BJT Amplifiers Problems & Solutions [PDF]

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Electronics-Problems



Second Semester 2018-2019



Problem-1 Determine IC(sat) for the transistor in Figure below. What is the value of IB necessary to produce saturation? What minimum value of VIN is necessary for saturation? Assume VCE(sat) = 0 V. Solution:



𝐼𝐶(𝑠𝑎𝑡) =



𝑉𝐶𝐶 5𝑉 = = 0.5 𝑚𝐴 𝑅𝐶 10 𝑘Ω



𝐼𝐶(𝑠𝑎𝑡) 0.5 𝑚𝐴 = = 3.33 𝜇𝐴 𝛽𝐷𝐶 150



𝐼𝐵(𝑚𝑖𝑛) =



𝐼𝐵(𝑚𝑖𝑛) =



𝑉𝑖𝑛(𝑚𝑖𝑛) − 𝑉𝐵𝐸 𝑅𝐵



𝑅𝐵 𝐼𝐵(𝑚𝑖𝑛) = 𝑉𝑖𝑛(𝑚𝑖𝑛) − 0.7𝑉 𝑉𝑖𝑛(𝑚𝑖𝑛) = 𝑅𝐵 𝐼𝐵(𝑚𝑖𝑛) + 0.7𝑉 = (1 𝑀Ω) (3.33𝜇𝐴) + 0.7𝑉 = 4.03 𝑉 Problem-2 The transistor in Figure below has a βDC of 50. -



Determine the value of RB required to ensure saturation when VIN is 5 V.



-



What must VIN be to cut off the transistor? Assume VCE(sat) = 0 V



Solution: 𝐼𝐶(𝑠𝑎𝑡) =



𝑉𝐶𝐶



𝐼𝐵(𝑚𝑖𝑛) = 𝑅𝐵(𝑚𝑖𝑛) =



𝑅𝐶



=



𝐼𝐶(𝑠𝑎𝑡) 𝛽𝐷𝐶



15 𝑉 1.2 𝑘Ω



=



= 12.5 𝑚𝐴



12.5 𝑚𝐴 50



𝑉𝑖𝑛(𝑚𝑖𝑛) − 𝑉𝐵𝐸 𝐼𝐵



=



= 250 𝜇𝐴



5𝑉− 0.7 𝑉 250 𝜇𝐴



= 17.2 𝑘Ω



𝑉𝐼𝑁(𝑐𝑢𝑡𝑜𝑓𝑓) = 0 𝑉



1



Problem-3 The transistor in Figure below has the following maximum ratings: PD(max) = 800 mW, VCE(max) =15 V, and IC(max) =100 mA. Determine the maximum value to which VCC can be adjusted without exceeding a rating. Which rating would be exceeded first? Solution: First, find IB so that you can determine IC 𝐼𝐵 =



𝑉𝐵𝐵 − 𝑉𝐵𝐸 𝑅𝐵



=



5𝑉−0.7𝑉 22𝑘Ω



= 195 𝜇𝐴



𝐼𝐶 = 𝛽𝐷𝐶 𝐼𝐵 = (100) (195𝜇𝐴) = 19.5 𝑚𝐴 IC is much less than IC(max) and ideally will not change with VCC. It is determined only by IB and βDC. The voltage drop across RC is VRC = IC RC = (19.5 mA) (1.0 kΩ) = 19.5 V Now you can determine the value of VCC when VCE = VCE(max) = 15 V VRC = VCC - VCE So,



𝑽𝑪𝑪(𝒎𝒂𝒙) = 𝑽𝑪𝑬(𝒎𝒂𝒙) + 𝑽𝑹𝑪 = (𝟏𝟓𝑽) + (𝟏𝟗. 𝟓𝑽) = 𝟑𝟒. 𝟓 𝑽 VCC can be increased to 34.5 V, under the existing conditions, before VCE(max) is exceeded. However, at this point it is not known whether or not PD(max) has been exceeded.



𝑃𝐷 = 𝑉𝐶𝐸(𝑚𝑎𝑥) 𝐼𝐶 = (15𝑉) (19.5𝑚𝐴) = 293 𝑚𝑊 Since PD(max) is 800 mW, it is not exceeded when VCC = 34.5 V. So, VCE(max) =15 V is the limiting rating in this case. If the base current is removed causing the transistor to turn off, VCE(max)will be exceeded first because the entire supply voltage, VCC, will be dropped across the transistor.



