Question

Problem: In the circuit shown in Figure 1, Vee = 1.2 V, Vcc = 20 V, Rp = 60 kN, Rc = 2 k. The input signal is a sinusoidal voltage given by Vin(t) = 0.2 sin(2000 ) V. The input and output characteristics of the transistor are provided on Page 2. (1) Find Ig, Ic and Vce. (30 points) Hint: Use the load line method. (Vor.) and (Vce: 1c) are the operating points of the transistor in the input circuit and output circuit, respectively, when Vin(t) = 0. (Vor.) and (Ver.) are also known as the quiescent points (i.e., Q-points) of the transistor circuit. (2) Estimate the current amplification factor, B, of the transistor. (10 points) Hint: B = Ic/1g. (3) Find the maximum and minimum values for vce. (30 points) Hint: vn causes the operating points of the transistor to swing around the Q-points. (4) Find the maximum and minimum values for v.. (30 points) Hint: Vomax = icmaxRc, Vomin = lomin Rc. Figure 1

Answer 1

For DC operating point

set vin=0

then

KVL at input loop

VBB = 18 Rg + VBE

this can be written as

$I_B=\frac{V_{BB}-V_{BE}}{R_B}=\frac{-1}{R_B}V_{BE}++\frac{V_{BB}}{R_B}$

substituting the given values

IB = -0.167 x 10-4VBE + 0.2 x 10-4

this equation is a equation of line .graph it on the input characteristics given

At vBE=0 Ib=20uA, at iB=0, vBE=VBB=1.2V

is (A) (0.6,10u) 0.4 0.6 0.8 1.0 (a) Input characteristic

therefore, the operating point is

1b = 10uA, VBE = 0.6V

Now, KVL at output gives

$V_{CC}=I_CR_C+V_{CE}$

therefore

$I_C=\frac{V_{CC}-V_{CE}}{R_C}=\frac{-1}{R_C}V_{CE}+\frac{V_{CC}}{R_C}$

substituting the given values

$I_C=-0.5mV_{CE}+10m$

graph this line on output characteristics

ic (mA) 25uA 10 20 MA 15uA (10V,5mA) 10 MA 5 A 1₂=0 20"ce(V) 16 (b) Output characteristics

therefore, the operating point is

Ic=5mA

Vce=10V

b)

$\beta=\frac{I_C}{I_B}=\frac{5m}{10\mu}=500$

c)

FRom output graph

Vce(sat)=~1

therefore

$v_{CE}|_{\max}=V_{CE}-V_{CEsat}=10-1=9V$

$v_{CE}|_{\min}=V_{CE}-V_{CE}|_{I_B=0}=10-18=-8V$

similarly

$i_{CE}|_{\min}=I_{CE}-I_{CEsat}=5m-9.5m=-4.5mA$

$i_{CE}|_{\max}=I_{CE}-I_{CE}|_{I_B=0}=5m-1m=4mA$

4)

therefore

$v_O_{\max}=i_{CE}|_{\max}R_C=4m\times2k=8V$

$v_O_{\min}=i_{CE}|_{\min}R_C=-4.5m\times2k=-9V$