Electronics - Bipolar Junction Transistors (BJT) - Discussion
Discussion Forum : Bipolar Junction Transistors (BJT) - General Questions (Q.No. 12)
12.
The Q point on a load line may be used to determine:
Discussion:
24 comments Page 1 of 3.
Diptendu Mitra said:
1 decade ago
Vcc-IcRc-Vc=0 [Vc=Vce+Ve+Ground voltage].
or, Vcc-IcRc-(Vce+Ve+0)=0.[since,Vce=Vc-Ve & '0' is Ground voltage].
or, Vcc-IcRc-Vce-Ve=0.
or, Vcc-IcRc-Vce-Vb+Vbe=0 [Vb=Vbe+Ve, or,Ve=Vb-Vbe].
or, (Vcc-IcRc-Vce+Vbe)-Vb=0.
or, Vb=Vcc-IcRc-Vce+Vbe [Vbe=0.7v(approx),Vce=0.2v(approx)].
Now if Vce becomes equal or almost equal to Vbe , Vb=Vcc-IcRc.
So bottom line It could depend on Vb in a way.
or, Vcc-IcRc-(Vce+Ve+0)=0.[since,Vce=Vc-Ve & '0' is Ground voltage].
or, Vcc-IcRc-Vce-Ve=0.
or, Vcc-IcRc-Vce-Vb+Vbe=0 [Vb=Vbe+Ve, or,Ve=Vb-Vbe].
or, (Vcc-IcRc-Vce+Vbe)-Vb=0.
or, Vb=Vcc-IcRc-Vce+Vbe [Vbe=0.7v(approx),Vce=0.2v(approx)].
Now if Vce becomes equal or almost equal to Vbe , Vb=Vcc-IcRc.
So bottom line It could depend on Vb in a way.
Rohit dang said:
1 decade ago
For CE,
Q point signifies the conditions for bjt to be operated in either active cutoff or saturation regions.
And from commonly used diagrams.
Vce=Vcc-IcRc. And this is also the equation of load line in output characterstics (Ic vs. Vce).
Hence from q-point Vce, Ic can be obtained graphically.
But we can also find Vb from relation Vce+Veb+Vbc=0.
Q point signifies the conditions for bjt to be operated in either active cutoff or saturation regions.
And from commonly used diagrams.
Vce=Vcc-IcRc. And this is also the equation of load line in output characterstics (Ic vs. Vce).
Hence from q-point Vce, Ic can be obtained graphically.
But we can also find Vb from relation Vce+Veb+Vbc=0.
Siddarthchaitanya said:
9 years ago
Load line(dc) between Ic& Vce used to bias the transistor.
Usually we consider CE configuration of transistor amplifier only.
In CE configuration input is Ib which is determined by Vb.
So answer is Vb.
Usually we consider CE configuration of transistor amplifier only.
In CE configuration input is Ib which is determined by Vb.
So answer is Vb.
Imran said:
1 decade ago
Hey guys answer is right. Vb means not base voltage it means biasing voltage. With the help of load line analysis we find the value of biasing voltage for maintain Ic at Q point. T.
Amit gupta said:
1 decade ago
@Satya.
Vb is the base voltage and IcRc is voltage across collector. So there is no relation of Vb in this formula Vce = Vcc-IcRc.
Vb is the base voltage and IcRc is voltage across collector. So there is no relation of Vb in this formula Vce = Vcc-IcRc.
Sandeep kr. gupta said:
1 decade ago
Load line drawn between Ic & Vcc at constant Ib. Ib is obtain from Vb hence Q point on a load line is used to determine Vb.
Saroj kumar mekap said:
1 decade ago
In CE configuration the general equation is:
Vce = Vcc - IcRc - IeRe.
So there is no relation to Vb. So the answer is Vc.
Vce = Vcc - IcRc - IeRe.
So there is no relation to Vb. So the answer is Vc.
Savi said:
8 years ago
We get the value Ib from VB so, the answer will be VB. Because we know that on the graph is already given the VCC and Ic.
Bittu said:
1 decade ago
To draw a load line we require two points (vce, ic). But considering any one of the points we cannot determine load line.
Harpal singh said:
1 decade ago
A load line is used to get quesent point which is Q(Vce,Ice).so in quetion ans. D is correct.............
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