2



Problem-4 a) For the transistor circuit in Figure below, what is VCE when VIN = 0 V? (b) What minimum value of IB is required to saturate this transistor if βDC is 200? Neglect VCE(sat). (c) Calculate the maximum value of RB when VIN = 5 V.



Solution: When vin = 0 V, the transistor is in cutoff (acts like an open switch) and



VCE = VCC = 10 V



Since VCE(sat) is neglected (assumed to be 0V), 𝑰𝑪(𝒔𝒂𝒕) = 𝑰𝑩(𝒎𝒊𝒏) =



𝑽𝑪𝑪 𝟏𝟎 𝑽 = = 𝟏𝟎 𝒎𝑨 𝑹𝑪 𝟏 𝒌𝛀



𝑰𝑪(𝒔𝒂𝒕) 𝟏𝟎 𝒎𝑨 = = 𝟓𝟎 𝝁𝑨 𝜷𝑫𝑪 𝟐𝟎𝟎



This is the value of IB necessary to drive the transistor to the point of saturation. Any further increase in IB will ensure the transistor remains in saturation but there cannot be any further increase in IC. When the transistor is on, VBE ≈ 0.7V. the voltage across RB is



𝑉𝑅𝐵 = 𝑉𝑖𝑛 − 𝑉𝐵𝐸 ≅ 5𝑉 − 0.7𝑉 = 4.3𝑉 Calculate the maximum value of RB needed to allow a minimum IB of 50 µA using ohm`s law as follows: 𝑹𝑩(𝒎𝒂𝒙) =



𝑽 𝑹𝑩 𝑰𝑩(𝒎𝒊𝒏)



=



3



𝟒. 𝟑𝑽 = 𝟖𝟔 𝒌𝛀 𝟓𝟎𝝁𝑨



Problem-5 The LED in Figure below requires 30 mA to emit a sufficient level of light. Therefore, the collector current should be approximately 30 mA. For the following circuit values, determine the amplitude of the square wave input voltage necessary to make sure that the transistor saturates. Use double the minimum value of base current as a safety margin to ensure saturation. VCC = 9 V, VCE(sat) = 0.3 V, RC = 220 Ω, RB = 3.3 k Ω, βDC =50, and VLED =1.6 V.



Solution: 𝑉𝐶𝐶 − 𝑉𝐿𝐸𝐷 − 𝑉𝐶𝐸 (𝑠𝑎𝑡)



𝐼𝐶(𝑠𝑎𝑡) =



𝑅𝐶



𝐼𝐵(𝑚𝑖𝑛) =



𝐼𝐶(𝑠𝑎𝑡) 𝛽𝐷𝐶



=



32.3 𝑚𝐴 50



=



9 𝑉−1.6 𝑉−0.3 𝑉 220 𝑘Ω



= 32.3 𝑚𝐴



= 646 𝜇𝐴



To ensure saturation, use twice the value of IB(min), which is 1.29 mA. Use ohm's law to solve for Vin. 𝐼𝐵 =



𝑉𝑅𝐵 𝑅𝐵



=



𝑉𝑖𝑛 − 𝑉𝐵𝐸 𝑅𝐵



=



𝑉𝑖𝑛 −0.7 𝑉 3.3 𝑘Ω



2𝐼𝐵(𝑚𝑖𝑛) 𝑅𝐵 = 𝑉𝑖𝑛 − 0.7𝑉 = (1.29 𝑚𝐴)(3.3𝑘Ω) 𝑉𝑖𝑛 = (1.29 𝑚𝐴)(3.3𝑘Ω) + 0.7𝑉 = 4.96 𝑉



4



Propblem-6 I.



Determine the following dc values for the amplifier in Figure below. (a) IE (b) VE (c) VB (d) IC (e) VC (f) VCE



II.



Determine the following ac values for the amplifier in Figure below. (a) Rin(base) (b) Rin (c) Av (d) Ai (e) Ap



III.



Assume that 600 Ω, 12μV rms a voltage source is driving the amplifier in Figure below. Determine the overall voltage gain by taking into account the attenuation in the base circuit. What is the phase relationship of the collector signal voltage to the base signal voltage?



