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1 Lecture 10b EE-215 Electronic Devices and Circuits Asst Prof Muhammad Anis Chaudhary BJT: Device Structure and Physical Operation The pnp Transistor figure shows a pnp transistor biased to operate in the active mode to achieve active mode of operation, EBJ is forward-biased while CBJ is reversebiased Unlike the npn transistor, current in the pnp transistor is mainly conducted by holes, injected from the emitter into the base as a result of the forward-bias voltage v EB These holes injected into the base diffuse through the base region into the depletion region of CBJ and are 1 of 22 10/16/ :55 PM

2 then swept away into the collector region under the influence of the electric field, resulting in A small number of base majority carriers (electrons) are injected into the emitter (forming i B1 ) or recombined with the holes in the base region (forming i B2 ), thus creating the base current ( ) i B = i B1 + i B2 thus the pnp transistor operates in a manner similar to that of the npn transistor = the current-voltage relations for the pnp transistor are identical to that of the npn transistor except that v BE has to be replaced by v EB = for a pnp transistor v / = I S e EB i B V T β = = I S β e v EB / V T α I S α e v EB / V T i E = = i E = + i B where α = β, β = α β+1 1 α Also the large signal operation of a pnp device in active mode can be modeled as 2 of 22 10/16/ :55 PM

3 3 of 22 10/16/ :55 PM

4 The npn Transistor The pnp Transistor The npn Transistor 4 of 22 10/16/ :55 PM

5 Exercise of 22 10/16/ :55 PM

6 Consider the model in Fig. 6.11(a) applied in the case of a pnp transistor whose base is grounded, the emitter is fed by a constant-current source that supplies a 2mA current into the emitter terminal, and the collector is connected to a 10V dc supply. Find the emitter voltage, the base current, and the collector current if for this transistor and. β = 50 I S = A Solution: Exercise 6.10 here β = 50, = 1 A, i E v E =?, i B =?, =? β 50 β+1 51 i E or i E i i C m B β 50 v / I S e EB V T I S 0 14 = 2mA α = = = α = = 2mA = = = = m = mA = = = μA = = mA m = v EB = V T ln = V I S = = 0 = V v E v B v E or v E = V 6 of 22 10/16/ :55 PM

7 Exercise 6.11 For a pnp transistor having and, calculate for = 1.5A. Solution: Exercise 6.11 here = 1 A,, I S 0 11 β = 100 = 1.5A v EB= =? v / ic I S e EB V T = 1.5 = v / e EB V T or e v EB / V T = take natural log on both sides v EB 1.5 = ln( ) V T BJT: Current-Voltage Characteristics I S = A β = 100 v EB 1.5 v EB = V T ln( )= 25m ln( )= V Circuit Symbols and Conventions the circuit symbols for an npn and pnp transistor are indicated in fig 7 of 22 10/16/ :55 PM

8 In both symbols, the emitter is distinguished by an arrowhead. this distinction is important as practical BJT is not a symmetrical device The polarity of the device i.e. npn or pnp is indicated by the direction of the arrowhead on the emitter. This arrowhead points in the direction of the conventional current flow. Recall that in npn BJT, current flows from the collector to emitter (as the electrons are flowing from emitter to the collector) in a pnp BJT, the current flows from emitter to the collector (as the holes flow from emitter to the collector) A summary of the BJT current-voltage relationships in the active mode of operation can be given as 8 of 22 10/16/ :55 PM

9 Note that an npn transistor, whose EBJ is forward-biased, will operate in the active mode as long as the collector voltage doesnot fall below that of the base by more than approximately 0.4V i.e. v CB > 0.4V for a pnp transistor whose EBJ is forward-biased will operate in the active mode as long as the collector voltage is not allowed to rise above that of the base by more than 0.4V i.e. < 0.4V or vbc v CB> 0.4V 9 of 22 10/16/ :55 PM

10 Example 6.2 β = 100 The transistor in the circuit of Fig. 6.14(a) has and exhibits a v BE of 0.7V at = 1mA. Design the circuit so that a current of 2mA flows through the collector and a voltage of +5V appears at the collector. 2.8cm Solution: Example 6.2 here npn, β = 100, v BE1 = 0.7V at 1 = 2mA, v C = 5V R C =?, R E =? v / As = I S e BE V T v / = 1 = I S e BE1 V T, = = 1mA v / 2 I S e BE2 V T dividing v 2 I = S e BE2 / V T = 1 I S e v BE1 / V T e ( v BE2 v BE1 )/ V T 10 of 22 10/16/ :55 PM

11 take natural log on both sides 2 ln =( )/ 1 v BE2 v BE1 V T i v BE2 v BE1 = V T ln C2 1 i v BE2 = v BE1 + V T ln C2 1 i v BE2 = v BE1 + V T ln C2 1 here v BE1 = 0.7V at 1 = 1mA, = = 2mA, v BE2 As the base is grounded or v E β 100 β i i C 2m E α By ohm's law 15 v R C C 2 = =? 2m = v BE = v BE2 = 0.7 +(25m)ln = 0.717V 1m = v BE = = v B v E = 0 v E = 0.717V α = = = = = = 2.02mA = = = = 5kΩ m 10 2m v BE 11 of 22 10/16/ :55 PM

12 also by ohm's law R E v E ( 15) ( 15) i E 2.02m m = = = = kΩ Exercise D6.12 Repeat Example 6.2 for a transistor fabricated in a modern integrated-circuit process. Such a process yields devices that exhibit larger v BE at the same because they have much smaller junction areas. The dc power supplies utilized in modern IC technologies fall in the range of 1V to 3V. Design a circuit similar to that shown in Fig except that now the power supplies are ±1.5V and the BJT has β = 100 and exhibits v BE of 0.8V at = 1mA. Design the circuit so that a current of 2mA flows through the collector and a voltage of +0.5V appears at the collector. 12 of 22 10/16/ :55 PM

