Vcc=10V, Vf(for LED)=1.5V. taking Vce=Ve=(10-1.5)/2=4.25V, Vbe=0.7V. Ic=If=Ie=50mA beta=200, Ib=Ic/beta=0.25mA. Re=Ve/Ie=85ohms. R2=Vb/Ib=19.8k . Vb=Vcc*R2/(R1+R2) voltage divider where R1=19.4k. if not correct need help from some one. Thanks |
by nurafalalu
July 05, 2013 |
Please sign in or create an account to comment.
CircuitLab is an in-browser schematic capture and circuit simulation software tool to help you rapidly design and analyze analog and digital electronics systems.