# Electronics - Bipolar Junction Transistors (BJT) - Discussion

### Discussion :: Bipolar Junction Transistors (BJT) - General Questions (Q.No.12)

12.

The Q point on a load line may be used to determine:

 [A]. VC [B]. VCC [C]. VB [D]. IC

Explanation:

No answer description available for this question.

 Krishn Mohan said: (Aug 10, 2011) Vb = Vcc - IcRc

 V.Nath said: (Sep 19, 2011) Since from load line we get values of Vcc and Ic, so ans is Ic, hence option D is correct.

 Sandeep Kr. Gupta said: (Sep 19, 2011) 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.

 Shrishail said: (Nov 4, 2011) We get Vce & Ic from dc load line not Vb.

 Sukhpreet said: (Jan 28, 2012) @ Mr. krishn mohan Vce= Vcc-IcRc not Vb = Vcc-IcRc

 Satya said: (Apr 1, 2012) Krisna Mohan is right

 Amit Gupta said: (Jun 13, 2012) @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.

 Rohit Dang said: (Aug 13, 2012) 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.

 Digvijay Pandey said: (Aug 17, 2012) Vc=Vcc-IcRc. SO CURRENT OPTION IS A

 Harpal Singh said: (Sep 2, 2012) A load line is used to get quesent point which is Q(Vce,Ice).so in quetion ans. D is correct.............

 Saroj Kumar Mekap said: (Oct 4, 2013) In CE configuration the general equation is: Vce = Vcc - IcRc - IeRe. So there is no relation to Vb. So the answer is Vc.

 Diptendu Mitra said: (Dec 23, 2013) 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.

 Md.Adil said: (Apr 2, 2014) I think answer is wrong it should be Q(vce, ic).

 Imran said: (Sep 16, 2014) 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.

 Bittu said: (Nov 23, 2014) To draw a load line we require two points (vce, ic). But considering any one of the points we cannot determine load line.

 Firas said: (Jun 18, 2016) Vc = Vcc - Vce. Therefore, the answer should be option A.

 Ramkamal said: (Sep 22, 2016) Ic and Vce denoted by Q point. By the Q point, we determined the value of Ic and Vcr.

 Siddarthchaitanya said: (Dec 15, 2016) 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.

 Heshmah said: (Jan 10, 2017) Q point represents current gain so, the correct option Ic that means D.

 Surendran Km said: (May 18, 2017) I think Q point is the zero signal condition of IC and Vce hence Q pont determine IC and VCE.

 Pradeep Sharma said: (Aug 12, 2017) The q points are Ic and Vce, not Vb so option X is right.

 Sankalp Gupta said: (Sep 16, 2017) We get the value of Ic and Vce due to Q point. Hence option is correct.

 Savi said: (Nov 21, 2017) 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.

 Rajneesh Ranawat said: (Jan 6, 2019) You are right @Krishn Mohan.