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Step of 6 8.050P Refer to Figure 7.4(a) in the text book for the common source amplifier circuit. Determine the overdrive voltage for transistor Here, is the early voltage of the transistor L is the length of the channel is the voltage gain of the amplifier Substitute 5x V/m for m for L and 10 V/V for = 10 = 0.18 V Step of 6 Determine the overdrive voltage of the transistor, A Here, is the early voltage of the transistor L is the length of the channel is the voltage gain of the amplifier Substitute V/m for m for L and 10 V/V for = 10 0.216 V Thus, the overdrive voltages of the transistors in the design, and are, 0.18 V and 0.216 V Step of 6 Determine the transconductances, 8ml and 8m2 when I = 50 8ml = 0.18 = 0.56 mA/V = 0.216 = 0.463 V Thus, the transconductances, 8ml and are, 0.56 mA/V and 0.463 mA/V Step of 6 Write the formula for transconductance, 8m for transistor 8m W W Substitute 0.56x10⁻³ A/V for for and A for W = 2 x387 50x 10⁻⁶ =8.1 W Thus, the ratio of the transistor is 8. L Step of 6 Write the formula for transconductance, 8m for transistor 8m W Substitute A/V for 86x10 for k', and A for REF W = 2 x86 50x 10⁻⁶ W Thus, the ratio of the transistor is 25 L Step of 6 The expression for the gain is, As gain is proportional to length, if the gain is increased by a factor of 2, L should also be increased by a factor of 2 Determine the length of the channel. 0.36 Thus, the required channel length for gain is increased by a factor of 2 is 0.36 As the width remains constant and the length is increased by a factor of 2 the total gate area is also increased by a factor of 2.