ki thuat dien tu bai 22

<span class='text_page_counter'>(1)</span>Bài 22.

<span class='text_page_counter'>(2)</span> Tính toán trong trường hợp. • R 1= R2= 18(KΩ) • RE= RL= 1(KΩ) • RS= 600 (Ω). • βDC= 150 • βac= 175 • VCC= 10 (V).

<span class='text_page_counter'>(3)</span> 1. Phân tích phân cực.

<span class='text_page_counter'>(4)</span> • RIN(base)= βDC.RE = 150x1= 150 (KΩ). • 10R2= 10.18= 180 (KΩ). • 10R2> RIN(base). => VB= (Vcc(R2∥ RIN(base))/(R1+R2∥ RIN(base))). • R2∥ RIN(base)=(18.150)/(18+150) = 16,07 (KΩ). • VB= = 4,7 (V)..

<span class='text_page_counter'>(5)</span> • VE= VB-0,7 = 4 (V). • IE= VE/RE = 4/103 = 4.10-3 (A). • IC=. -3 /( +1)).I = 4.10 (A). DC DC E. • VCE=VCC-IE.RE=VCC-VE=10-4=6 (V). • VCB=VCE-VBE=6-0,7=5,3 (V)..

<span class='text_page_counter'>(6)</span> 2. Phân tích xoay chiều.

<span class='text_page_counter'>(7)</span> • AV= (Vout/Vin)= (Ve/Vb)= (Ie.Re)/(Ie. (Re+r’e)). • Re=RE∥RL=(1/(1+1)) = 0,5 (KΩ). • r’e=(25mV)/IE = 0,25/(4.10-3)= 62,5.10-3 (KΩ). => AV= Re/(Re+r’e) = = 0,89 (V)..

<span class='text_page_counter'>(8)</span> • Rin(base)=Vin/Ib= (Ie(re’+RERL))/Ib=(βac+1) (re’+RERL). = (βac+1) )( re’+Re)= 176.0,5625= 99 (KΩ). • Rin=Vin/Iin= R1∥R2+ Rin(base)= (18.18/36)+99 = 108 (KΩ). • R= R1∥R2∥Rs => 1/R= (1/R1)+(1/R2)+(1/Rs).

<span class='text_page_counter'>(9)</span> => R= 0,5625 (KΩ). • Rout= RE∥(r’e+R/(βac+1)). • r’e+R/(βac+1)= (62+ (0.5625/176) ).10-3 = 65.10-3 (KΩ). • Rout= 0.065/(1+0,065)= 0,06 (KΩ)..

<span class='text_page_counter'>(10)</span> • AV=Vout/Vin => Vout= AV.Vin. • Vin= Iin(R1∥R2∥Rin(base)) = VS.(R1∥R2∥Rin(base))/(RS+ (R1∥R2∥Rin(base))) • AVS= Vout/VS. => AVS = AV.(RS+(R1∥R2∥Rin(base)))/ (R1∥R2∥Rin(base)).

<span class='text_page_counter'>(11)</span> • Với: • R1∥R2∥Rin(base)= 1/ R1+1/ R2+1/ Rin(base) =1/(1/18+1/18+1/99) =8,25(KΩ). • AVS= 0,89.((0,6+8,25)/8,25) = 0,95. • Ai=Ie/Iin= (Ve/Vin).(Rin/Re)= AV.(Rin/Re) => Ai= 0,89. (108/0,5)= 192,24 • Ap= Ai.AV= 171,09..

<span class='text_page_counter'>(12)</span>