HW7 Solution

# 6 a the short circuit transconductance g m is g m i

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6. (a) The short-circuit transconductance G m is G m = i out v in = g m 1 . The output impedance R out is given by R out = r O 1 || 1 g m 2 || r O 2 . Therefore, the voltage gain is A v = - G m R out = - g m 1 r O 1 || 1 g m 2 || r O 2 . (b) The short-circuit transconductance G m is G m = i out v in = - g m 1 . The output impedance R out is given by R out = r O 1 || 1 g m 2 || r O 2 . Therefore, the voltage gain is A v = - G m R out = g m 1 r O 1 || 1 g m 2 || r O 2 . (c) The small-signal equivalent is shown in Figure 1. By KCL, we have v in - v X r π 1 + g m 1 ( v in - v X ) - v X r O 1 = v X R E v X = 1 r π 1 + g m 1 1 r π 1 + g m 1 + 1 r O 1 + 1 R E v in . By KCL again, we have i out = g m 1 ( v in - v x ) - v x r O 1 G m = i out v in = g m 1 R E - 1 r O 1 r π 1 1 r π 1 + g m 1 + 1 r O 1 + 1 R E . Since g m 1 = β/r π 1 1 /r π 1 , we can approximate the short-circuit transconductance G m as G m g m 1 R E g m 1 + 1 r O 1 + 1 R E . The output impedance R out is given by R out = { [1 + g m 1 ( r π 1 || R E )] r O 1 + ( r π 1 || R E ) } || r O 2 . Finally, the voltage gain A v = - G m R out . Figure 1 3

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(d) When the output is shortened, the emitter voltage of Q 1 is zero and Q 1 does not affect the output current. Therefore, the short-circuit transconductance G m is G m = g m 2 . The output impedance R out is given by R out = { [1 + g m 1 ( r π 1 || R E )] r O 1 + ( r π 1 || R E ) } || r O 2 . Finally, the voltage gain A v is A v = - G m R out = - g m 2 ( { [1 + g m 1 ( r π 1 || R E )] r O 1 + ( r π 1 || R E ) } || r O 2 ) . (e) The small-signal equivalent circuit is shown in Figure 2(a). The emitter voltage of M 2 is zero and no output current goes through M 2 (please verify this by yourself!). Then by KVL, we have - v 1 - v in = i out R S v 1 = - v in - i out R S . By KCL, we have i out = g m 1 v 1 + v 1 r O 1 = ( g m 1 + r O 1 )( - v in - i out R S ) G m = i out v in = - g m 1 + 1 r O 1 1 + R S g m 1 + 1 r O 1 .

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