Unformatted text preview: Typically we do this at the node that the load is connected to. The load is associated with the voltage Vout. 2. We need to follow Thévenin / superposition rules when it comes to independent sources. We short the 5 V source and then calculate the equivalent resistance. We see two parallel paths to ground: 400 ohms and 2 50 ohm resistors in series. 3. Therefore g ³´µ = (50 Ω + 50 Ω)400 Ω = 80 Ω 4. If there is no load, Vout will equal 4 V. This is the max value of Vout! What happens to Vout as g ² decreases? From here it can be seen the ideal load is no load! Or a very large load.Unfortunately this is an unavoidable problem....
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This note was uploaded on 09/09/2011 for the course EECS 40 taught by Professor Changhasnain during the Summer '08 term at Berkeley.
 Summer '08
 ChangHasnain

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