Solution: I 𝜷𝑹𝑬 ≫ 𝟏𝟎 𝑹𝟐 𝑉𝑡ℎ = ( 𝑅



𝑅2



1 + 𝑅2



???



) ∗ 𝑉𝑐𝑐 = 3.66 𝑉



𝑅 ∗𝑅



𝑅𝑡ℎ = ( 1 2 ) = 9.56 𝑘Ω 𝑅 +𝑅 1



2



|𝑉𝑡ℎ + 𝑉𝑅𝑡ℎ + 𝑉𝐵𝐸 + 𝑉𝑅𝐸 | = 0 |𝑉𝑡ℎ + 𝐼𝐵 𝑅𝑡ℎ + 𝑉𝐵𝐸 + (𝛽 + 1) 𝐼𝐵 𝑅𝐸 | = 0 𝐼𝐵 =



𝑉𝑡ℎ − 𝑉𝐵𝐸 𝑅𝑡ℎ + (𝛽+1)𝑅𝐸



= 34.6 𝜇𝐴



𝐼𝐸 = (𝛽 + 1) 𝐼𝐵 = 2.63 𝑚𝐴 𝑉𝐸 = 𝑉𝑅𝐸 = 𝐼𝐸 ∗ 𝑅𝐸 = 2.6 𝑉 𝑉𝐵 = 𝑉𝐸 + 𝑉𝐵𝐸 = 3.3 𝑉 𝐼𝐶 = 𝛽 ∗ 𝐼𝐵 = 2.59 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 − 𝐼𝐸 𝑅𝐸 = 6.8 𝑉 𝑉𝐶 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 = 9.45 𝑉



5



Note: 𝑅𝑖𝑛(𝑏𝑎𝑠𝑒) =



𝑉𝑖𝑛 𝐼𝑖𝑛



=



𝑉𝑏 𝐼𝑏



𝑎𝑛𝑑 𝑠𝑖𝑛𝑐𝑒 𝐼𝑒 ≅ 𝐼𝑐 ≫≫≫ 𝑉𝑏 = 𝐼𝑒 𝑟𝑒, 𝐼 𝐼𝑏 ≅ 𝛽 𝑒 𝑎𝑐



Substituting for Vb and Ib, 𝑅𝑖𝑛(𝑏𝑎𝑠𝑒) =



𝑉𝑏 𝐼𝑏



=



𝐼𝑒 𝑟,𝑒



𝐼𝑒 /𝛽𝑎𝑐



Cancelling Ie, 𝑅𝑖𝑛(𝑏𝑎𝑠𝑒) = 𝛽𝑎𝑐 𝑟𝑒,



II 𝑅𝑖𝑛(𝑏𝑎𝑠𝑒) = 𝛽𝑎𝑐 𝑟𝑒, = 70 [



25𝑚𝑉 𝐼𝐸



] = 665Ω



𝑅𝑖𝑛(𝑡𝑜𝑡𝑎𝑙) = 𝑅1 ∥ 𝑅2 ∥ 𝑅𝑖𝑛 (𝑏𝑎𝑠𝑒) = 622 Ω 𝐴𝑣 = 𝐴𝑖 =



𝑅𝐶 ∥ 𝑅𝐿 𝑟𝑒, 𝐼𝑐 𝐼𝑏



= 261



Note: 𝐴𝑣 =



= 𝛽𝑎𝑐 = 70



𝑉𝑐 𝑉𝑏



𝑉𝑐 = 𝛼𝑎𝑐 𝐼𝑒 𝑅𝐶 ≅ 𝐼𝑒 𝑅𝐶 𝑉𝑏 = 𝐼𝑒 𝑟𝑒, 𝐼 𝑅 𝐴𝑣 = 𝐼𝑒 𝑟 ,𝐶 ≫≫≫ 𝐴𝑣 =



𝐴𝑝 = 𝐴𝑣 ∗ 𝐴𝑖 = 18281



𝑒 𝑒



𝑅𝑐 = 𝑅𝐶 ∥ 𝑅𝐿



III 𝑅𝑖𝑛(𝑡𝑜𝑡𝑎𝑙)



𝑉𝑏 = ( 𝑅 +𝑅 𝑠



𝑖𝑛(𝑡𝑜𝑡𝑎𝑙)



