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Question No. 4 |Marks 10, CLO 41 Determine the following DC values for the amplifier in Figure 4. (a) V. (b) VE () LED (d) I

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Answer #1

For a DC signal capacitor becomes open circuit.

The transistor in the given circuit is NPN transistor.

For an NPN transistor Vbe=0.7v

Collect current = Beta × Base current

Emitter current= Base current+ Collector current

ie=ib+ic

TOV 4752 louf lokar Boc=75 lout Bac=70 12 ur for a DC signal capacitor becomes open circuit. so, ci, Cz, ez got open and į is

from the circuit VE 12k - 12K B 12k+47k X 18 = 3-66 V C voltage division dule) VB-3-66V b) The given transistor is an NPN tra

d) for a BJT Тc - във and IB t I = It Is the FIE Ic (6 + 1) = 16 IC (AP) = 16 Je = TE CA pays 72-96(75) = 2.92mn Ic= 2.92mA 1f) UCE = Ve-VE Nc = 8. 364 NE=2-96 0. 264 - 2-96 = 5.404 v VCF = 8.364 -2-96 CS Scanned with CamScanner

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