Lecture Week 5 Quiz #2 Ohm s Law Homework Power Review Shorthand Notation Active Components Ideal Op-amps
Quiz 2 Ohm s Law (20 pts.) Please clear desks and turn off phones and put them in back packs You need a pencil, straight edge and calculator 10 minutes Keep eyes on your own paper Follow the same format as for homework
Homework- P10 (a) Determine the circuit current for the simplified circuit in P9 if R1 R4 are equal to 100 Ω and Vn is equal to 9 V, (b) Use ohm s law to confirm the voltage drop at V A, and (c) Use the expression for P9 part (b) to determine V A using the voltage divider method. Show every step, units, and unit conversions for full credit. Pick Random Group to Turn in HW Analyzed by Professor on Board
Power and Passive Sign Convention Power can be supplied/delivered to a circuit or it can be absorbed by a circuit component. POWER GENERATED - + - + - POWER CONSUMMED + If power is supplied by a circuit component (battery), the power is negative - CURRENT FLOWING FROM TO + If power is absorbed by a circuit component, the power is positive CURRENT FLOWING FROM + TO
Sources: Engineering Circuit Analysis, by Hayt, Kemmerly and Durbin,2007 and Circuits by Ulaby and Maharbiz, 2010 While voltage is related to Potential Energy and is equal to V = IR The Power is equal to the voltage across a device times the current entering through its positive terminal. or Power and Passive Sign Convention When one joule of energy is required to transfer one coulomb of charge through a device in one second, this produces a rate of energy transfer of one Watt. P = IV The algebraic sum of all power in a circuit must equal zero based on the LAW OF CONSERVATION OF POWER.
Power and Passive Sign Convention Visualizing Power
Calculating Power Unit Conversions V 2 R = V2 J VC A V C As Ws J = W (Watts) I V = A V J VC C As Ws J = W (Watts) R I 2 = A2 V A C As J VC Ws J = W (Watts) V = J C 1 Joule is equal to the energy associated with moving 1 coulomb charge through 1 volt of potential Image Source: http://www.sengpielaudio.com/formulawheelelectronics.gif
Exercise: Calculating Power Determine the power consumed by R4 for homework P10. For P10, R1 R4 are equal to 100 Ω and Vn is equal to 9 V. You have already determined V A and I 4.
Recall Voltage Divider Method: Can help analyze circuits The voltage divider relates the total voltage to the potential drop across one of the resistors. R eq = R 1 + R 2 I I = V s R eq = V s R 1 +R 2 V s V 1 = I R 1 = V s R 1 + R 2 R 1 = Vs R 1 R 1 + R 2 v i = v s R i R eq V 2 = I R 2 = V s R 1 + R 2 R 2 = Vs R 2 R 1 + R 2 VOLTAGE DIVIDER METHOD VERY POWERFUL TOOL USED TO ANALYZE CIRCUITS!!!
Equivalent Circuit and Shorthand Notation Req = R1 + R2 + R3
Equivalent Circuit and Shorthand Notation Req = R1 R2 R3
Equivalent Circuit and Shorthand Notation R = R2 R3 LEARN TO RECOGNIZE RESISTORS CAN BE COMBINED IN SERIES OR IN PARALLEL! Req = R1 + R2 R3 + R4
Example: Equivalent Circuit and Shorthand Notation Simplify using parallel/series and voltage divider to determine V A and I 3. SKETCH the equivalent circuit and label resistor values using SHORTHAND NOTATION.
Exercise: Equivalent Circuit and Shorthand Notation Simplify using parallel/series and voltage divider to determine V A, V B and I 2. SKETCH the equivalent circuit and label resistor values using SHORTHAND NOTATION. Note: before you determine V B, you need to determine V A. To be completed on the board
Solutions: Voltage Divider and Shorthand Notation
Active Components Unlike passive components, active components require power (their own power connected to the component) to make them work. Examples to follow
Active Components Examples for Intro Instrumentation Amplifier: AD623 Op-Amp: LMC6484 AND Gate: SN74ATC081
Active Components Examples: Instrumentation Amplifier AD623 WHAT S THE GAIN IF USE RG = 100? R G =100 kω/(g 1) The chip provides an output which is simply, Vout=Vref+ G(Vin+ Vin )
Active Components Examples: Instrumentation Amplifier AD623 RG INPUT 1 5V Ch1+ INPUT 2 Ch1- Grd 2.5 V Ref Grd R G =100 kω/(g 1) The chip provides an output, Vout=Vref+ G(Vin+ Vin )
5 V Active Components Examples: Op-Amp: LMC6484 Ground - + + - 4 Op-Amps - 1 inverting input/op-amp - 1 non-inverting input/op-amp - 1 output/op-amp - + + -
Active Components Examples: AND Gates Channel DIO 0 Channel DIO 1 Output 5 V 4 AND Gates - 2 inputs/gate - 1 output/gate
Basic Op-Amp Theory An Operational Amplifier (Op-Amp) is a versatile device capable of: Amplifying signals Differentiating signals Integrating signals Inverting signals Adding/Subtracting signals It is an extremely important device for analog signal processing, and it is used everywhere in electronics.
