I was given the problem for homework to calculate Ib for this circuit.
Vcc=9V
βdc=300
I am happy with collector feedback bias, emitter feedback bias and voltage divider bias but this seems to combine collector and voltage divider boas and i am stuck with combining the working out how to combine or derive the formula for it.
So i thought if i could work out Ie i could then divide that by βdc to get Ib.
I tried this by removing the Re term and subtracting R1, but it doesn't seem right i think the subtracting R1 bit is wrong, but i'm not sure how to make it work.
Ie=(Vcc-Vbe)/(Rc-R1+(Rb/βdc)
Ie=(9-0.7)/(2700-1300+(2400/300)
Ie=(8.3)/(1400+8)
Ie=5.89mA and then Ib would be 19.65uA
Have i missed something when trying to combine these two types of bias? Is there a better way to think about this so i can look at the circuit and work out the right method?
Best,
Bernie
Vcc=9V
βdc=300
I am happy with collector feedback bias, emitter feedback bias and voltage divider bias but this seems to combine collector and voltage divider boas and i am stuck with combining the working out how to combine or derive the formula for it.
So i thought if i could work out Ie i could then divide that by βdc to get Ib.
I tried this by removing the Re term and subtracting R1, but it doesn't seem right i think the subtracting R1 bit is wrong, but i'm not sure how to make it work.
Ie=(Vcc-Vbe)/(Rc-R1+(Rb/βdc)
Ie=(9-0.7)/(2700-1300+(2400/300)
Ie=(8.3)/(1400+8)
Ie=5.89mA and then Ib would be 19.65uA
Have i missed something when trying to combine these two types of bias? Is there a better way to think about this so i can look at the circuit and work out the right method?
Best,
Bernie