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Step of 5 9.083P The base emitter junction area of one transistor is twice the other. The emitter bias current splits between the two transistors according to their area ratio. Consider as the larger device. Find in terms of Obtain in terms of Step of 5 Find the expression for Since, The modified expression for Vc is, = 1 Step of 5 The expression for the differential gain is, = Find the dc offset voltage, Ad = = Simplify further. 16.7 mV Thus, the input voltage that would restore current balance to the differential pair in small-signal is 16.7 mV mV Hence, in small signal analysis, if a dc signal of 16.7 mV is applied as then it would restore the current balance in the transistors and reduce Vc to zero. Step of 5 Consider the large-signal analysis. The expression for the current is, The expression for the current is, Find the ratio Step of 5 Obtain the value of VT VT 2 Simplify further. = Thus, the input voltage that would restore current balance to the differential pair in small-signal is 17.3 17.3 mV Hence, similar results are obtained in both large signal analysis and small signal analysis.