For a transistor, the current amplification factor is 0.8. The transistor is connected in common emitter configuration. The change in the collector current when the base current changes by 6 mA is (A) 6 mA (B) 4.8 mA (C) 24 mA (D) 8 mA
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- Re 14.4 V 11.4 V Rg 8 kohm 12 kohm 12 V 20 V le(mA) A !g-1.425 mA Vee(Volt) Answer the questions given below, since they are taken from the circuit next to the output chart given above a) What common transistor configuration is the given circuit diagram? b) what type is the given transistor? NPN or PNP? SpecifyAn apa transistor is accidentally connected with col-lector and emitter leads interchanged. The resulting currents in the normal emitter and base leads are 0.5 mA and I mA, respectively. What are the values of as and fa?In a junction diode, a) The charge carriers do not flow by diffusion from the p-type side to the n-type side and vice versa. b) The depletion capacitance decreases with increase in the reverse bias. c) The depletion capacitance increases with increase in the reverse bias. d) The diffusion capacitance increases with increase in the forward bias.
- In the transistor circuit on the sideMaximum values are given below. Normalmaximum VCC voltage of the BJT under conditionscalculate. Pd max = 07/100 W Vce max = 20V Ic max = 100mA Bdc = 150A Pentium IV microchip has dimensions of 217 mm x 217 mm. Suppose each transistor on the chip is made up of a 10 x10 square of atoms (that is 100 atoms per transistor). How many transistors could fit on the Pentium IV chip if the spacing between the silicon atoms is 0.54 nm?4 Q: A- For the transistor circuit shown in figure bellow 20 v Determine: 1- Ig , Ic , le and Vce · 2- Sketch the load line (Q-point) of the transistor. 470 2 270 kQ B=125 تمت الاجابة وسارفع الملف لاحقا
- 1) A Si p-n-p transistor has the following properties at room temperature: Tn = Tp 0.1 us NE 1019 сті Emitter concentration — 10 ст2/s -3 Dn = Dp NB 3D 1016 ст Base concentration Nc 1019 ст -3 = Collector concentration WE 3 µm Emitter width W 1.5 um Metallurgical base width, i.e. the distance between base-emitter junction and base-collector junction A = 10-5 cm² = Cross-sectional area If VCB = 0 V and VEB = 0.6 V, calculate the following: ЕВ a) Neutral base width (WB) b) Base transport factor c) Emitter injection efficiency d) a, ß and y. e) Ic, Ig and Ig.An abrupt uniformly doped silicon pn junction is reversed biased by Vg= 20 V. If Na(in n-side)=10" cm, N,(in p-side)=10" cm then the junction capacitance is 20 pF. The junction capacitance if Na(in n-side) increased to 3x10" cm' is equal to ....pF. a) 9 b)21 c)35 d) 52 e) 87When the current gain of a transistor is 200 and the base current is 50 μA, it leads to a collector current of: Ο ) 4. 0 mA 250 μ.Α Ο 4.0 μ.Α Ο 10 mA
- The transfer ratio B of a transistor is 50. The input resistance of the transistor when used in common – emitter configuration is IkQ. The peak value of the collector AC current for an AC input voltage of 0.01 V peak is (A) 100 µA (B) 250 µA (C) 500 µA (D) 800 µAIn Fig. 14, the transistor Q₁ has ß = 50, and VBE,on = 0.7 V. 5V 5V 1200 ic 2.2ΚΩ www iB in Fig. 14. NPN transistor circuit no. 2. NPN State Conditions Cutoff VBE 0, BiB > ic>0 Forward Active iB > 0, VCE > 0.2 V PNP State Conditions Cutoff VBE> -0.5 V, VBC > -0.5 V Saturation İB > 0, BiB > ic>0 Forward Active iB > 0, VCE < -0.2 V 37. Using simple model transistor analysis, determine the value of ic in mA, to 1 decimal place. (Hint: guess the state of the transistor, calculate the relevant parameters, check calculated parameters against the conditions for the guessed state to be valid.) ic = mA. Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If for an n-type semiconductor, the density of electrons is 10¹⁹ m-³ and their mobility is 1.6 m² (V-s), then the resistivity of the semiconductor 2 semiconductor (since, it is an n-type contribution of holes is ignored) is close to