This quiz is being assigned on Wednesday 4 / 27 / 11. It is due Friday 4 / 29 /

Question

This quiz is being assigned on Wednesday 4 / 27 / 11. It is due Friday 4 / 29 / 11 by can work in groups consisting of no more than 2 members for the quiz. Each member of a group will receive the same grade for this assignment. This quiz will consists of demonstration of your results as well as a written report ( one page maximum ) . The demonstration will be worth 30 points and the report will be worth 70 points Additionally, there will be 15 bonus points given to each member of the group with the best result 10 bonus points given to each member of the group with the second best score, and 5 points for the group member with the third best score. Design and constant a BJT amplifier to have the following specification: V = 12 V, beta = 200, A = - 20, R = 10 k ohm and RL = 1 k ohm ( Assume all capacitors associated with the amplifier are large enough each that they are short circuits for all Ac signals ) Information that may come in hand y: To design a common emitter amplifier, we usually start with the value for the DC voltage Vcc , the current gain beta , voltage gain A the input resistance R and the load resistor R knowing these values, we will use the following steps to find resistor values R , R , R1 and R we will choose RC = RL ( this will maximize the power delivered to the load ) Then, use equation A lambda = Rc||RL / r to find a value for lambda Then find IC = 26mV / r From the value of R use R = to find R If R is not given, then use equation R to find R ( this equation from the bias stability derivation ) . Then use equation V = to find V Finally, R and R are found from R = and R2 =

Explanation / Answer

Follow the given steps to design the circuit Given Vcc = 12V , =100 , Av = -20 , Rin = 10k , RL = 1k Let RC = RL RC = 1k AC load resistance rL = RC || RL = 1k || 1k = 500 Voltage gain Av = - rL /re -20 = - 500 /re re = 25 Emitter current IE = 25mV/re = 25mV / 25 = 1mA Input resistance Rin = RB || re 10k = RB (200)(25) / [ RB + (200)(25) ] 10000RB + 5000000 = 5000RB 5000RB = 5000000 RB = 1k For stiff biasing using voltage divder biasing RB = 0.1RE RE = 1000 / (0.1)(200) RE = 50 From the input section VBB = VBE + IBRB + IERE VBB = VBE + (IE / )RB + IERE VBB = 0.7 + (0.001/200)(1000) + (0.001)(50) VBB = 0.755V Also VBB = VCCR2 / (R1 + R2) RB = R1R2 / (R1 + R2) VBB = 0.7 + (0.001/200)(1000) + (0.001)(50) VBB = 0.755V Also VBB = VCCR2 / (R1 + R2) RB = R1R2 / (R1 + R2) R2 = RB / [ 1 - (VBB/VCC) ] R2 = 1000 / [ 1 - 0.063] R2 = 1.07k R1 = RB VCC / VBB R1 = (1000)(12)/ 0.755 R1 = 16k R1 is the resistance from Vcc to base terminal R2 is the resistance from base to ground terminal

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