Velec741 New Member Jan 19, 2012 #1 hi there... have any idea on how to solve this one? View attachment 60302
ericgibbs Well-Known Member Most Helpful Member Jan 19, 2012 #2 Velec741 said: hi there... have any idea on how to solve this one? View attachment 60302 Click to expand... hi, Give you a few hints. Its a current mirror, that means the same current flows thru both resistors, the left side transistor is connected collector to base which makes it a diode. What do you think the answer is and why.?
Velec741 said: hi there... have any idea on how to solve this one? View attachment 60302 Click to expand... hi, Give you a few hints. Its a current mirror, that means the same current flows thru both resistors, the left side transistor is connected collector to base which makes it a diode. What do you think the answer is and why.?
Velec741 New Member Jan 19, 2012 #3 is it 27.8V? Ir=(12-0.7)/10k = 1.13mA ........ (1) Ir=Ic+IB1+IB2 ; Ir=Ic+(2*Ic/Bo)------------->(beta) Ir=Ic(1+2/Bo) from (1), 1.13=Ic(1+2/100) => Ic=1.107mA Ic of trans1 = Ic of Trans2 =Ic' =(50-V2)/20k = 1.107 => V2=27.86V that ok?
is it 27.8V? Ir=(12-0.7)/10k = 1.13mA ........ (1) Ir=Ic+IB1+IB2 ; Ir=Ic+(2*Ic/Bo)------------->(beta) Ir=Ic(1+2/Bo) from (1), 1.13=Ic(1+2/100) => Ic=1.107mA Ic of trans1 = Ic of Trans2 =Ic' =(50-V2)/20k = 1.107 => V2=27.86V that ok?
ericgibbs Well-Known Member Most Helpful Member Jan 19, 2012 #4 Velec741 said: is it 27.8V? Ir=(12-0.7)/10k = 1.13mA ........ (1) Ir=Ic+IB1+IB2 ; Ir=Ic+(2*Ic/Bo)------------->(beta) Ir=Ic(1+2/Bo) from (1), 1.13=Ic(1+2/100) => Ic=1.107mA Ic of trans1 = Ic of Trans2 =Ic' =(50-V2)/20k = 1.107 => V2=27.86V that ok? Click to expand... hi, That looks fine to me. Well done.
Velec741 said: is it 27.8V? Ir=(12-0.7)/10k = 1.13mA ........ (1) Ir=Ic+IB1+IB2 ; Ir=Ic+(2*Ic/Bo)------------->(beta) Ir=Ic(1+2/Bo) from (1), 1.13=Ic(1+2/100) => Ic=1.107mA Ic of trans1 = Ic of Trans2 =Ic' =(50-V2)/20k = 1.107 => V2=27.86V that ok? Click to expand... hi, That looks fine to me. Well done.