the base current I C = β DC I B These curves are called collector characteristic curves. By setting I B to other constant values, you can produce additional I C versus V CE curves as shown.

14 EXAMPLE 3-2: Sketch the family of collector curves for the circuit in figure 3-10 for I B = 5 A to 25 A in 5 A increments. Assume that β DC = 100. Solution: Calculate I C = β DC I B I B I C 5 A 0.5 ma 10 A 1.0 ma 15 A 1.5 ma 20 A 2.0 ma 25 A 2.5 ma

15 Cutoff and Saturation When I B = 0, the transistor is in cutoff and there is essentially no collector current except for a very tiny amount of collector leakage current, I CEO, which can usually be neglected.

16 Cutoff and Saturation According to Kirchhoff s voltage law, if the voltage across R C increases, the drop across the transistor must decrease. Ideally, when the base current is high enough, the entire V CC is dropped across R C with no voltage between the collector and emitter. This condition is known as saturation.

17 Cutoff and Saturation Once the base current is high enough to produce saturation, further increases in base current have no effect on the collector current, and the relationship I C = β DC I B is no longer valid.

18 Before the actual collector current can be found, the value of the base current, I B, needs to be established. V RB = V BB - V BE = 12 V 0.7 V = 11.3 V With Ohm s law we find the base current I B = V RB / R B = 11.3 V / 1.0 M = 11.3 A DC Load Line

19 First, the cutoff point on the load line is determined. V CE(cutoff) = V CC = 12V and I C(cutoff) = 0 ma Next, the saturation point on the load line is determined. I C(sat) = V CC / R C = 12 V / 2.0K = 6.0 ma Now, draw the load line Q-Point

20 Chapter 3 Section 3.2 BJT Bias Circuits In this section, methods for biasing a bipolar junction transistor are presented. The choice of biasing circuits depends largely on the application. You will learn about four biasing methods and see the advantages and disadvantages of each method.

21 The simplest biasing circuit is base bias. The collector current, for a linear operation, is I C = β DC I B The base resistor, R B, has base current, I B, through it. Base Bias From Ohm s law, you can substitute for I B and obtain I C = β DC (V RB / R B )

22 EXAMPLE 3-3: If a 2N3904 transistor datasheet shows that the β DC has a range from 100 to 300 and assuming it is used in the base-biased circuit as shown, compute the minimum and maximum collector current based on this specification. Solution: If the voltage drop across base-emitter is 0.7 V V RB = V CC V BE = 12V 0.7V = 11.3V Using Ohm s Law to find the base current I B = V RB / R B = 11.3V/1.0M = 11.3 A The collector current is β DC times I B I C(min) = β DC I B = (100)(11.3 A) = 1.13 ma I C(max) = β DC I B = (300)(11.3 A) = 3.39 ma

23 Collector-Feedback Bias Another type of bias is the collector-feedback bias. The collector feedback uses an important idea in electronics called negative feedback to achieve stability. Negative feedback returns a portion of the output back to the input in a manner to cancel changes that may occur. Current for collector-feedback bias is derived from Kirchhoff s voltage law: I C = (V CC V BE ) / ( R C + R B /β DC )

24 EXAMPLE 3-4: Assume a 2N3904 is used in the collector-feedback biased circuit shown. Compute the minimum and maximum collector current when β DC = 100. Solution: Substituting values I C = (V CC V BE ) / ( R C + R B /β DC ) = (12V 0.7V) /( 2.0k +150k /100) = 3.2 ma Repeat for calculation for βdc = 300. I C = (V CC V BE ) / ( R C + R B /β DC ) = (12V 0.7V) /( 2.0k +150k /300) = 4.5 ma

25 Voltage-Divider Bias An even higher degree of stability can be obtained with voltage-divider bias. V out = ( R 2 / (R 1 +R 2 ) ) V in

26 An even higher degree of stability can be obtained with voltage-divider bias. V B = ( R 2 / (R 1 +R 2 ) ) V CC V E = V B - 0.7V V C = V CC I C R C V CE = V C - V E I E = V E / R E I C I E Voltage-Divider Bias

27 EXAMPLE 3-5: Find V B, V E, I E, V CE for the circuit shown. Solution: Use formulas and substitute values V B = ( R 2 / (R 1 +R 2 ) ) V CC = ( 3.9k / (27k + 3.9k ) ) 18V = 2.27 V V E = V B - 0.7V = 2.27V 0.7V = 1.57 V I E = V E / R E = 1.57V / 470 = 3.34 ma I C I E 3.34 ma V C = V CC I C R C = (3.34mA)(2.7k ) = 8.98V V CE = V C V E = 8.98V 1.57V = 7.41 V

28 Practice Exercise: Find V B, V E, I E, V CE for the circuit if V CC was set for 12V. Solution: Use formulas and substitute values V B = ( R 2 / (R 1 +R 2 ) ) V CC = ( 3.9k / (27k + 3.9k ) ) 12V = 1.51 V 12V V E = V B - 0.7V = 1.51V 0.7V = 0.81 V I E = V E / R E = 0.81V / 470 = 1.72 ma I C I E 1.72 ma V C = V CC I C R C = (1.72mA)(2.7k ) = 4.64V V CE = V C V E = 4.64V 0.81V = 3.84V

29 Voltage-Divider Bias A voltage-divider bias for pnp transistors. Review Example 3-6 to find parameters of pnp transistors. (uses same formulas and procedure as what we just covered)

30 Emitter bias is a very stable form of bias that uses both positive and negative power supplies and a single bias resistor that puts the base voltage near ground potential. (Used in most integrated circuit amplifiers) The emitter current is computed by applying Ohm s law to the emitter resistor and assuming that I C I E. V C = V CC I C R C Emitter Bias

31 EXAMPLE 3-7: Find V E, I E, I C, V CE for the emitter bias circuit shown. Solution: Start with the approximation V E -1V, that implies R E 11V. I E = V RE / R E = 11V / 15k = 0.73 ma I C I E 0.73 ma V C = V CC I C R C = 12V (0.73 ma)(6.8k) = 7.0 V V CE = V C V E = 7.0V (-1V) = 8.0 V

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