Q-Point in Voltage Divider Biasing – Bipolar Junction Transistor – Analog Electronics

Now in the exam they have asked you to calculate Q point for following circuit the very first step is to observe the circuit so we will observe the circuit while observing we come to know that certain points in the circuit first we will identify whether this is voltage divider bias or not now as there are two resistors at the input side r1 and r2 definitely it is voltage divider bias then there is a resistor at the collector and the emitter so it is definitely voltage divider bias but there is a slight modification in the circuit diagram you can see that now there is a battery DC battery at the collector called as VCC is present but along with that another battery is present at the ground of minus 6 volt so this is a new variation in the voltage divider bias circuit so now in this particular case before that what steps we are following first step is always redraw the circuit diagram second step is to get base current IB step three is to gate collector current IC and step four is to gate vc collector to emitter voltage these are four different steps in order to find out Q point am i right yes but now in the first step in order to get Vth and rth equation of v th will change let’s see what is change in the equation so step 1 v th is given as VCC multiplied by r2 normally the Vth is given as VCC multiplied by r2 divided by r1 plus r2 when this battery is absent but whenever these batteries present equation of V th will become VCC plus V EE multiplied by r2 divided by r1 plus r2 and from whole term minus or subtract the second see once again the Vth will become V CC Plus Vee multiplied by r2 divided by r1 plus r2 and from this whole term subtract second battery I will write that this is r2 multiplied by VCC plus ve divided by r1 plus r2 minus Vee let’s see the equation if we substitute the values it is r2 is 23.5 ke VCC is 6 and V is also 6 so it will become 12 divided by twenty three point five kilo plus 68 kilo minus six so value is equal to minus 2.9 manuals now this is the value of VT h rth is given as R 1 parallel R 2 which is nothing but 68 kilo parallel with twenty three point five kilo and value is equal to seventeen point four six kilo ohms this is the value of rth so from step one we have got value of Vth and rth the next step is to redraw the circuit diagram and get value of base current and collector current and VC so now step two the permanence equivalent circuit will become as the Vth was minus two point nine volts so it will be like this this is v th battery along with that there will be a rth now there will be a transistor here there will be the re here there will be the VCC and RC value of RC is two point five kilo value of arrays 0.94 kilo value of RT edges seventeen point four six kilo this value is nothing but minus six volt value of VT edges now see here the battery is positive to negative and hence I will write value of this is plus 0.9 one holds value while calculating we have got minus 2.9 the voltage source cannot be minus so we need to change the polarity so we have changed the polarity between this point there will be VB and here there will be VC this current will be IB this current will be beta plus one time so now in step two we will try to find out base current IB for which apply KVL to input if we apply KVL to the input we will get that this equation will become the current is flowing from my plus to minus so it is minus vth minus IB into rth minus VB minus beta plus one times of IB into re minus minus of minus will become plus vee is equal to zero substitute the corresponding values or before the through the values let’s write base current on the left hand side and all the voltages on the right hand side so base current will become IB is equal to ve e minus vth minus VB divided by RT H plus beta plus 1 times of re this is very simple what we have done is we have taken all voltages on one side where space equations on the other side base current equation on the other side in that particular case the base equation become the base current equation becomes Vee minus vth plus vbe divided by RT h plus beta plus 1 times of re now substitute the values 6 volts minus 2.9 1 minus 0.6 divided by rth is seventeen point four six below beta is 180 so beta plus 1 becomes 1 81 x 0.9 for K the value of base current is equal to 12 point two micro amperes now step three to find I see we know that IC is equal to beta times of IB and hence value of IC becomes two point two nine milli amperes so we have got first Q point IC o—- or one parameter in the Q point called as IC step 4 is to find VC for that apply kvl to output loop equation will become VCC minus vc minus IC into RC minus IC into re plus ve is equal to zero hence hence VCE is equal to VCC plus ve minus IC into bracket RC plus re substitute the values we will get VC VC C is 6 volts plus VB depend VCC is 6 volts plus ve a 6 volts so it is 6 plus 6 minus IC is 2.2 nine milli amperes multiplied by RC is 2.5 k plus re is 0.9 for K if we solve this VCE is equal to 4 point 1 2 volt hence Q point is 4 point 1 2 comma 2 point 2 9 milli amperes this is how we can calculate Q point in voltage divider bias thank you

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