Lecture 3 Resistors. A note on charge carriers Physics origin of resistance Ohm s Law Power dissipation in a resistor Combinations of resistors

Transcription

1 Lecture 3: esistance is futile ECEN 400 ntroduction to Analog and Digital Electronics Lecture 3 esistors A note on charge carriers Physics origin of resistance Ohm s Law Power dissipation in a resistor Combinations of resistors Series voltage divider The potentiometer Parallel current divider nput and output resistance Clicker quizzes An alternative to the hydraulics analogy obert. McLeod, University of Colorado 5

2 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics Charge carriers Due to an unfortunate choice by Ben Franklin, electrons, which typically carry charge, have the negative sign. This leads to a physically correct but confusing physical view of current: Negatively charged electrons moving opposite direction. Current The physically incorrect but perfectly acceptable view used by most electrical engineers is instead, this: Positively charged mystery particles moving same direction. Current recommend you think of positive charge carriers traveling in the direction of the current in this class. obert. McLeod, University of Colorado 6

3 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics The esistor Current Package + Potential - Physics Drude model Positive charge Symbol Charge path Points of collision Note how symbol is similar to charge path. obert. McLeod, University of Colorado 7

4 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics Ohm s Law The definition of resistance Friction of the charge carrier travelling through the solid slows it down. An empirical observation that holds for many materials under a large range of conditions is Ohm s Law: G esistance in ohms [Ω] is proportionality between and. Occasionally useful to use the inverse quantity, called Conductance, G, measured in mhos [ ] (Formerly. Now siemens [S] is recommended, but is not as much fun). obert. McLeod, University of Colorado 8 Ω

5 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics Power dissipated in a resistor There is a potential drop of 0 0 J/C across the resistor. 5 A, so 5 C of charge is experiencing this loss of energy per second (AC/s) The amount of electrical energy lost to heat in the resistor per unit time is thus P [ J ] [ J ] [ C W ] 50[ W] s esistors have maximum power ratings: C s obert. McLeod, University of Colorado 9

6 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics Combining Ohm s Law and power equation P P ( ) ( ) The quadratic dependence of dissipated power on voltage is a major motivation for low-voltage logic families. Power on the diagram: Providing Dissipating P P [Watts] 4 ½ Dissipating Providing obert. McLeod, University of Colorado 30

7 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics esistors in series Equivalent and voltage division Q: What is voltage across individual resistors in series? Step : Find equivalent resistance Eq + [ Ω] + 8 [ Ω] 0 [ Ω] Step : Find common current via Ohm s law Source Eq [ ] 0 [ Ω] [ A] 0 Step 3: Find voltage drop on each resistor via Ohm s law [ A] [ Ω] [ ], [ A] 8 [ Ω] 8 [ ] n general, combine all three to get voltage divider equations ( ) source Eq obert. McLeod, University of Colorado 3 source +

8 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics Package Potentiometer A variable voltage divider Operation A W B Symbol Equivalent circuit A B W + AB AW WB obert. McLeod, University of Colorado 3

9 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics esistors in parallel Equivalent and current division Q: What is the current through individual resistors in parallel? Step : Find equivalent resistance or Eq ( ) + ( + ).6 [ Ω] Step : Find common current via Ohm s law Source Eq 0 [ ].6 [ Ω] 6.5 [ A] 0 G Eq G + G Step 3: Find current through each resistor via Ohm s law [ ] [ Ω] 5 [ A ], 0 [ ] 8 [ Ω].5[ A] source 0 source n general, combine all three to get current divider equations source ( ) + Eq obert. McLeod, University of Colorado 33 G G + G

10 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics Bonus nput and output resistance You design a great voltage divider to produce 0 from your 5 power supply. but when you hook up the next component, the voltage drops to 6. The problem is that the load resistance is too low, pulling current from your voltage divider. A high load resistance works well. obert. McLeod, University of Colorado 34

11 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics Quiz 3. Simple Ohm s Law Q: The voltage across and current through are: A: 5 m, 5 ma B: 5 m, 5 A C: 5, 5 ka D: 5, 5 ma E: 5 k, 5 A / 5/000 obert. McLeod, University of Colorado 35

12 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics Quiz 3. Power Q: The battery burns out on your night-light. You replace the battery with one whose voltage is twice that of the previous battery. The optical power emitted by the light bulb A: is a quarter what it was before. B: is half what it was before. C: is the same as it was before. D: is twice what it was before. E: is 4x what it was before. P obert. McLeod, University of Colorado 36

13 Lecture 3: esistance ECEN 400 ntroduction to Analog and Digital Electronics Quiz 3.3 Parallel resistors Q: A resistor of unknown resistance is added in parallel with. The current drawn from the battery A: goes down. B: goes down or stays the same. C: goes up or stays the same. D: goes up. E: There is not enough info to say. Eq ( ) + so t could stay the same if is infinite. Yes, it s a fine point, but a better answer than D. obert. McLeod, University of Colorado 37 Eq