13 Solution: Exercise D6.12 here β = 100, v BE1 = 0.8V at = 2mA, = 0.5V By ohm's law As v C R C or v E By ohm's law v C = 1mA = = = 500Ω m i v BE = v BE1 + V T ln C 1 2m = = 0.8 +(25m)ln( )= V v BE v BE = = v B v E = 0 v E = v E = = V R E = ( 1.5) v E i E 1m 13 of 22 10/16/ :55 PM

14 As i E α = = 2m β 100 α and α = = = β m 2m i E α v ( 1.5) R E ( 1.5) E i E 2.02m = = = = 2.02mA m = = = = = Ω Exercise 6.13 In the circuit shown in Fig. E6.13, the voltage at the emitter was measured and found to be. If, find,,, and. 0.7V β = 50 I E I B I C V C 14 of 22 10/16/ :55 PM

15 Solution: Exercise 6.13 here v E = 0.7V, β = 50 I E =?, I B =?, I C =?, and V C =?. As v E = 0.7V and v B = 0V = v BE = v B v E = 0 ( 0.7)= 0.7V by ohm's law as i E i B = β i E α v E ( 10) 10k k = = = 0.93mA β 50 i E where α = = = β+1 51 i E = = α = α = m = mA m 50 = = = μA 15 of 22 10/16/ :55 PM

16 thus i E = 0.93mA, = mA, i B by ohm's law v C or v C = μA 10 =(5k) =(5k) m 10 v C = = v C = 5.441V Exercise 6.14 In the circuit shown in Fig. E6.14, measurement indicates V B to be +1.0V and V E to be +1.7V. What are α and β for this transistor? What voltage V C do you expect at the collector? 16 of 22 10/16/ :55 PM

17 Solution: Exercise 6.14 here v B = 1V, v E = 1.7V α =?, β =?, =? by ohm's law, v i B 1 B 100k 100k 10 v and i E E 5k As as as i E v C = = = 10μA k = = = 1.66mA i E = + i B = = i E i B = 1.66m 10μ = 1.65mA i i B = C β = β = i B = = β = = = 165 α 1.65m i B 10μ = α = = = m i E 1.66m 17 of 22 10/16/ :55 PM

18 by ohm's law v C ( 10)= 5k( )= 8.25V v C + 10 = 8.25 = = 1.75V or v C BJT: Current-Voltage Characteristics Graphical Representation of Transistor Characteristics for an npn transistor, v / = I S e BE this relation can be represented graphically as I also i B = S and β e v BE / V T I i E = S α e v BE / V T V T 18 of 22 10/16/ :55 PM

19 = i B v BE and i E v BE characteristics are also exponential but with different scale currents I S I for i S and for β B α i E As the constant of the exponential characteristic is = 1 1 V T 25m the curve rises very sharply (you can verify this in matlab/octave) for v BE smaller than about 0.5V (the cut-in voltage of EBJ), the current is negligibly small Note that for most of the normal range, v BE lies in the range of 0.6V to 0.8V In performing rapid first order dc calculations, = = = 40 we can utilize the constant-voltage drop model by assuming v BE 0.7V v / = I S e EB V T = v EB vbe v EB for a pnp transistor, characteristics will be identical to that of the npn transistor with replaced with 19 of 22 10/16/ :55 PM

20 Note that for a BJT, the voltage across the emitter-base junction decreases by about 2mV for each rise of 1 o C in temperature, provided the junction is operating at a constant current Exercise 6.15 Consider a pnp transistor with v EB = 0.7V at i E = 1mA. Let the base be grounded, the emitter be fed by a 2mA constant-current source, and the collector be connected to a 5V supply through a 1kΩ resistance. If the temperature increases by 30 o C, find the changes in emitter and collector voltages. Neglect the effect of I CBO Solution: Exercise 6.15 here pnp, v EB1 = 0.7V at i E1 = 1mA change in v E and v C are? if BO = v C is neglected will not change with temperature = change in v C with temperature =0 as v EB decreases by 2mV for every 1 o C rise in temperature = v EB decreases by 4mV for 2 o C rise in temperature = v EB decreases by 6mV for 3 o C rise in temperature 20 of 22 10/16/ :55 PM

21 = v EB decreases by 60mV for 30 o C rise in temperature decreases by 60mV for 0 o rise in temperature v EB 3 C as v B = 0 = v EB = v E v B = v E if v EB decreases by 60mV = v E will also decrease by 60mV exact value of v EB can be determined as follows, As I i E = S α e v EB / V T I = i E1 = S α e / v EB1 V T, = I i S E2 α e v EB2 / dividing i v E2 e = EB2 / V T = i E1 e v EB1 / V T e ( v EB2 v EB1 )/ V T for v EB1 = 0.7V, i E1 = 1mA, i E2 = i E = 2mA, v EB = v EB2 =? i E2 2m = = = i E1 e ( v EB v EB1 )/ V T e ( v EB 0.7)/25m 1m e ( v EB 0.7)/25m = 2 ( v EB 0.7)/25m = ln 2 V T 21 of 22 10/16/ :55 PM

22 v EB = 0.7 +(25m)ln 2 = V v EB = V when temperature increases by 30 o C = v EB decreases by 60mV i.e. v EB, (at T+30 o C) = m = V v EB (at T+30 o C) v EB = = = 60mV 22 of 22 10/16/ :55 PM

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