Question

All questions refer to the "simplified" CE amplifier shown below that has no RE at all, and no Cg. There is also no CL. The bias circuit gives Ic 1.5 mA The transistor has B 120, ro 50 k, Cu 1.5 pF and C 9 pF. You can neglect r. VCC RC 5k C2 +Vo C1 2uF RI R3 Vi- 10k 100 2uF RB 200k a) Find the short-circuit time constants Ti and Ta for capacitors C1and Ca, and find the approximate low-frequency cutoff fi using the time constants b) Find the approximate high-frequency cutoff f. You can find the open-circuit time constants on the input and output sides as we did in class, or you can use the single equation as in the book c) You should find that this problem has a pretty high f for a CE amplifier. Give two reasons why we get a high fa in this problem d) Approximately, how do you think fH would change if there was a 100 Q unbypassed RE in the ac circuit? You don't need to do the calculations, but make an estimate such as stay about the same, or increase by a factor of, or decrease by a factor of

Answer 1

fest fnd Semall ignal parameters of given Ans transistor 5 7-69 mS 1.S Ic Vr gm 2-08 K2 120 57.69 X10-3 - 1 gm (a) Signal egteivalent circuit for low Now for 8mall freguem y Valtage and replace ground transistor with mall signal modal bias CI RT C2 loo RuF Rc Om Va Vi RB 200 K 2.08K Ey? 5 K time Conotant T, due to (short -circuit) and) C2 So whe Reg=eglivalent resistance actoss G ,Rez (Relln) Ren=Rz C(RT+Rel/n) + Rooxlo8o 2X/O 200+2.08 T,4.32x0-3 Simi larly Tz du to C2 Sox SX)0.029 So+s T-C2 CRs+ /IRe) = ex- ( lo

4.32x/0-3 a-029 Now 1 = 42.31 H2 Cut-off draus igh forguonoy madel S for high freyuenay Cu Vi Vo RS ISPF t 3%ろRe Sk Omlr Re 2 JOK 2.BK 9pf 260k -Fm ( 7// Rclls) gm Re mid band gain Vi Vo 50 XSXIO Av - Vo Vi 57.69 X1 - 50xS+ Soxlo +Sxla Av --180.28 Now By milby thsem the egivalont high frzusncy cirauit is Rェ EF Vo Ro JmVA T7 TO W..

=9+1.5(1+ 180. 28) Are Ci=Cr+ CuCI-Av) Ci280.92 PF = / 5pF Cu CI-) Cu Co = The open -circuit tim Constant due t Ci and Co Now are 20.92x/o XRaax 08 3 = a.X2 aaa.IX28+2a. て = 28092xja/2 X 9S 367 -8 2-68 X/0 て」 -12 /SXIO Tell Rel/ ( てっ= SoxSXlu SoXs+SoNa +SXI T2=4.5875 Xl0-j So 2K(T,) 5.656 MH3 () foom f erprersiom itis cheay that R-Rrl Ra/] R what RC is hih du to dow R So R will domi nant C should Smlarly high tH C2 GGC1-Av)

and Av should be Small be C and eu So Emall So forgiven cis cui thee pava mater ax Small So tt increare (d) if resistance Re add Than loo 2 AV (19m RE) from part (b) gain decveare by factor t So 1+ 9m RE factor Samu C dicreas by (1Av) dthure forefincrean by +9mRE) fctor apoimaty