Question

Electronics-common emmitter amplifier. Pls be try to be helpful

Im trying to study here in this book example..

However im kinda lost and i cant follow the correct answers they got.. could someone pls help me..

I just wanna learn this, i hope u can help me learn by showin clear and corect solutions on how to properly got the answer..

Also especially on finding the values on the laod line analysis, im lost.

Pls help

Any helpful help would be apprciated..

polar Junction Transistors Rc RB UCE vino) UBE VBB Figure 4.10 Common-emitter amplifier. Graphical Determination of O-Point and Peak Signal Swings Example 4.2 Suppose that the circuit of Figure 4.10 has Vcc 10 V, VBB 1.6 V, RB 40 k2, and Rc = 2kS2. The input signal is a 0.4-V peak, 1-kHz sinusoid given by Vin (t) 0.4 sin(2000π). The common-emitter characteristics for the transistor are shown in Figures 4.12a and b. Find the maximum, minimum, and Q-point values for UCE SOLUTION First, we must find values for ig. The load lines for vin 0 (to find the Q-point), Vin 0.4 (positive extreme), and vin0.4 (negative extreme) are shown in Figure 4.12a. The values for the base current are found at the intersections of the load lines with the input characteristic. The (approximate) values are IB max-35 μΑ, IBQ ~ 25μΑ, and IB min ~ 15 μΑ. Next, the load line is constructed on the output characteristic, as shown in Figure 4.12b. The intersection of the output load line with the characteristic for IBQ 25 11A establishes the Q-point on the output characteristics. The values are Junction Transis i, (HA) 50 30 2 point 20 la' 바 10 -0.4 2.0 1.5 1.6 1.2 0.5 (a) Input ic (mA) 45 μΑ 40μΑ 30 HA 0 point 18,-25 μΑ 20pA 10pA 14 618 vec-10V 12 -5v CE min (b) Output Figure 4.12 Load-line analysis for Example 4.2. Section 4.2 Load-Line Analysis of a Common-Emitter Amplifier 227 %(mV) UCE (V) 10 400 +- Vamax =가 200 CEO VCEmin t (ms) 2.0 t (ms) 0 2.0 0.5 1.0 1.5 1.0 1.5 0.5 (b) Output (a) Input Figure 4.13 Voltage waveforms for the amplifier of Figure 4.10. See Example 4.2 Ico 2.5 mA and VCEO 5V. Similarly, the intersection of the load line with the characteristic for IB max-35 pA yields VCE min ~ 3 V. The opposite extreme b/B min 15 μΑ, resulting in VCE max-TV. If more points are found as vin varies with time, we can eventually plot the VCE waveform against time. The waveforms for Vin() and vcEt) are illustrated in Figure 4.13. Notice that the ac component of UcE(1) is inverted compared with the input signal. The peak-to-peak value of the input voltage is 0.8 V, and the peak-to-peak value of the ac component of vcE is 4 V. Thus, the voltage gain magnitude is 5 (i.e., the ac component of vCE is five times larger in amplitude than we would state the gain as-5 to emphasize the fact that the amplifier inverts the input signal. A PSpice circuit file for this example can be found on the is name is Fig4.10. You can run the simulation to verify the results found in the example and to examine what happens when various parameters are changed, The transistor model parameters are β = BFs 100 and is-S-2x website. The file

Answer 1

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