hw08sol - ECE 3150 Homework 8 Solution Spring 2009 1....

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ECE 3150 Homework 8 Solution Spring 2009 1. (Cascode amplifier derivation) Use Fig. 5.29 for the MOSFET cascode amplifier considering the Early effect but ignore the body effect, (a) Use the small-signal KCL to derive R out = r o2 + A vo2 r o1 and R in2 = 1/g m2 + R L /A vo2 . Confirm this result with the intuitive observation in Fig. 5.29. (4 pts) 0 = = Vin out out out i v R At the output node, not including R L : ) ( 1 2 2 1 o m o o out out v g r v v i - + - = All current i out flows through the bottom resistor r o1 , since the ideal drain has a vgs-dependent-only current source. 1 1 1 1 o out o o o out r i v r v i = = . Plug this equation into the current equation at the output node, we have 2 2 1 1 2 1 o out o o o m out r v r r r g i = + + 1 2 2 1 2 2 2 1 2 2 2 1 ) 1 ( o vo o o o m o o o m o o out out out r A r r r g r r r g r r i v R + = + + = + + = = where A vo2 = 1 + g m2 r o2 To find R in2 , we will look upwards into v o1 , not including transistor Q 1 R in2 = 1 1 o o i v ; 2 1 1 2 1 ) ( o out o o m o r v v v g i - + - - = All of i o1 flows upward through R L L o out L out o R i v R v i 1 1 = = If we plug this back into the above equation 2 2 2 2 2 2 2 2 2 1 1 2 2 2 2 1 2 1 1 1 1 1 1 vo L m vo L o m o o m L o o o in o mb m o o L o A R g A R r g r r g R r i v R r g g v r R i + + + = + + = = + + = + The last step assumes that g m2 r o2 >> 1 These results are easily confirmed by looking at Figure 5.29. (b) If Q 1 and Q 2 are identical, derive A v = -A 0 2 R L /(R L + A 0 r o ) with A 0 defined as g m r o . (4 pts) Q 1 and Q 2 are identical, meaning that r o1 = r o2 = r o , g m1 = g m2 = g m A v = -G m R o , where G m is the small-signal transconductance for the entire amplifier and R o is the output resistance looking into the output port . In this case, all the small-signal current produced is due to a change in v in , because V BIAS is a DC large-signal voltage, meaning that the cascode transistor does not “produce” any small- 1
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signal current of its own, it just flows what was created by the input transistor, or in other words, i o1 = i o2 , and hence i o1 / v in = i o2 / v in . Therefore G m = g m R o = R out ||R L , o L L o o vo o L L o vo o out L L out o r A R R r A r A r R R r A r R R R R R 0 0 1 2 2 1 2 2 ) ( ) ( + + + + = + = , assuming that A vo2 r o1 >> r o2 and that A vo2 = 1+g m2 r o2 ≈ g m2 r o2 = g m r o = A 0 since g m2 r o2 >> 1 Then A v = -g m R o o o L L o o o L L o o m v r A R R A r A R R A r g A + - = + - = 2 (c) If Q 1 and Q 2 have different W but the same L (still both are in saturation), if we want a larger R out , which transistor should be larger? (4 pts) We know that 1 2 2 2 1 1 2 2 o o m o o o vo o out r r g r r r A r R + + = + = Since the L’s are the same for both transistors, we know that r o1 = r o2 But the W’s are different, meaning that (W/L) 1 ≠ (W/L) 2 We see that R out depends on g m2 but not g m1
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hw08sol - ECE 3150 Homework 8 Solution Spring 2009 1....

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