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Transcript
Example First step: Find iB corresponding to the Q-point and the two extreme values of the applied voltage , i.e. vin = +0.4 V, and -0.4 V The values come out to be IBQ = 25 µA, IB, max = 35 and IB, min = 15 µA, respectively. 1 Example Thus, from the Q-point of the o/p circuit, we have, ICQ = 2.5 mA, and VCEQ = 5V corresponding to iB = 25 µA . We also have, ICQ,max = 3.5 mA, and VCEQ, min = 3 V corresponding to iB, max = 25 µA . ICQ,min = 3.5 mA, and VCEQ, max = 7 V corresponding to iB, min = 15 µA . Thus, the peak-to-peak value of ac component of VCE = 4 V, and peak-to-peak value of vin = 0.8 V Thus the gain is 5, but with a negative sign, as seen from Fig. 4.13. Confirm that when the base voltage VB is higher the o/p voltage is lower. 2 Distortion – Cutoff and Saturation regions Assume that in the previous example the vin has amplitude of 1.2 V. Find the output. When IB reduces to zero, the IC also reduces to zero, and the voltage VCE becomes constant at the maximum value (10 V here). This is called CUTOFF region. When IB increases beyond a certain limit, the IC fails to increase any more, and the voltage VCE becomes constant at a minimum value, usually of 0.2 V. This is called the SATURATION region. 3 BJT large signal models 1 – the three regions of operation VCE = 0.2 V IB = 0, IC = 0 4 BJT large signal models 2 5