Taller Shirley

Solución ejercicio 3: Malla colector- emisor -Vcc+Vrc+Vce+Vre=0v -Vcc+IcRc+Vce+IeRe=0V -Vcc+3.08Ma*2.7Kh+7.3V+3.08Ma*0.68Kh=0V 8.31V+7.3V+2.09V= Vcc 17.7V=Vcc Malla base-emisor -Vcc+VRb+VBe+VRe=0V -17.7V+IbRb+VBe+IeRe=0V -17.7V+20µA*RB+0.7V+3.08Ma*0.68Kh=0V RB=17.7V-0.7V-2.09V / 20 µA RB=14.91 / 20 µA RB=745.5 Kh IC=β*IB IC=154*20 µA IC=3.08mA β =IC/IB β=3.08Ma/20 µA β=154 VE=IE*RE VE=( 3.08Ma) * ( 0.68Kh) VE=2.09V VC=VCE+VE VC=7.3V+2.09V VC=9.39V VB=VBE+VE VB=0.7V+2.09V VB=2.79V VBC=VB-VC VBC=2.79V-9.39V VBC=-6.6V IE=Ve/Re IE=2.1V/0.68Kh IE=3.08Ma

Solución ejercicio 2: -Vcc+VRc+VCe+VRe=0V -Vcc+IC*RC+VCE+IE*RE=0V -12V+2Ma*3.8Kh+2V+2Ma*RE=0V RE=12V-7.6V-2V / 2Ma RE=2.4V / 2Ma RE=1.2Kh Vce=Vcc-Ic ( Rc+Re) Vce=12V-2Ma (3.08Kh+1.2k) Vce=12V-7.6V-2.4V Vce=2V -Vcc+VRb+VBe+VRe=0V -Vcc+Ib*Rb+VBe+IE*RE=0V -12V+25µA*Rb+0.7V+2Ma*1.2Kh=0V Rb=12V-0.7V-2.4V /25µA Rb=8.9V / 25µA Rb=356Kh -Vcc+VRc+VCE+VRe=0v -Vcc+Ic*Rc+VCE+Ie*Re=0V -12V+2Ma*RC+2v+2Ma*1.2Kh=0V RC=12V-2V-2.4 / 2Ma RC=3.8Kh VB=VBE+VE VB=0.7V+2.4V VB=3.1V VE=IE*RE VE=(2Ma)*(1.2Kh) VE=2.4V VC=VCE+VE VC=2V+2.4V VC=4.8V VBC=VB-VC VBC=3.1V-4.8V VBC= -1.7V IB=IC/β IB=2Ma/80 IB=25µ