القوانين اللي استخدمناها فالمسألة اللي فاتت :-
IBQ = (VBB - 0.7) / RB
ICQ = IBQ β
VCEQ = VCC - ICQ RC
IBQ = (VBB - 0.7) / RB
ICQ = IBQ β
VCEQ = VCC - ICQ RC
As shown in Figure a dc load line .....
Anonymous Quiz
13%
cutoff point
13%
saturation point
61%
Both of them
12%
Active point
In between cutoff and saturation along
the load line is the active region of the
transistor’s operation
the load line is the active region of the
transistor’s operation
Anonymous Quiz
95%
T
5%
F
At ... , BCJ becomes Fwd-biased
and there is no more increase for IC
and there is no more increase for IC
Anonymous Quiz
18%
cutoff
71%
saturation
11%
active
من الاسئلة اللي فاتت هنلاحظ ان
cut off => IC = 0 VCE = VCC- IC RC
saturation => IC = IC(sat) VCE = VCE(sat)
cut off => IC = 0 VCE = VCC
At saturation , BCJ becomes ...-biased
and there is no more increase for IC
and there is no more increase for IC
Anonymous Quiz
83%
Forward
17%
Reverse
القوانين اللي استخدمناها المسالة اللي فاتت
IC(sat) = (VCC - VCE(sat)) / RC
IB = (VBB - VBE) / RB
IC = IB β
VCE(sat) = VCE
لاحظ اول قانونين هتلاقيهم زي بعض الفرق اني ببدل الC بالB ويطلعلي القانون التاني
IC > IC(sat) => saturated
IC(sat) = (VCC - VCE(sat)) / RC
IB = (VBB - VBE) / RB
IC = IB β
VCE(sat) = VCE
لاحظ اول قانونين هتلاقيهم زي بعض الفرق اني ببدل الC بالB ويطلعلي القانون التاني
IC > IC(sat) => saturated