Inside an Op-Amp The insides of an Op-Amp consists of active and passive devices such as transistors, resistors and sometimes capacitors. For EE1305/1105 we do not analyze or worry about the internal components and behavior of an Op-Amp. You will study Op-Amps in depth in EE3338 and EE3340 (Electronics I and II)
Ideal Op-Amp Behavior For this class we will only deal with external Op-Amp behavior. We will also only cover ideal Op- Amp Theory It is important to note that by ideal, we mean that the real behavior of an Op-Amp will not perfectly match your theorical results, but it is pretty close.
Ideal Op-Amp Behavior An ideal Op-Amp has the following characteristics: I in+ = 0 I in- = 0 R in = R out = 0 V in+ = V in- A ol = Current at positive input is zero Current at negative input is zero Input resistance is infinite, that is why input currents are zero Output resistance is zero Voltage at positive input equals Voltage at negative input Open loop gain is infinite
Op-Amp Linear Range The Op-Amp can only amplify signals up to the voltage supply range. This means that if we power the Op-Amp with a 5V power supply, WE CANNOT get more than 5V at the output. If we measure (with oscilloscope Ch1 and Ch2) relative to 2.5 V, then we cannot get more than (5V 2.5V) or 2.5 V. If the input exceeds the voltage supply we enter the Saturation Region, where the output voltage signal will not change or will get heavily distorted. See negative and positive saturation regions.
LMC6484 14 pin Op-Amp For this class we will be using the LMC6484 Quad 14 pin Op-Amp. It contains 4 independent Op-Amps. As you can observe from the image, all Op-Amps share the same power supply given by V+ and V-
LMC6484 Op-Amp: 4 Op-Amps Location of the 4 Op-Amps on your LMC6484 chip.
LMC6484 Op-Amp Power Connections Notice where the voltage connections are (V+ will be connected to 5V, and V- will be connected to ground). Therefore, orient the Op-Amp so you can connect easily to 5 V and ground.
LMC6484 Op-Amp: Pins for each Op-Amp + - 4 Op-Amps - 1 inverting input - 1 non-inverting input - 1 output/op-amp - + Ground - + + - 5 V
LMC6484 Op-Amp Circuit Design Notice that for each Op- Amp, the outputs are all at the corners of the chip. It does not matter which Op-Amp you use, as long as it is practical for your circuit design.
LMC6484 Op-Amp: Inverting and Non- Inverting Pins - + Notice where the inverting (-) and non-inverting (+) inputs are for each Op- Amp on the chip, and that the inverting (-) input is always next to the output pins. + - - + + -
Breadboard + Integrated Circuits (IC) Remember that we have a separation in the middle of the breadboard. The breadboard is purposely built this way so that we can place Integrated Circuits (ICs) easily. ICS WILL ALWAYS BE CONNECTED TO A BREADBOARD THIS WAY
LMC6484 Op-Amp Power Connections An Op-Amp is an active device since it needs a power supply to operate. The Op-Amp has 2 inputs and 1 output. How many pins are on the Op-Amp chip that we use? - 3 pins per Op-Amp, 12 pins total, plus 2 pins for +Vss and Vss, so 14 total pins Image source: http://elektronikadasar.info/wp-content/uploads/2013/01/op-amp.png
Can analyze with Complex Impedance Can analyze with KCL/KVL Filter OpAmp: Filters the signal: low, high or band pass LMC6484 Op-Amp: Types of Op-Amps Covered in This Course Voltage Follower or Buffer: Sets the voltage at that point in the circuit Inverting Amplifier: Inverts the signal horizontally, and amplifies the signal Non-Inverting Amplifier: Amplifies the signal
LMC6484 Op-Amp: Voltage Follower of Buffer No Current Voltage = Input Voltage Input Voltage Output Voltage Input Voltage Used to isolate two circuits
LMC6484 Op-Amp: Inverting Amplifier Power Supply (Vcc not shown) R2 The inputs (non-inverting (+) and inverting (-) to the opamp draw no current (i.e. i = 0 into both inputs). The two input voltages values are equal to each other (Vn = Vp). Ex. If Vref is ground, then Vn = Vp = 0 V. Vin R1 i=0 Vn i=0 + Vp - Vout One output, Vout. OpAmp can saturate if Vout is greater than Vcc!!!! - more about this later Vref
Inverting Amplifier LMC6484 Op-Amp: Inverting Amplifier WHAT IS THE APPROXIMATE RELATIONSHIP BETWEEN R1 AND R2? V out V in = R 2 R 1 where the Gain = R 2 R 1
Homework- Due 10/2 (Najera), Due 10/5 (Martinez-Acosta) P11 and P12
What s Next in Week 6? Will introduce LECTURE Quiz 3 Voltage Divider Kirchhoff s Voltage and Current Laws (KCL and KVL) LAB Module I Angle Sensor Please bring laptops to all lectures and labs.
Questions?