1. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 1 Today: Ideal versus Real elements: Models for real elements Non-ideal voltages sources, real voltmeters, ammeters Series and parallel capacitors Charge sharing among capacitors, the paradox

2. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 2 MODELING NON-IDEAL VOLTAGE AND CURRENT SOURCES V BB i +v+v + Real battery Voltage drops if large current i v V BB A model of a device is a collection of ideal circuit elements that has the same I vs V characteristic as the actual (real) device (and is therefore equivalent). What combination of voltage sources, current sources and resistors has this I-V characteristic? Example: A real battery Two circuits are equivalent if they have the same I-V characteristic.

3. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 3 MODELING NON-IDEAL VOLTAGE AND CURRENT SOURCES Current-voltage characteristic of a real battery Approximation over range where i > 0: v = V BB  iR i = (V BB  v)/R (straight line with slope of -1/R) V BB i +v+v + Real battery Voltage drops if large current i v V BB Model of battery +-+- +v+v i R Simple resistor in series with ideal voltage source “models” real battery V BB

4. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 4 REAL VOLTMETERS Concept of “Loading” as Application of Parallel Resistors How is voltage measured? Modern answer: Digital multimeter (DMM) Problem: Connecting leads from a real voltmeter across two nodes changes the circuit. The voltmeter may be modeled by an ideal voltmeter (open circuit) in parallel with a resistance: “voltmeter input resistance,” R in. Typical value: 10 M  Real Voltmeter Ideal Voltmeter R in Model

5. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 5 REAL VOLTMETERS Concept of “Loading” as Application of Parallel Resistors Example:  + V SS R 1 R 2 - +  V 2 But if a 1% error Computation of voltage (uses ideal Voltmeter) - + V SS R 1 R 2 R in Measurement of voltage (including loading by real VM) - +

6. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 6 MEASURING CURRENT Insert DMM (in current measurement mode) into circuit. But ammeters disturb the circuit. Ammeters are characterized by their “ammeter input resistance,” R in. Ideally this should be very low. Typical value 1 . Real Ammeter Ideal Ammeter R in ? Model

7. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 7 MEASURING CURRENT Potential measurement error due to non-zero input resistance: R in _ + V I meas R 1 R 2 ammeter with ammeter _ + V I R 1 R 2 undisturbed circuit Example: V = 1 V: R1 + R2 = 1 K , Rin = 1 

8. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 8 IDEAL AND NON-IDEAL METERS DMM amps MODEL OF REAL DIGITAL AMMETER C + R in Note: R in may depend on range Note: R in usually depends on current range R in typically > 10 M  R in typically < 1  C + IDEAL DMM amps C + IDEAL DMM volts DMM volts MODEL OF REAL DIGITAL VOLTMETER C + R in

9. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 9 CAPACITORS IN SERIES Clearly, C1C1 V1V1 i(t) C2C2 | ( V2V2 +  +  CeqCeq i(t) | ( V eq +  Equivalent to

10. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 10 CAPACITORS IN PARALLEL C1C1 i(t) C2C2 | ( +  V Equivalent capacitance defined by C eq i(t) | ( +  V(t) Clearly,

11. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 11 CHARGE REDISTRIBUTION Pre-charged capacitor C A is connected to C B at t = 0 Let C A = C B = 1mF C A C B t =0 Find v A (t = 0 + ). From conservation of charge: Q A (t>0) = Q B (t>0) = ½ Q A (0) Thus v A (t>0) = ½ V From conservation of energy: ½ C A v A 2 (t>0) = ½ C B v B 2 (t>0) =½ [½ C A v A 2 (0)] so v A 2 (t>0) = [½ V A 2 (0)] Or v A 2 (t>0) =½ These answers are inconsistent. What is wrong with this circuit? Hint : We set up a paradox : Capacitor V jumps (infinite current so we dare not ingore the wire resistance) Initial Voltage = 1V Initial Voltage = 0

12. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 12 DIGITAL CIRCUIT EXAMPLE (Memory cell is read like this in DRAM) For simplicity, let C C = C B. If V C = V 0, t < 0. Find V C (t), i(t), energy dissipated in R. t/  V C / V 0 t/  +  initially uncharged -

13. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 13 ENERGY DISSIPATION IN R TWO FACTS: (1) 1/2 of initial E lost (for C C = C B ) (2) Independent of R!

14. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 14 Simple Proof of Energy Division For simplicity, let C C = C B. If V C = V 0, t < 0. Find V C (t), i(t), energy dissipated in R. Thus initial Energy Stored in Capacitors is 1/2C C V 0 2 + 0 Final Energy is 1/4 C C V 0 2 so clearly the resistor dissipated the rest, independent of the value of the resistance. So even if the resistance is very, very, very small, it still dissipates half the energy in this example (where C C =C B ). +  initially uncharged -

15. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 15 THE BASIC INDUCTOR CIRCUIT +  v i (t)R L vXvX KVL: V1V1 v(t) t t=0 i t 0 t 0  =L/R Solution has same form as RC!

16. EECS 40 Fall 2002 Lecture 13 Copyright, Regents University of California S. Ross and W. G. Oldham 16 TRANSIENTS IN SINGLE-INDUCTOR OR SINGLE- CAPACITOR CIRCUITS - THE EASY WAY 1) Find Resistance seen from terminals of L or C (short voltage sources, open current sources). 2) The circuit time constant is L/R or RC (for every node, every current, every voltage). 3) Use initial conditions and inductor/capacitor rules to find initial values of all transient variables. (Capacitor voltage and inductor current must be continuous.) 4) Find t=  value of all variables by setting all time derivatives to zero. 5) Sketch the time-behavior of all transient variables, based on initial and final values and known time constant. 6) Write the equation for each transient variable by inspection.