HW8_Sol - Homework 8 Solns 1 KCL implies that the same current Iout must flow through both transistors Assuming they are all operating in

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Homework 8 Solns 1. KCL implies that the same current, I out , must flow through both transistors. Assuming they are all operating in saturation, it follows from the EKV model that , V V V c in - = κ which implies that ( 29 in c V V V - = regardless of whether the transistors are operating in weak, moderate, or strong inversion. 2. To keep the bottom transistor saturated, we must have that DSsat V V . Substituting how V depends on c V and in V into this inequality, we have that / DSsat in c V V V + . The minimum allowable c V would thus be k V V DSsat in / + . If c V were set at this value, then we must have that DSsat DSsat DSsat out V V V V 2 = + . If 0 T in V V , we have that ( 29 0 T in DSsat V V V - = . If 0 T in V V < , we have that T DSsat U V 5 = . 3. When V in is at its smallest value, I out is at its smallest value, and V will adjust itself to its highest value, which will be just far enough below V c so that the cascode transistor just passes the I out set up by the bottom transistor. Initially, the bottom transistor will be
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This note was uploaded on 09/15/2007 for the course ECE 3150 taught by Professor Spencer during the Spring '07 term at Cornell University (Engineering School).

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HW8_Sol - Homework 8 Solns 1 KCL implies that the same current Iout must flow through both transistors Assuming they are all operating in

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