622 Ω



) 𝑉𝑖𝑛 = (600 Ω+ 622 Ω) 12 𝜇𝑉 = 6.1 𝜇𝑉



Attenuation of the input network is 𝑅𝑖𝑛(𝑡𝑜𝑡𝑎𝑙)



(𝑅 + 𝑅 𝑠



𝑖𝑛(𝑡𝑜𝑡𝑎𝑙)



) = 0.51



𝐴,𝑣 = 0.51 ∗ 261 = 133.6 Vb / Vs



6



𝑅𝐶 𝑟𝑒,



𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶 (𝑅𝐶 + 𝑅𝐸 ) Decrease



Increase



Constant



𝜃 = 180𝑜



7



Propblem-7 Determine the exact voltage gain for the unloaded emitter-follower in Figure below.



Solution: 𝑉𝐵 = ( 𝑅



𝑅2



4.7 𝑘Ω



1 + 𝑅2



𝐼𝐸 = 𝑟𝑒, ≅ 𝐴𝑣 =



𝑉𝐵 − 𝑉𝐵𝐸 𝑅𝐸 25𝑚𝑉 1.06 𝑚𝐴 𝑅𝐸 𝑅𝐸 + 𝑟𝑒,



) 𝑉𝐶𝐶 = (14.7 𝑘Ω) 5.5 𝑉 = 1.76𝑉 =



1.76𝑉−0.7𝑉 1.0 𝑘Ω



= 1.06 𝑚𝐴



= 23.6 Ω =



1.0 𝑘Ω 1.0 𝑘Ω+23.6 Ω



= 0.977



Note: 𝐴𝑣 =



𝑉𝑜𝑢𝑡 𝑉𝑖𝑛



𝑉𝑜𝑢𝑡 = 𝐼𝑒 𝑅𝑒 𝑉𝑖𝑛 = 𝐼𝑒 (𝑟𝑒, + 𝑅𝑒 ) 𝐼 𝑅 𝐴𝑣 = 𝐼 (𝑟𝑒, +𝑒𝑅 ) ≫≫≫ 𝐴𝑣 = 𝑒



𝑒



𝑒



𝑅𝑒 = 𝑅𝐸 ∥ 𝑅𝐿 𝑅𝑒 ≫ 𝑟𝑒, 𝐴𝑣 ≅ 1



8



𝑅𝑒 , 𝑟𝑒 + 𝑅𝑒



Propblem-8 Find the overall current gain Ai in below.



Solution: 𝑅𝑖𝑛(𝑏𝑎𝑠𝑒) = 𝛽𝑎𝑐1 𝛽𝑎𝑐2 𝑅𝐸 = (150)(100)(1.5 𝑘Ω) = 22.5 𝑀Ω



R in = R1 ∥ R 2 ∥ R in(base) = 33kΩ ∥ 22kΩ ∥ 22.5 MΩ = 13.2 kΩ Note:



𝐼𝑖𝑛 =



𝑉𝑖𝑛 𝑅𝑖𝑛



=



𝐼𝑖𝑛(𝑏𝑎𝑠𝑒1) =



1𝑉 13.2 𝑘Ω



= 75.8 𝜇𝐴



𝑉𝑖𝑛 𝑅𝑖𝑛(𝑏𝑎𝑠𝑒1)



