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a. Using the characteristics of Fig., determine Bac at IB 60 mA and VCE z4 V b. Using the characteristics of Fig., determine Bac at IB 30 mA and VCE -7 V c. Using the characteristics of Fig., determine Bac at IB 10mA and VCE = 10 V d. Reviewing the results of parts (a) through (c), does the value of Bac change from point to point on the characteristics? Where are the high values located? Can you develop any general conclusions about the value of Bac on a set of collector characteristics? e. The chosen points in this exercise are the same as those employed in Problem. If Problem was performed, compare the levels of Bdc and Bac for each point and comment on the trend in magnitude for each quantity FIG. Characteristics of a silicon transistor in the common- emitter configuration: (a) collector characteristics lc (mA) 80μΑ 70μΑ 60μΑ Saturation region) 5 50μΑ 40 HA (Active region) 20μΑ IO HA 10 15 20 V V) (Cutoff region) Jazz 46.1171% 12:51 AM X Step 1 of 5 Refer to Figure 3.13 from the textbook. Represent the values provided on the characteristics curve. Consider the following for the common emitter CE and the base current 1,-60 μΑ The collector current 'c ranges from 1c,-5.9 mA to IC1-4.6 mA and The base current 's has the range of 1B2 = 70 μΑ to 181-50 μΑ Calculate the common-emitter forward-current amplification factor Pac. ΔΙ ΔΙ 5.9 mA-4.6 mA 70 μΑ-50 μΑ 65 Hence, the ac parameter B is 65 Jazz 念".1171% 12:51 AM X Step 2 of 5 Consider the following for the common emitter =7V CE- and the base current 1,-30 μΑ The collector current ranges from 'c is 1C2 = 4.1 mA to 1C1-2.3 mA and The base current B has the range of ,2-40 μΑ to 1B1-20pA Calculate the common-emitter forward-current amplification factor ac Bac ΔΙ ΔΙ 4.1 mA-2.3 mA 40 μΑ-20 μΑ = 90 Hence, the ac parameter pac is s 90 Jazz 46.1171% 12:51 AM X Step 3 of 5 Consider the following for the common emitter CE 10and the base current 1,-10 μΑ The collector current ranges from 'c is 1C2 = 2.4 mA to 1C1-0.3 m A and The base current' B has the range of Calculate the common-emitter forward-current amplification factor ac ac ΔΙ 2.4 mA-0.3 mA 105 Hence, the ac parameter P is T05 Bae is 105 Jazz 46.11 71% 12:51 AM X Step 4 of 5 Yes, the ac parameter changes from point to point and the highest value of P is at the lowest levels of collector current. Jazz 46.11 71% 12:51 AM X Step 5 of 5 The following table illustrates the comparison between the ac parameter Pand the dc parameter Píc corresponding to the provided collector currents: Table 1 Collector emitter Ratio voltage AC Collector DC current parameter parameter 5.25 mA 87.5 3.25 mA 108.3 1.35 mA 135 65 90 105 CE 4 V 7 V 10 V 1.35 1.20 1.29

Answer 1

The number written beside each curve is value of Ib current (micro amperes)..for that curve you have to see it's vertical limits..that becomes variation of Ic..infact the exact plot has not been recreated on my graph..but I have tried to resemble the graph in question.

the number beside each curve is value of Ib current (micro amperes)..for that curve you have to see it's vertical limits..that becomes variation of Ic..infact the exact plot has not been recreated on my graph..but I have tried to resemble the graph in question.

The output characteristics of silicone transistor are shown i.e IC vs VCE..In the output characteristics each curve is obtained from fixing input at a level i.e fixing Ib at a level ..the current amplification factor is given by beta ac = delta Ic divided by delta Ib

a)it is given in that Ib = 60 micro amp

So go to Ib= 60 curve and now calculate range of IC for this curve (vertical variation of the curve) ...you can see that Ic ranges from 4.6 to 5.9 ma..and Ib ranges from 50 to 70 micro amp

Now as per definition beta ac= delta Ic divided by delta Ib

You can see that beta ac= (5.9-4.6)micro amp divided by (70-50 )milli amp

I.e beta ac = 65

b) Similar to above process now refer to Ib = 30 curve

On calculation we get beta ac =90 as shown in the answer you

Have shown..

Note: Here input Ib is reduced but still amplification factor or the current gain beta ac is increased .this is because Vce is increased ..outp ou current also depends on Vce...Ic is directly proportional to Vce..

C) Here Ib is further reduced to 10 micro amp and Vce is increased to 10 V..on calculation we got beta ac= 105

d) so from a to c we see beta acis changing from point to point and high when Vce is high or IC is low which are evident from parts a to c

e) From the table it is understanding that beta dc > beta ac ..we have kept decreasing Ib