hapter 9 ipolar Junction Transistor hapter 9 - JT ipolar Junction Transistor JT haracteristics NPN, PNP JT D iasing ollector haracteristic and Load Line ipolar Junction Transistor (JT) JT is a three-terminal device, which consists a collector (), an emitter () and a base (). There are two types of JTs: NPN type and PNP type. Fig. 1 shows the symbols of both NPN and PNP types JTs. i i N i i P P N N P i NPN i PNP The terminal currents are illustrated in the above figure. It should be noted that i is always greater than i and i. 1. haracteristics of JT JT is a current-controlled device; the values of i and i are determined primarily by i. i = β i, where β is the ratio between the collector and base currents. i = i +i = (1 + β) i. For some cases, β can be written as h F. In practice, β is much greater than 1. So, i i., and are the voltages across collector-to-emitter, base-to-emitter, and collector-to-base, respectively. -1-
hapter 9 - JT ipolar Junction Transistor mitter is heavily doped and collector is relatively lightly doped with the same material. The base is with the smallest size and lightest doped in a JT. 2. Operation of a NPN JT Fig. 1 shows a NPN JT, which is connected to two external dc source and they are and. This circuit configuration is so called common emitter configuration. (1) > (2) is kept y using KL, two equations constant can be obtained I I = + I = + I = 0.7 I Increasing causes decreasing Fig. 1 ommon mitter onfiguration When I increases, I and also increase. So, decreases. onversely, decreasing I causes increasing of. So, if is an ac signal, e.g. sinusoid, will be an inverted sinusoid. Such arrangement is rarely used for actual application. 3. D load line and ollector haracteristic urve D load line a graph that represents all the possible combinations of I and. D load line is constructed from the equation = + I is a fixed value When I = 0, (off)=. When = 0, I (sat)= /. -2-
hapter 9 - JT ipolar Junction Transistor ollector haracteristic urve D load line ollector haracteristic urve illustrates the relationship among I, I and. When I is fixed, a constant amount of I is drawn from the supply source. The characteristic curve shown in the last page illustrates that different values of give different. Fig. 2 ombination of dc load line and collector characteristic curve The graph shown in Fig. 2 is used to determine the operating point (Q point) of a JT. Once is determined, the dc load is kept unchanged. The intersection between the dc load line and the collector characteristic curve is the operating point (Q-point). That means, when the JT is subject to an input current I, -3-
hapter 9 - JT ipolar Junction Transistor respective and I can be obtained at the collector terminal of JT. With the use of this graph, we can determine the desired operating condition for a JT. In practice, Q-point is selected at the mid-point of and I. 4. D iasing The ac operation of an amplifier depends on its initial values of I, I, and. The function of dc biasing is to set the initial value of I, I, and. Two dc biasing methods are introduced: base bias and voltage divider bias. Fig. 3(a) and 3(b) show the circuit configurations of base bias and voltage divider bias, respectively. The primary goal of circuit analysis is to determine the Q-point values of I and for a given I. Fig. 3(a) ase bias Fig. 3(b) oltage divider bias ase bias Fig. 3(a) shows a base-biased JT circuit. After applying KL, two equations can be formulated: = I +, = I + The base current I can be obtained as, I = So, can be obtained as, = I where I I β = β. -4-
hapter 9 - JT ipolar Junction Transistor We can say that the obtained I and equal to I Q and Q. The subscript Q means the Q-point. Appropriate choosing the value of can adjust the location of Q-point at the midpoints of and I. Disadvantage: The value of β is temperature dependent. I could be changed under different operating temperature such that the Q-point could shift along the dc load line. Such problem is called Q-point shift. When an ac signal is injected to the base terminal, the resultant will be either saturated or cutoff. oltage-divider bias oltage-divider bias circuit is by far the most commonly used. Fig. 3(b) shows the circuit configuration. The analysis procedures are slightly different for different ratios between 2 and. ase 1: 0.1 2 β From Fig. 3(b), we can formulate 2 = where is the voltage drop across 2. (1) + 1 2 The voltage of can be found as = = 0.7 (2) Using ohm s law, the emitter current can be obtained as I = (3) As mentioned before, I Q I I, Q can be found as = I ( + ) (4) Q Q It can be shown that the β term is not involved. That means, Q-point shift is no longer presence. -5-
hapter 9 - JT ipolar Junction Transistor ase 2: 0.1 2 β alculate the parallel equivalent resistance from the base of the JT to ground. This resistance is found as eq 2 β = (5) Solve for the base voltage as follows: = 1 eq + eq (6) After solving for, substituting (6) into (2) to continue the sequence of calculations to find the values of I Q and Q. xample 1: onsider the circuit of Fig. 3(a), determine the Q-point values of I and when = 2kΩ, = 360kΩ, = 8 and β = 100. Solution: I can be found as I = (8 0.7) = 360kΩ = 20.28µ A Next, I is found as I = β I = 100 20.28µA = 2.028mA Finally, is found as = I = 8 (2.028mA)(2kΩ) = 3.94 The dc load line can be obtained as = I + 8 = 2000I + When I = 0, (off) = 8. When = 0, I (sat) = 8/2000 = 4mA. I (sat) = 4mA I I Q = 2.028mA I = 20µA Q = 3.94 (off) = 8-6-
hapter 9 - JT ipolar Junction Transistor xample 2: onsider the circuit of Fig. 3(b), determine the Q-point of I Q and Q when 1 = 18kΩ, 2 = 4.7kΩ, = 3kΩ, = 1.1kΩ, = 10, and β = 50. Solution: heck the value 2 and (0.1 β ), So, the base voltage can be found as 0.1 50 1.1kΩ = 5.5kΩ > 4.7kΩ = 10 = 2.07 18kΩ + 4.7kΩ 2 is found as I Q is then found as = = 2.07 0.7 = 1.37 I Q = = 1.37 1.1kΩ = 1.25mA Q = I Q ( + ) = 10 (1.25mA)(4.1kΩ) = 4.87 The dc load line can be obtained as = I ( + ) + 10 = 4100I + When I = 0, (off) = 10. When = 0, I (sat) = 10/4100 = 2.44mA I (sat) =2.44 ma I I Q = 1.25 ma I = 25µA Q = 4.87 (off) = 10 xample 3: onsider the circuit of Fig. 3(b), determine the Q-point of I Q and Q when 1 = 68kΩ, 2 = 10kΩ, = 6.2kΩ, = 1.1kΩ, = 20, and β = 50. -7-
hapter 9 - JT ipolar Junction Transistor Solution: heck the value 2 and (0.1 β ), 0.1 50 1.1kΩ = 5.5kΩ < So, the equivalent resistance eq can be found as = β = (10kΩ) (55kΩ) = 8.46kΩ eq Then, can be found as 2 2 = eq + eq 1 8.46kΩ = 20 68kΩ + 8.46kΩ = 2.21 is found as = = 2.21 0.7 = 1.51 I Q is then found as I Q = = 1.51 1.1kΩ = 1.37mA Q = I Q ( + ) = 20 (1.37mA)(7.3kΩ) = 9.99 The equation of dc load line is When = 0, I (sat) = 2.74mA. When I = 0, (off) = 20. 5. Interfacing = 20 = + I ( + 7300I + ) Apart from the application of amplification, JT can be used as a ON/OFF switch. With the aid of dc load line, two phenomena can be observed: (1) If I = 0, I will be zero and will be equaled to. (2) If I is sufficient large, I will reach to its maximum value and will become zero. That means, if a JT is subject to a square pulse train with sufficient large in magnitude, the measurement of will be an inverted square pulse train. -8-