B= 0 VCC=10V The emitter current of Q1 is: Rc 315 A) 365 A B) 965...
2. In the following current mirror circuit, Vcc -10V, and the three transistors, Q1, Q2, Q3, have the same saturation current (i.e.,IssIs), and with V for active mode is 0.7V. Then, the three beta values are given by: β91-100, ßQ2-50, and ßQ3-200. The thermal voltage is ντ-25mV. Assuming that you need an output current of i1mA: ref db (a) Find the collector, base, and emitter current for all three Q1 O2 transistors when ia 1mA. (b) Find the refern ie....
Problem 2. T Vcc=10v 4R3 Rc=5k12 Ko Yout C=0 Q(8=100, VBE=0.7, V=25mV, VA ) IRB IC=0 RB Fig. 3 In the common emitter amplifier shown in Fig. 3. a) Neglecting the base current (18 = 0), determine Rei and Rez to set Ice = 1 mA and the small-signal voltage gain |Apl= |Vout/Vin) = 20. [45 pt.) b) Base on the Ico given and ß = 100, choose Rp so that lg is 5% of IRB (15 pt.)
FIND THE VALUES OF Rb1, Rb2, Re,Rc, rin , rout, overall gain and open circuit gain First, design a common emitter BJT amplifier Second, analyze the amplifier.( Avo, Gv, Rin, Rout) Third, compare your calculation with Multisim. Report must include comparison between your calculation & simulation results overall voltage gain, open circuit voltage gain, input resistance, and output resistance. This design project is not group work, must be done individually. Type your report. Design a discrete common emitter BJT amplifier.(Determine...
Problem 5 Given the following circuit, assume the following parameters VBB IV, RB 220 k, RC = 2 k, VCC= 10 V, VBE(on)-0.7 V, and B 200 a) Calculate the base, collector, and emitter currents and the C-E voltage, also, calculate the transistor power dissipation b) What transistor configuration does the circuit resemble? Vcc=10V RC=2k Rg=220 kQ VCE VBB= 1V o + VRE IB
1) Calculate the value of the base current IB. (in μA) 2) Calculate the value of the base collector current IC. (In mA) 3) Calculate the value of the collector-emitter voltage VCE. (In V) Required information In the circuit below: RB = 820 kN, Vcc = 12 V, RC = 3 kN2, and Bdc = 100. Note: The transistor is silicon. Vcc w Rc Re Bdc
could you please answer these 8 questions ? thnx VCC 15V Ic 1kQ RC R1 33KQ VC C B 2N 5210 VB VBE VE R2 4700 RE ·10kΩ Refer to the data table for each of the following symptoms, choose one of the following faults as the most probable cause and write it in the space provided. POSSIBLE FAULTS R1 open Rc open RE decreased RE open R1 decreased R2 decreased R2 open Rc decreased Trouble in dc supply Transistor...
The 1 mA. V, ls -VE -15 15 V, in the following differential amplifier circuit, Vcc parameters are given as β, 100, VBE# 0.7 V, pr-25 mV, K.-100 V. transistor Rc-10 kΩ For: RE-150 Ω Rc Rc REE-200 kΩ a) What is the input differential resistance, Rid b) What is the overall voltage gain vV? You c) What is input common mode resistance, d) What is the worst case common mode gain that appear across the two input terminals? (4...
Consider the BJT common-emitter amplifier in Figure 1. Assume that the 2N3904G transistor has the following parameters: β-206, VBE-0.TV and the Early voltage VAT 1000V. vCC RB1a I multiple resistors RC want n Vload 22HF Rload 01 2N3904G V1 6302 4.7HF RE2CE 0.01Vpk 1kHz maliple esistons lue you available in the ki Figure 1 BJT CE amplifier 0.5 V and VC-3 V (a) Design the DC biasing circuit so that lc-2 mA, VCE = 2.5 V, VE Consider the BJT...
EXERCISES 8.12 For the circuit in Fig. 8. 19, let 1-1 m1A, Vcc-15 VR-|0 kQ, with α 1, and let the input voltages be: t'ai = 5 + 0.005 sin 2π × 1000t. volts, and = 5-0.005 sin 2π × 10001, volts. (a) If the BJTs are specified to have gr of 0.7 V at a collector current of 1 mA. find the voltage at the emitters. (b) Find g, for each of the two transistors. (c) Find ic for...
Use Exact Analysis Use 4 decimal places, no commas. a. VCC = 15V b. R1 = 55 kΩ c. R2 = 15 kΩ d. RC = 2.5 kΩ e. RE = 1.5 kΩ f. RS = 15 Ω g. RL = 500 Ω h. βDC = 50 i. βac = 100 Single-stage, voltage-divider biased, common-emitter with **C1 and C2 are coupling capacitors. C3 is bypass capacitor Given: A. VB = [a] V ; B. VE = [b] V ; C....