A CE amplifier utilizes a BJT with ? = 100 is biased at IC =...

A CE amplifier utilizes a BJT with ? = 100 is biased at IC = 1 mA and has a collector resistance RC = 5 k?. Find Rin, Ro, and Avo. If the amplifier is fed with a signal source having a resistance of 5 k?, and a load resistance RL = 5 k? is connected to the output terminal, find the resulting Av & Gv. If v?? is to be limited to 5 mV, what are the corresponding v?sig and v?o with the load connected?

Transcript

00:01 Hello students here we have a bjd transistor that means bipolar transistor is connected with the source signal and from this we have to calculate parameters that is we have to calculate from the first circuit we have to calculate r in and r out and g gain in output so for that we know that formula for them emeter current that is i equal to i b plus i see because we are are dealing with the transistor so we have to know the all three current values so for that we have i .e equal to i b plus ic so we know the ic value that is one milliamper so in order to calculate the ib so we know the beta value that is 100 so from this we can calculate the ib value so i b value is 1 by 100 into 10 power minus 3 ampere substituting this value into the equation i .e that is emitter grant we will get the equation for answer for the i that is the emitter current is 0 .51 millimeters so from this we have to calculate the emitter current emitter resistance so the bjt has a inbuilt specific resistance of it won't so we have to calculate that value so for that the formula is r e dash divided by 2 .5 millimeter divided by i b because the 2 .5 millimeter wood is constant for all type of bjt transistor so from this we can calculate the emitter resistances 2 .5 kilo -oam so from the after these calculations we will calculate the rn v not a b a 0v that means our gain so the value for rn that is input resistance is beta into r e resistance of the emitter so for that 2 .5 into 10 power minus 3 10 power 3 into 100 will give you the r in value that is 250 kilo so the v out is the output voltage across the emitter is emitter is ie into because we are getting output from the emitter so that means we have to calculate the output at that point so we are substituting the values for ie and r i see we will get that vot as 0 .1 word and from these values we can calculate our gain because we know the v out and we know the v in so dividing these two things we will give you the gain of the circuit the first circuit as 0 .04 and in second we connect this bj tree bgt amplifier circuit through the signal so the signal has some gain and losses so we have to calculate the gain of third signal circuit when the load is connected as 5 kilo -oom so the connected load in the signal circuit is 5 kilo -oom so for that the gain is calculated as rl divided by rl plus re so substituting these values we get the gain as 0 .666 so from that we have to calculate the gv that is the signal source gain so the formula can be written as r & divided by r in plus r signal and rl divided by rl plus r0 into av0.

04:01 We have, we calculate av0 and we know the rl value and r signal is, the rest of the signal is pi.

04:16 So from this we can calculate the gv as 0 .013...

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