MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.viviVI00a v o l t a g e s o u r c ea c u r r e n t s o u r c ev o l t a g e i s i n d e p e n d e n t o ft h e c u r r e n t s u p p l i e d t o t h el o a d .c u r r e n t i s i n d e p e n d e n t o ft h e vo l t a g e r e q u i r e d t om a i n t a i n t h e c u r r e n t .+-V ( t ) s y s t e mV+I ( t )Is y s t e m+-b a t t e r yRLivRLvv� Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 81 1 Electrical System Modeling (continued) Modeling Real Sources: Up to this point we have considered only ideal voltage and current sources: However, practical (real) sources do not have an ideal characteristic – they are power limited. For example we may think of a battery as an ideal voltage source - say with a 9 volt output, but if we were to measure the terminal voltage under increasing load currents we would find that at high currents the voltage decreases in a nonlinear manner: 1copyright cD.Rowell 2008 8–1+-s y s t e mviRoss o u r c e m o d e lv = V - i RosviV0iR = = d vd ia c t u a l s o u r c ec h a r a c t e r i s t i covio cs co cs cT h e v e n i n s o u r c e m o d e lviV0" l i n e a r " r e g i o no p e n - c i r c u i t v o l t a g ew h e n i = 0 ( R = ¥) .s h o r t - c i r c u i t c u r r e n tw h e n v = 0 ( R = 0 ) .F o r a r e a l s o u r c e , t h e v o l t a g ed e c r e a s e s a s t h e c u r r e n ti n c r e a s e sThe voltage–current relationship is determined by the electrochemical reaction within the battery. It may be possible to define a region of operation at low currents in which there is an approximately linear relationship between voltage and current v(i) ≈ voc − Rsi where voc is the open-circuit voltage (when i = 0), and Ro is a resistance. How can we model this linear part of the characteristic? One way to do this is to model the real source(the battery) as an ideal voltage source Vs in series with a resistor Ro (the value of which is found experimentally). The voltage drop vRo = iRo across the resistor accounts for the “droop” in the source characteristic. This is known as the Th´evenin equivalent source model. 8–2i n d i c a t e d v o l t a g eiRLvo u t1 . 2 3 41 . 2 3 4RLa c t u a lv o l t a g ea m p l i f i e r5 0 Wvo u tv1v o l t m e t e rF u n c t i o n G e n e r a t o rv = o u t R R + 5 0 LLv1T h e m e t e r i s c a l i b r a t e d t o r e a dc o r r e c t l y o n l y w h e n a 5 0 W l o a dr e s i s t o r i s c o n n e c t e d !Note that (1) The open-circuit (i = 0) voltage is Vs. (2) The short-circuit (v = 0) current is isc = Vs/R. Example 1 Electronic instruments are often designed to have a specific output impedance (resistance) Ro. For example when the Tektronix function generator in the 2.004 laboratory is connected to an arbitrary load resistance RL the measured output voltage vout will not necessarily be equal to the value set on the front panel: (a) If RL = ∞ (no load connected) the true voltage at the terminals is twice the indicated value. (b) If RL = 0 (short circuit) the actual voltage is zero but the indicated voltage is unchanged. (c) If RL = 50 Ω the indicated and measured output voltages are the same. Why? Answer: The function generator is specified as having an output impedance of 50Ω. In fact, if you look at the circuit schematic, you will find that there is a 50Ω resistor in series with the output terminal: 8–3s y s t e mvRoNN o r t o n s o u r c e m o d e l( a )NlRoIn the lab we have 50Ω terminator resistors connected across the output terminals so that the voltage reads correctly. A second approach to modeling the real source characteristic viV0iR = = d vd iovio cs co cs cv = V - i Ro cLLois to use a current source IN in parallel with the resistor Ro This is known as the Norton equivalent source model. If no load resistor is connected (open-circuit) vos = IN Ro. The short-circuit current is isc = IN . If the load is a resistor RL RLRo RL v/i v = = voc = RL + Ro RL + Ro vı + Ro and rearranging v = voc − iRL. The Norton and Th´evenin source models are equivalent and cannot be dis-tinguished by measurements at the terminals. 8–4R = 5 WoV = 2 0 0 vt h+-Example 2 Find Thev´Thev´enin Model: Measurements on an industrial dc power supply showed it to have an open-circuit voltage of 200 v, but when 10 Ω resistor is connected, the voltage drops to 133v. enin and Norton equivalent source models. RoRLVt h+-INRoRLR = 1 0 WLRoV = 2 0 0 vt h+-Vth = voc = 200 v. Using the voltage divider relationship 10 vo = vRL = = 133volts 10 + Ro which gives Ro = 5Ω. The Thev´enin model is: Norton Model: From above, the short-circuit current is Vth 200 isc = = = 40 amps Ro 5 and the Norton model is: 8–5R = 5 WNI = 4 0
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