=



1𝑉 22.5 𝑀Ω



= 44.4 𝑛𝐴



𝐼𝑒 ≅ 𝛽𝑎𝑐1 𝛽𝑎𝑐2 𝐼𝑖𝑛(𝑏𝑎𝑠𝑒1) = (150)(100)(44.4𝑛𝐴) = 667𝜇𝐴 𝐴,𝑖 =



𝐼𝑒 𝐼𝑖𝑛



=



667 𝜇𝐴 75.8 𝜇𝐴



= 8.8



9



Propblem-9 Find power gain Ap for the unloaded amplifier in Figure below



The same circuit



Solution: 𝑉𝐸 = ( 𝑅



𝑅2



1 + 𝑅2



𝐼𝐸 =



6.8 𝑉 620Ω



10𝑘Ω



) 𝑉𝐶𝐶 − 𝑉𝐵𝐸 = (32𝑘Ω) 24𝑉 − 0.7𝑉 = 6.8𝑉



= 10.97𝑚𝐴



𝑅𝑖𝑛(𝑏𝑎𝑠𝑒) = 𝑟𝑒, ≅



𝑨𝒗 =



𝑹𝑪 𝒓,𝒆



=



25 𝑚𝑉 25𝑚𝑉 = = 2.28 Ω 𝐼𝐸 10.97 𝑚𝐴



𝟏.𝟐 𝒌𝛀 𝟐.𝟐𝟖 𝛀



Note: 𝐼𝑓 𝑅𝐸 ≫ 𝑟𝑒, , then 𝑅 𝐴𝑣 ≅ 𝑟𝐶,



= 𝟓𝟐𝟔



𝑒



𝑅𝑐 = 𝑅𝐶 ∥ 𝑅𝐿



𝐴𝑖 ≅ 1 𝑨𝒑 = 𝑨𝒊 𝑨𝒗 ≅ 𝟓𝟐𝟔



10



Propblem-10 For the two-stage, capacitively coupled amplifier in Figure below, find the following values: (a) voltage gain of each stage (b) overall voltage gain (c) Express the gains found in (a) and (b) in dB.



Solution: 𝑉𝐸 = ( 𝑅



𝑅2



8.2𝑘Ω



1 + 𝑅2



𝐼𝐸 =



𝑉𝐸 𝑅𝐸



=



) 𝑉𝐶𝐶 − 𝑉𝐵𝐸 = (33𝑘Ω+8.2 kΩ) 15𝑉 − 0.7𝑉 = 2.29𝑉



2.29 𝑉 1.0𝑘Ω



𝑟𝑒, ≅



= 2.29𝑚𝐴



25 𝑚𝑉 𝐼𝐸



=



25𝑚𝑉 2.29 𝑚𝐴



= 10.9 Ω



R in(2) = R 5 ∥ R 4 ∥ 𝛽𝑎𝑐 𝑟𝑒, = 8.2kΩ ∥ 33kΩ ∥ 175(10.9) = 1.48 kΩ 𝐴𝑣1 =



𝐴𝑣2 =



𝑅𝐶 ∥ 𝑅𝑖𝑛(2) 𝑟𝑒, 𝑅𝐶 𝑟𝑒,



=



=



3.3 𝑘Ω 10.9 Ω



3.3𝑘Ω ∥1.48𝑘Ω 10.9 Ω



= 93.6



= 302



A,v = Av1 Av2 = (93.6)(302) = 28267 Av1(dB) = 20log(93.6) = 39.4 dB Av2(dB) = 20log(302) = 49.6 dB A,v(dB) = 20log(28267) = 89 dB



11



Propblem-11 Figure below shows a CE power amplifier in which the collector resistor serves also as the load resistor. Assume βDC = βac = 100, Determinea. The dc Q-point (ICQ and VCEQ). b. The voltage gain and the power gain. c. The power dissipated in the transistor with no load d. the total power from the power supply with no load e. the signal power in the load with a 500mV input



Solution: (a) 𝑉𝐸 = ( 𝑅



𝑅2



330Ω



1 + 𝑅2



) 𝑉𝐶𝐶 = (1.0𝑘Ω+330 Ω) 15𝑉 = 3.72𝑉



𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 = 3.72𝑉 − 0.7𝑉 = 3.02𝑉 𝐼𝐶𝑄 ≅ 𝐼𝐸 = 𝑟𝑒, ≅



25 𝑚𝑉 𝐼𝐸



𝑉𝐸 𝑅𝐸1 + 𝑅𝐸2



=



=



25𝑚𝑉 68.3 𝑚𝐴



3.02𝑉 8.2Ω+36Ω



= 68.3mA



= 0.37 Ω



𝑉𝐶𝐸𝑄 = 𝑉𝐶𝐶 − (𝐼𝐶 )(𝑅𝐸1 + 𝑅𝐸2 + 𝑅𝐿 ) = 15𝑉 − (68.3𝑚𝐴)(8.2Ω + 36Ω + 100Ω) = 5.15𝑉



(b) 𝐴𝑣 =



𝑅𝐿 𝑅𝐸1 + 𝑟𝑒,



=



100 Ω 8.2Ω+0.37 Ω



= 11.7



R in = R1 ∥ R 2 ∥ 𝛽𝑎𝑐 (𝑅𝐸1 + 𝑟𝑒, ) = 330Ω ∥ 1.0kΩ ∥ 100(8.2Ω + 0.37Ω) = 192Ω 𝑅



192 Ω



𝐴𝑝 = 𝐴2𝑣 ( 𝑖𝑛 ) = 11.72 ( ) = 263 𝑅 100Ω 𝐿



The computed voltage and power gains are slightly higher if 𝑟𝑒, 𝑖𝑠 𝑖𝑔𝑛𝑜𝑟𝑒𝑑. (c) If RL is removed, there is no collector current; hence, the power dissipated in the transistor is zero.



