Problem 1 on Fixed Bias Circuit in DC Analysis – Bipolar Junction Transistor – Analog Electronics


Hello students now we will try to find out solution of fixed bias circuit the solution means to find out DC points for fixed bias circuit so let’s have circuit this is a simple circuit based upon fixed by a circuit and they have asked you to find out Q points so very first question in my mind is what is Q point so answer is Q point is nothing but VCE and I C so we are supposed to find out VCE and I C we will log this or getting this point what I’m supposed to find out VCE and IC okay we have already performed DC analysis of fixed bias circuit whether it is normal fixed bias or modified fixed bias we know that in order to find out Q point we need to apply three steps step one is to find out base current IB step two to find out collector current IC and step three to find out voltage between collector to emitter VC and from state two and three three we will get Q point so let us apply state one to find IB for which apply KVL to input that is base to emitter so I mock direction of the current from this there will be a base current IB whereas here there will be a collector current IC from these two points we will get base to emitter voltage VB and here we will gate collector to emitter voltage we see now apply KVL to the input the equation will be VCC minus ib into RB minus VB minus IE into re is equal to zero we know that I is equal to beta plus one times of IB if we solve this we will get that IB is equal to VCC minus VB divided by RB plus beta plus one times of re where vbe is equal to 0.7 volts I will repeat vbe which will have standard value of 0.7 volts if it is not mentioned in the problem if it is mentioned in the problem we will take that particular value so in this problem they are only mentioned about beta they are not mentioned about VB and hence we will take value of VB as 0.7 volts so I will salute the values so IB is equal to VCC is 20 minus 0.7 volts divided by RB is equal to four thirty kilo plus beta is equal to 50 hence beta plus 1 will become fifty one so I will write that it is fifty one you understand 51 because it is beta plus one next is re which is nothing but one kilo if we solve this we will get IB is equal to 40 point one micro amperes state two is to find I see where IC is equal to beta times IB if we multiplied IB that is forty point one micro amperes with beta that is 50 then we will get collector current as two point zero one milli amperes so from state one and state two we have got base current and collector current respectively now from step three we will get collector to emitter voltage VCE step three is to find VC which is in which we need to apply KVL from output that is collector to emitter hence the equation will become VCC minus IC into RC minus VC minus IE into re is equal to zero but we know that IC is approximately equal to IE hence VCE is equal to VCC minus IC into bracket RC plus re where VCC is equal to 20 volts minus IC is two point zero one milli amperes into bracket RC is two kilo ohms plus re is one kilo ohms and hence VCE is equal to thirteen point nine seven volt so from Step three we have got VC so at the end we can write Q point equal to thirteen point nine seven volt comma two point zero one milli amperes this will be your final answer thank you

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