assess the biasing requirements for transistor amplifiers

1 INTODUTION In this lesson we examine the properties of the bipolar junction transistor (JT) amd its typical practical characteristics. We then go on to devise circuits in which we can take best advantage of its properties. YOU AIMS At the end of this lesson, you should be able to: use transistor data assess the biasing requirements for transistor amplifiers design biasing circuits. STUDY ADIE This topic assumes a knowledge of the basic operation of a bipolar transistor. It is important to get clear in your mind the direction of current flows in both an npn and a pnp transistor, and the way in which the basic circuit works. See the module Digital and Analogue Devices and ircuits for such basic knowledge.

2 TJ HAATEISTIS The basic characteristic of a bipolar transistor is that a small base-emitter current controls a much larger collector-emitter current. FIGUE 1 shows the typical input and output characteristics of a transistor. I I E E I (ase current) I (ollector current) I 5 I4 Increasing I I 3 I 2 I 1 olts E 0 0.5 0.7 E Linear reakdown E ase-emitter voltage E ollector-emitter voltage Saturation FIG. 1 There are several important points to note. 1. The base-emitter voltage is essentially that of a forward-biased diode. For mid-range values of base current, the voltage level is typically about 0.7. For E less than about 0.5, the collector current is effectively zero, i.e. cut off.

3 2. The collector current characteristic shows three regions of operation. (a) At low E (below about 0.25 ), the collector current depends strongly on E and the current gain β I falls sharply. This is I the fully ON state of the transistor, sometimes called the saturation condition. (b) For E greater than about 0.25, the collector current is essentially constant for a given I. This is the normal linear operating region, where β is large and fairly constant, and does not depend on E. However, β does depend on base current to some extent. (c) As E is increased beyond a certain value, current begins to grow quickly leading to excessive dissipation and probable destruction of the transistor. 3. Transistors which are nominally the same type can have a wide range of β values, typically 50 to 300. Therefore, the published collector current characteristics for a particular type of transistor are only an indication of the typical shape. 4. The reverse breakdown voltage of the base-emitter junction is low (for a planar transistor) compared with an ordinary diode (typically 6 ). EMINDE: I E I + I and β I I

4 ATINGS The ratings for a transistor are more complex than those for a diode because there are 2 junctions. The most important are: E E I M P TM T JM ollector-emitter voltage with base open circuit everse emitter-base voltage Peak collector current Maximum power dissipation Maximum junction temperature Exceeding any of these ratings can seriously damage the health... of your transistor. IDENTIFIATION Transistors are manufactured in a wide range of packages, as can be seen by a cursory inspection of easily available catalogues. For example, metal can TO 18 types are common, as are plastic encapsulated TO 92 types. The most important thing when building circuits is to connect the transistor correctly. Many transistors have met an unfortunate destiny at the hands of students. When you try to identify the collector, base and emitter leads, do not make assumptions. Examine the outline diagram and, most important, determine which side of the transistor is being shown. Usually, it is the lead side.

5 Identification of semiconductors may appear to be a random jumble, but it is not quite so bad as it initially appears. There are two generally recognized systems. (1) JEDE (JOINT ELETON DEIE ENGINEEING OUNIL : USA) Devices have the identification pattern nnnnnn, e.g. 1N4004. The initial number, in general, defines the number of terminals: 1 2 terminal Diodes, Zener diodes 2 3 Transistors 3 4 MOSFETS with substrate connection. The four figure number defines a registered specification, so that many manufacturers can supply nominally the same device (although some parameters may be better than the minima specified). (2) PO ELETON (Europe) Devices have the identification pattern L 1 L 2 N 1 N 2 N 3, e.g. 107 The initial letter, L 1, defines the semiconductor material. A Germanium Silicon Gallium-arsenide D ompound material The second letter, L 2, defines the function. Some examples are listed below.

6 A Signal diode Transistor: low power, low frequency D Transistor: high power, low frequency F Transistor: low power, high frequency L Transistor: high power, high frequency N Photo-coupler S Transistor: Low power, switching U Transistor: High power, switching Y ectifier diode Z egulator (Zener) diode The next three characters are three type numbers or a letter and two numbers. E.g. A ZY88 56 is a silicon Zener diode with a Zener voltage of 5.6. What would a FY50 be? an you tell if it is npn or pnp? It is a silicon low power, high frequency transistor. Unfortunately, the polarity of the transistor is not specified by the type number.

