Are you sure about that result? And we don't need to know Beta here. Because the BTJ will be saturated.How did you determine Vo without Beta ??
I got Vb=2.273 V and Ib=0.1573mA. Without Beta how did you find Ic?
Yes I am sure. Vb(max) = Vin(Max)*10/11 and Ib =Vb/10k.Are you sure about that result? And we don't need to know Beta here. Because the BTJ will be saturated.
You can do this by assuming that Beta is larger than 10.
Not quite right about Ib and Vb.Yes I am sure. Vb(max) = Vin(Max)*10/11 and Ib =Vb/10k.
Even if the BJT is saturated then Vo=5-(300*Ic).
And using KVL : -5+300Ic+0.2(Vce sat)=0
This results in Ic= 16mA.
And Vo= Vce sat =0.2V.
Also BJT does not act as amplifier in sat it is just an switch.
Am I right here?
Thanks Jony130 about Vb and Ib. Your post was helpful.Not quite right about Ib and Vb.
Try read this
http://forum.allaboutcircuits.com/threads/bjt-npn-pull-down-resistor.56690/#post-377931
Yes, if we assume Vce_sat = 0.2V. Then your answer is right. Vo = Vce_sat for Vin = 2.5V.Thanks Jony130 about Vb and Ib. Your post was helpful.
But is my output right ??
Well in that case the BJT enters into Cut-off mode and it is an open circuit which implies Vo= 5V.Yes, if we assume Vce_sat = 0.2V. Then your answer is right. Vo = Vce_sat for Vin = 2.5V.
But what about the case when Vin = 0V.
by Jake Hertz
by Jake Hertz
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by Jake Hertz