12



(d) Power is dissipated only in the bias resistor plus a small amount in RE1 and RE2. Since the load resistor has been removed, the base voltage is altered. The base voltage can be found from the Thevenin’s equivalent draw for the bias circuit in figure below 𝑹𝒕𝒉 =



𝑹𝟏 ∗ 𝑹𝟐 𝑹𝟏 + 𝑹𝟐



= 𝟐𝟒𝟖 𝑽 Rth = 248 V



0.7 V RE1 + RE2 = 44.2 Ω



Applying the voltage –divider rule and including the base-emitter diode drop of 0.7 V result in a base voltage of 1.2 V. The power supply current is then computed as 𝑰𝑪𝑪 =



𝑽𝑪𝑪 − 𝟏. 𝟐 𝑽 𝟏𝟓𝑽 − 𝟏. 𝟐𝑽 = = 𝟏𝟑. 𝟖𝒎𝑨 𝑹𝟏 𝟏. 𝟎𝒌𝜴



Power from the supply is then computed as 𝑷𝑻 = 𝑰𝑪𝑪 𝑽𝑪𝑪 = (𝟏𝟑. 𝟖 𝒎𝑨)(𝟏𝟓𝑽) = 𝟐𝟎𝟕 𝒎𝑾 (e) 𝐴𝑣 = 11.7 𝑽𝒊𝒏 = 𝟓𝟎𝟎 𝒎𝑽𝒑−𝒑 =



𝑽𝒑−𝒑 𝟐∗ √𝟐



=



𝟓𝟎𝟎 𝑽 𝟐∗ √𝟐



= 𝟏𝟕𝟕 𝒎𝑽𝒓𝒎𝒔



𝑉𝑜𝑢𝑡 = 𝐴𝑣 𝑉𝑖𝑛 = (11.7)(177𝑚𝑉) = 2.07𝑉 𝑃𝑜𝑢𝑡 =



2 𝑉𝑜𝑢𝑡



𝑅𝐿



=



(2.07)2 100Ω



= 42.8 𝑚𝑊



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Problem-12 For the BJT amplifier of Fig. below, if vin (t) = 0.05 sin (ωt) determine: a) Zi b) Zo c) Av d) Draw the total collector voltage waveform and the total output voltage waveform. Solution: 𝑉𝐵 =



𝑉𝐶𝐶 ∗ 𝑅2 𝑅1 + 𝑅2



=



15 𝑉∗ (5∗ 103 𝛺) 10∗103 𝛺 +5∗ 103 𝛺



=5𝑉



𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 = 5 𝑉 − 0.7 𝑉 = 4.3 𝑉 𝐼𝐸 = 𝑟𝑒 =



𝑉𝐸 𝑅𝐸



=



26 𝑚𝑉 𝐼𝐸



4.3 𝑉



= 4.3 𝑚𝐴



1000 𝛺



=



26 𝑚𝑉 4.3 𝑚𝐴



= 6.05 𝛺



𝑅1 ∗ 𝑅2



𝑅𝐵 = 𝑅1 ∥ 𝑅2 =



𝑅1 + 𝑅2



= 3.33 𝑘𝛺



𝛽𝑟𝑒 = 1815 𝛺 𝑍𝑖𝑛 = 𝑅𝐵 ∥ 𝛽𝑟𝑒 = 1.12 𝑘𝛺 𝑍𝑜𝑢𝑡 = 𝑅𝐶 = 1 𝑘𝛺 𝐴𝑣 = −



𝑅𝐶 𝑟𝑒



= −



1000 𝛺 6.05 𝛺



= −165



𝑉𝑜𝑢𝑡 = 𝐴𝑣 ∗ 𝑉𝑖𝑛 = − (165) ∗ (0.05) 𝑠𝑖𝑛 (𝜔𝑡) = −8.25 𝑠𝑖𝑛 (𝜔𝑡)



14