7 D.. OPEATION The basic operation of the transistor is that a small base current causes a large collector-emitter current to flow. Most amplifier circuits will therefore apply an input current change to the base and generate a large collector current change. This collector current change is converted to a voltage by a resistor in the collector circuit. See FIGUE 2. s o I s o 1 2 s I E(SAT) 0 0 I FIG. 2 If I is zero, then the collector current is zero and o s. As I increases, I increases and the voltage o falls. o s I... ( 1) As I is increased further, o eventually falls to the saturation level and no further significant change occurs. There are therefore three operational situations.

8 (a) When I 0, the transistor is in the cut-off state and o s. (b) o > E (SAT) defines the linear amplification range. (c) When o E (SAT), the transistor is fully on. onditions (a) and (c) are used in switching circuits and in binary logic circuits to represent logic 1 and 0. However, in this lesson, we are mainly concerned with linear conditions. In linear circuits, it is important that the output can provide undistorted amplification of the input. In the circuit of FIGUE 2, the output changes from s to E (SAT). Therefore, we should try to bias the circuit so that a zero input signal gives a d.c. level about midway between these limits, i.e.. 2 Then the amplifier has an approximately equal voltage +ve and ve excursion of s about the operating point. 2 iasing the transistor either towards s or E (SAT) may result in distortion of the output. See FIGUE 3. s o 0 I FIG. 3

9 It is therefore necessary to provide a d.c. bias circuit to provide some base current to set the quiescent condition of the output. Several circuits are developed and compared in the next section. IAS IUITS LOAD LINES I Load line: slope 1 Q E s 2 s FIG. 4 Load Line FIGUE 4 shows a set of typical collector circuit curves for a JT. It also 1 shows a load line whose slope is, where is a collector resistance. The intercept with the voltage axis is the supply voltage s, and the intercept with the current axis is s. Given the values of s and, the load line is defined and the operating point Q is defined. It is where o is about 1 /2 s. This tells us the required collector current, I, and the corresponding base current, I.

10 Alternatively, if the operating current I is defined as well as s, a suitable can be determined. We shall now investigate several methods of biasing the transistor. In each case, we shall assume a constant E. ASE ESISTO IASING I s I o E 0 FIG. 5 emember that the current gain β I I Analysing the circuit of FIGUE 5, o s I... ( 2) So, β I o s s ( ) β s E... ( 3a) β s 1 + β E... ( 3b)

11 WOKED EXAMPLE 1 Given that s 10, I 5 ma, E 0.7 and β 100, we must select and to give o 5. Equation (2) can be used to determine. 3 ( ) 5 10 5 10 1kΩ Use of equation (3a) permits a value to be determined for. 5 10 103 100 10 0. 7 ( ) 100 10 3 9. 3 5 186 kω So, select 180 kω as the nearest E12 standard value. For these values, o +4.83, which is close to the design value. Now suppose that β can in fact vary (for the type of transistor used) from 50 to 250. What are the limits of o, calculated from equation (3)?

12 When β 50, o 7.42. This bias condition may be acceptable even though well off-centre. When β 250, o 2.92? E(SAT). This negative answer is not possible. The value calculated implies that the amplifier will be biased in the saturation region, the gain β will have fallen to a low value, and the circuit will be useless as an amplifier! Hence, unless is chosen or adjusted for each transistor, the desired bias point will not be achieved. A further difficulty is that both E and β are temperature dependent. Typically, and Δ E 22. m/ ΔT Δβ + 04. %/ ΔT These effects both act to cause o to fall. If temperature increases, E falls, so I increases and hence I increases; if β increases with temperature, I increases. For a 10 increase in temperature, how much would the bias voltage change in the previous example?

13 If E 0.7 then, for a 10 increase, E is modified as follows. 0. 7 10 2. 2 10 3 0. 678 If β 100 then, for a 10 rise, β is modified as follows. 100 1 + 10 04. 104 100 Substitution of these values of E and b into equation (3b) gives the new bias voltage. 104 o + 1 10 1 104 0. 678 4.61 180 180 So the voltage has fallen a little.

14 EMITTE ESISTO STAILIZATION s 1 o I I E IE 2 E E 0 FIG. 6 A better biasing circuit is shown in FIGUE 6. The following equations define the circuit. I β I... (4) o s s s + I E E E + I + I E E + (β + 1) I... ( 5) E E ( ) I +... ( 6) 1 2

15 We want to eliminate I and from these equations. Substitute I from equation (4) into (5). s o E + E( + ) ( ) β 1 β... ( 7) earrange equation (6). 1 s 1 I + + 1... ( 8) 2 Now substitute I from equation (4) and from equation (7) into equation (8). s 1 ( s o) + + 1 β 1 2 E + E ( β + ) ( ) s o 1... ( 9) β Gather terms in o and simplify. o s β s E + 1 + E β + 1 1 1 1 ( ) + 2 1 2... ( 10) This is rather tiresome and quite complex. If β is very large, equation (10) simplifies by dividing the right hand term above and below by β, assuming β 1 cancels with (β + 1) and neglecting. β

