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Engineering 43. Chp 2.8 Dependent Sources. Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. A Convention About Dependent Sources Unless Otherwise Specified The Current And Voltage Variables Are Assumed In SI Units Of Amps And Volts.
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Engineering 43 Chp 2.8DependentSources Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu
A Convention About Dependent Sources Unless Otherwise Specified The Current And Voltage Variables Are Assumed In SI Units Of Amps And Volts Ckts With Dependent Sources • For This Example The Multiplier, ,Must Have Units Of V/A or Ω
Other Dependent Sources • Three More Types of Dependent Sources • VD = VX ( a Unitless Scalar) • ID = VX ( in Units of A/V, or Siemens) • ID = IX ( a Unitless Scalar)
General Strategy Treat Dependent Sources As Regular Sources And Add One More Equation For The Controlling Variable Plan for Ckt at Rt Single Loop Circuit Use KVL To Determine Current KVL: KVL Analysis Strategy • 1-Equation, TWO-Unknowns: I1 and VA • Controlling Variable Provides the Extra Equation
Analysis Example cont. • The Dependent Source Control Equation • Substitute VA Relation Into KVL • Use Ohm’s Law to Find VO
Find VO Example a • Strategy • If VS Is Known, Then VO Can Be Determined Using Voltage Divider • To Find VS We Note That We Have A Single Node-Pair • Mark Node-A
Apply KCL (out POS) & Ohm at Node-A Example cont.1 • This Single Eqn Contains TWO Unknowns • Use The Equation For The Dependent Source Controlling Variable To Eliminate IO In Favor Of VS • Sub For IO (Be Sure Maintain Consistent Units)
Solving for VS (mult. Both sides by LCD of 6 kΩ) Example cont.2 • Now Apply The Voltage Divider Equation to Find VO
Find VO The Plan A One Loop Problem Find The Current Then Use Ohm’s Law The Dependent Source Is Just One More Voltage Source Apply KVL KVL TO THIS LOOP Example
Example cont. • Use the Dependent Source Control Equation to Eliminate VO • Solving for the Current, I • Finally, Solve for VO Using Ohm’s Law
Find Voltage Gain or Amplification KVL FET Amp Example KCL • The Plan • One Loop On The Left → KVL • One Node-pair On Right → KCL • KVL
FET Amp Example cont.1 • Ohm’s Law • Determined by a Voltage Divider • Load Resistor & vo • Apply KCL At Large Node • Then vo • And from Above have vg as a function of vi
Finally the Gain FET Amp Example cont.2 • Note That the Common-Source FET is an INVERTING Amplifier
WhiteBoard Work • Let’s Work Problem This Problem • If the OutPut Current Io is 2A • Then Find the Supply Current IS