walid Posted October 24, 2005 Report Posted October 24, 2005 I write the following prog to simulate the circuit shown in Fig.C-E below:common E Amp freq responseVCC 7 0 DC 12VS 1 0 AC 1Q1 4 3 5 Q_2N2222A_NRS 1 2 1KC1 2 3 2uFR1 3 0 10KR2 7 3 30KRC 7 4 4.3KRE 5 0 1.3KC3 5 0 10uFC2 4 6 0.1uFR3 6 0 100K.MODEL Q_2N2222A_N NPN( IS=11.9F NF=1 NR=1 RE=649M RC=1 + RB=10 VAF=83.5 VAR=41.7 ISE=350F ISC=350F + NE=1.58 NC=1.58 BF=204 BR=5 IKF=149M + IKR=149M CJC=13.1P CJE=30P VJC=2.87 VJE=678M + MJC=330M MJE=330M TF=531P TR=69N EG=1.11 + KF=0 AF=1 ).op.endThe .op command results are:NAME Q1 MODEL Q_2N2222A_N IB 1.25E-05 IC 1.71E-03 VBE 6.66E-01 VBC -1.74E+00 VCE 2.40E+00 BETADC 1.37E+02 GM 6.53E-02 RPI 2.35E+03 RX 1.00E+01 RO 4.90E+04 CBE 8.44E-11 CBC 1.12E-11 CJS 0.00E+00 BETAAC 1.54E+02 CBX 0.00E+00 FT 1.09E+08 And I need to discuss some of these results with you:(1) How far or close these results to the real life, can i depend on it initially?(2) RPI is the i/p impedance looking into the base of Q1, so, the total i/p impedance is: R1//R2//RPI = 1.8K ohm. so we can look at the i/p portion of the circuit as: Vs connected in series with Rs then to C1 then connected inseries to that 1.8k. if we look at Vs as a MIC and Rs as the impedance of the MIC and C1&Rtot as a high pass filter, is the value of C1 suitable and why?(3) What exactly these parameters mean: GM, RX, RO, CJS, CBX.(4) FT is (as I remember) is the freq at which the gain =1, that is at this freq there is no amplification, Is this OK?(5) In the MODEL Q_2N2222A_N, BF = 204 is this the upper value that this Q can reach?(6) In the MODEL Q_2N2222A_N, what these mean: NF=1 NR=1 NE=1.58 NC=1.58 IKF=149M IKR=149M VJC=2.87 VJE=678M MJC=330M MJE=330M TF=531P TR=69N EG=1.11 KF=0 AF=1 thats allthank you in advancewalid Quote
audioguru Posted January 14, 2006 Report Posted January 14, 2006 (1) How far or close these results to the real life, can i depend on it initially?Hi Walid,Doesn't your sim program have built-in spec's for common transistors? A typo can mess it up.The sim program works with "typical" spec's. The actual results with "real" transistors having a wide range of current gain and Vbe will be much different. Change the current gain in the program to see if the circuit still works.(2) RPI is the i/p impedance looking into the base of Q1, so, the total i/p impedance is: R1//R2//RPI = 1.8K ohm. so we can look at the i/p portion of the circuit as: Vs connected in series with Rs then to C1 then connected inseries to that 1.8k. Quote
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