16 o s s E 1+ 1 E 1 + 2 1 2... ( 11) So o is practically independent of β. In practice, we can ensure this by observing the following design rules. 1. hoose emitter voltage E I E E 0.1 s. 2. To give equal positive and negative output changes, choose o to be midway between s and E ; i.e. o 0.55 s. 3. hoose the current which flows through 1 to be 10 I. WOKED EXAMPLE 2 Taking the example of the previous section, in which 10, I 5 ma, 0. 7and β 100, s E choose the resistors for the emitter resistor stabilization circuit of FIGUE 6, using the rules above.

17 1. hoose I 01. 01. 10 1 E E E s Using this value of E and taking I E I, we may determine E. E 1 5 10 3 200Ω Select E 180 Ω as the nearest value in the E12 series. Again taking I E I and using this value of E, the actual value of E may be determined. E 180 5 10 3 0. 9 2. hoose o 0. 55 0. 55 10 5. 5 s Hence, 55. I s 3 ( ) 5. 5 10 5 10 900Ω Select 820 Ω as a suitable approximation from the E12 series.

18 3. Use the values of β and I to determine I. I I β ( 5 10 3) 50 μ A 100 Make the current through 1 10 I 500 μa The current through 1 may now be used to determine its value as follows. ( s ) 500 10 1 + 09. + 07. 16. E E 6 1 10 1. 6 ( ) 500 106 16 8. kω Select 1 18 kω as the nearest value in the E12 series. Actual current through 1 Hence, 10 1. 6 ( ) 3 18 10 467μA current required through 2 467μA I 417μA This current is now used to determine 2.

19 2 10 417 6 16. 10 417 6 384. kω Select 2 3.9 kω as the nearest value in the E12 series. We have chosen the resistors for the circuit. Now we should check the voltage, o, calculated from equation (10). o 10 18 100 820 + 10 0. 7 1 39. 18 103 18 180( 100 1) 1 + 39. ( ) + + 586. As a further check, let us see how close equation (11) is to the exact analysis. o 10 18 820 + 10 0. 7 1 39. 18 180 1 + 3.9 508. So this is correct to within 20%. As a further check, calculate o from equation (10) for β 50 and 250.

20 β 50 o 6.42 β 250 o 5.42 So this design mostly eliminates the effects of large changes in β. FEEDAK IAS s o F I I E 0 FIG. 7 Analysis of FIGUE 7 gives the following equations. ( ) ( + ) I + I β 1 I... o s s ( 12) o E + F I... ( 13) Eliminate I from equations (12) and (13). ( ) β + 1 o s o E ( ) F

21 earrange and gather terms in o. o s ( ) 1+ β + 1 F + ( ) E β + 1 β + 1 + F... ( 14) WOKED EXAMPLE 3 Let us once more choose the same numerical example, in which 10, I 5 ma, 0. 7and β 100. s E We should bias o at +5. Use equation (12) to determine the value of. I s o s o 10 5 + I I 5 10 3 1kΩ This is a standard value in the E12 series. Use this value of in equation (13) to determine F. 5 0. 7 + I 07. + β F F 5 10 100 3

22 Therefore, F 86 kω Select F 82 kω as the nearest value in the E12 series. Now calculate o for β 50, 100 and 250 from equation (14). β o 50 6. 43 100 4. 87 250 2. 99 So this circuit is also effective in stabilizing the bias point.

23 SUMMAY OF IAS ONSIDEATIONS The table of FIGUE 8 summarizes the results of the design process, applied to our worked examples. The design is centred on a β of 100. ircuit ias voltage o β 50 β 100 β 250 ase esistance 7.41 4.83 E(SAT) Emitter esistance 6.37 5.82 5.40 Feedback esistance 6.43 4.87 2.99 FIG. 8 It is clear that both emitter resistance and feedback resistance improve stability. In fact, both are examples of negative feedback. With an emitter resistance, suppose β is increased, then I E increases as does E. Therefore increases, but then I E decreases because more current flows through 2 and less through 1. So the circuit is stabilized. In the case of the feedback resistor, suppose again that β increases. The value of I increases and o falls, so reducing I and hence I. The biasing rules discussed in this lesson apply also to pnp transistors, but all voltages and currents are reversed.

24 SELF-ASSESSMENT QUESTIONS 1. The following table shows the test results for an npn JT. arefully plot the following characteristics on A4 graph paper. (a) I against E for constant I (b) I against E N.. etain these curves, since we will use them in a future lesson! ase current I (μa) ase -emitter oltage () ollector current I (ma) at E () 0.5 1.0 2.0 3.0 4.0 10 0.66 3.0 3.4 4.2 5.0 5.8 50 0.69 12.5 14 15 15.5 16 100 0.72 20 23 26 29 31 200 0.74 27.5 37.5 46 51.5 58 300 0.75 40 48 60 70 80 500 0.76 53 65 82 94 110 Draw a load line for s 5 and a current of 100 ma. Decide on a suitable biased operating point for o for base resistor biasing. Deduce the required base current and collector resistance. hoose a suitable base resistance.

25 2. For a transistor with gain β 125 and E 0.65, design the following bias circuits for a supply voltage of s 15 and I 15 ma at the bias point. (a) ase resistor bias (b) Emitter resistor bias (c) Feedback bias. hoose resistors from the E12 range. In each case, calculate the collector voltage from the chosen resistor values and estimate the error between the values obtained and the target bias.

26 ANSWES TO SELF-ASSESSMENT QUESTIONS 1. The required curves will not be plotted here. However, using these curves and the load line for a E of 2.5, there is a close intercept with the I 200 μa curve. If we take this as our design target, then I 48 ma. For I 200 μa, E 0.74 (from the curves). 5 The inverse of the slope of the load line 01. 50 Ω. hoose the nearest E12 value. 47Ω The base resistor value is calculated as follows. ( s ) ( E 5 074. ) I 200 10 6 21. 3kΩ hoose 22 kω as the nearest E12 standard value. 2. (a) ase esistor ias For s 15, then o 7.5. So, 75. 15 I ( ) 15 15 10 3

27 Hence, 500Ω Select 470 Ω as the nearest value in the E12 series. Now, I I β 15 10 125 3 120μA Therefore, s E 15 0. 65 I 120 10 6 119. 6 kω Select 120 kω as the nearest value in the E12 series. alculate o from equation (3). o s 1 β + β E 15 1 125 470 470 125 0. 65 120 103 + 120 103 797. Error 7.97 7.5 0.47. (b) Emitter esistor Stabilization hoose E 01. 01. 15 15. s

28 So, E I E E I E 15. 100Ω 0. 015 This value, E 100 Ω, is a standard value of E12 series. hoose o 055. 055. 15 825. s Since I o s s o 15 8. 25 I 15 10 3 450Ω Select 470 Ω as the nearest value in the E12 series. Now, I I β 15 10 125 3 120μA Make current through 1 10 I 1.2 ma. Therefore, ( ) + 12. 10 12. 10 s s E E 1 3 3 ( ) 15 1.5 + 065. 12. 10 3 10. 7 kω Select 1 10 kω as the nearest value in the E12 series.

29 Actual current through 1 s 12. 85 10 10 3 1 1. 285 ma So current through 2 1 285 10. 3 I 1 285 10. 3 6 1. 165 ma ( 120 10 ) Hence, 2 3 1. 165 10 215. 1. 165 10 3 184. kω Select 2 1.8 kω as the nearest value in the E12 series. Now check target o from equation (10). o s β s E + 1 + E β + 1 1 1 1 ( ) + 2 1 2 15 o 10 125 470 15 0. 65 1 + 1. 8 10 10 103 100( 125 1) 1 + 18. ( ) + + 819. Error 8.25 8.19 0.06. This is very close!

30 (c) Feedback Stabilization We should bias o at 1 /2 s 7.5. is the same as for base resistor biasing, i.e. 470 Ω. An expression for F can be obtained from equation (13) of the lesson text. I I o E + F E + F β F β ( o E) I Put values into the above expression. F (.. ) 125 75 065 0. 015 57. 1kΩ Select F 56 kω as the nearest value in the E12 series. Now check the target o from equation (14). o s ( ) 1+ β + 1 F + ( ) E β + 1 β + 1 + F o 15 1 + ( 125 + 1) 047. 56 + 0. 65( 125 + 1) 56 125 + 1 + 047. 762. Error 7.5 7.62 0.12. Again, this is a good result!

31 SUMMAY To be able to make the transistor operate as an amplifier, we need to provide base bias. The three ways in which this can be done are: base resistor bias emitter resistor stabilization feedback bias. The first is not very effective because transistor characteristics vary so much, but the other two give reasonable stability against changes in current gain, β. It is important to be able to design bias circuits using the rules of thumb provided.