Document Sample

Resistance Characteristic Remember what happened when we plotted the current flowing through a resistor versus the voltage across the resistor, while varying the voltage? We got a straight line, passing through the origin (If VR = 0, then IR = 0), with slope = 1/R. This is called the characteristic curve of the resistor. IIR slope 1 R IR + + E R VR - VR - VR E Resistance Characteristic The slope-intercept form of the equation for a line is y = mx + b, where y is the vertical coordinate, x is the horizontal coordinate, m is the slope and b is the y-intercept. In the case of the characteristic curve of a resistor R, m = 1/R, b = 0 (because the line passes through the origin), y = IR, and x = VR The resulting equation is: IIR The I-V characteristic 1 maps a particular slope 1 R I R VR 0 voltage to a particular current. R or, VR IR VR R where VR and IR are in Volts and Amperes, respectively, and R is the resistance in ohms. Resistance Characteristic Intuitively, this makes sense: Increasing the battery Voltage E increases VR (according to KVL), and is like increasing the pressure at the inlet to a pipe. Increasing pressure increases the flow rate, which is analogous to current IR. Reducing the resistance is analogous to increasing the pipe’s diameter, which also increases the flow rate. IIR slope 1 R Increasing Voltage IR increases current + + E R VR - - VR E VR IR R Ohm’s Law VR The equation IR R is called Ohm’s Law, and it’s true for any resistance: The resistance of an ideal resistor, the dynamic resistance (the slope of the characteristic curve at a particular point) of an element with a nonlinear characteristic curve (e.g., a diode or transistor), or the resistance of a wire (which we’d rather IIR was zero). if we know R and IR, but not slope 1 VR, we can solve the equation R VR IR R for VR: VR I R R VR if we know VR and IR, but not R, we can solve for R: VR R IR Ohm’s Law Let’s take a simple example: A battery (Voltage = E) in series with a resistor R1. KVL tells us that E = V1, and KCL tells us that the same current, I, flows through both the battery and the resistor. Ohms law tells us V V1 IR1 or I 1 R + + E R1 V1 - I - Ohm’s Law Let’s add another resistor in series, as shown. Now, E V2 V1 E IR1 IR2 E I R1 R2 E So, I R2 R1 R2 and + V - E 2 R R R1 V1 IR1 + + 1 2 E R1 V1 - R1 - V1 E R1 R2 Ohm’s Law Turning our attention to R2, E R R R2 V2 IR2 1 2 R2 V1 E R2 R1 R2 + V - 2 Ohm’s law, KVL and KCL + + allowed us to fully analyze this E R1 V1 circuit. We were able to find - - the current flowing around the loop, and the Voltages across each of the resistors. Nonlinear Resistance The characteristic of an ideal resistor is a straight line. In other words, a resistor has a linear I-V characteristic. Some devices, such as transistors and diodes, have nonlinear I-V characteristics. This means plotting current versus voltage does not result in a straight line. diode II resistor characteristic characteristic reverse bias Forward Bias The diode’s characteristic can V be divided into three regions, reverse each of which is approximately breakdown linear: reverse breakdown, reverse bias, and forward bias. Nonlinear Resistance Pick a voltage in the forward-bias region. Now, if the voltage is increased by a small amount, the current increases by a large amount. If the Voltage is reduced slightly (but stays in the forward-bias region) the current diminishes by a small amount. Within the forward-bias region, the characteristic can be approximated by a line with a large slope. II Recall that the slope of the characteristic is 1/R, so a large (steep) slope means small V resistance. We can say that a linear forward-biased diode has a approximation to small (ideally, zero) dynamic diode in forward- resistance. In other words, the bias region forward-biased diode conducts current easily. Nonlinear Resistance Next, consider a point on the diode characteristic in the reverse-bias region. A large increase or decrease in voltage (but not so large as to leave the reverse-bias region) results in a very small increas or decrease in current. The line approximating the characteristic in this region has very small slope, so the dynamic resistance of a reverse- biased diode is very large – II ideally infinite, like an open circuit. V linear approximation to diode in reverse- bias region Nonlinear Resistance Finally, consider a point on the diode characteristic in the reverse- breakdown region. A small increase or decrease in voltage (staying within the reverse-breakdown region ) results in a large increase or decrease in current. The line approximating the characteristic in this region has large slope, so the dynamic resistance of a diode in reverse- breakdown is very small. II V linear approximation to diode in reverse- breakdown region Nonlinear Resistance Finally, consider a point on the diode characteristic in the reverse- breakdown region. A small increase or decrease in voltage (staying within the reverse-breakdown region ) results in a large increase or decrease in current. The line approximating the characteristic in this region has large slope, so the dynamic resistance of a diode in reverse- breakdown is very small. II V linear approximation to diode in reverse- breakdown region Nonlinear Resistance Here’s an example. We’d like to light an LED. We can increase the LED’s brightness by increasing the current flowing through it, until the current becomes so large it destroys the LED. We need to make sure the current is great enough to make the LED as bright as we want, but not so great that the LED destroys itself. W do this with a resistor in series with the LED. II VR + - R I 12V + + VLED VLED - - 2.5 V Nonlinear Resistance The LED’s I-V characteristic shows that its forward-bias voltage is about 2.5 V. The LED must be forward biased to light the LED. Only a very large (large enough to destroy the LED) increase in current could produce an increase beyond 2.6 V or so, so it’s a reasonable approximation to say VLED 2.5 V II VR + - R I 12V + + VLED VLED - - 2.5 V Nonlinear Resistance Examining the manufacturer’s data sheet, we find that an LED current of 20 mA should make the LED sufficiently bright. Applying KVL around the loop, 12 V VR VLED 12 V VR 2.5 V 9.5 V VR II VR + - R I 12V + + VLED VLED - - 2.5 V Nonlinear Resistance Applying Ohm’s law, 9.5 V VR IR 475 is not a standard VR 9. 5 V resistor value, but 470W R 475 I 20 mA and 510 are. II VR + - R I 12V + + VLED VLED - - 2.5 V Power Rule We’ve already seen that the power consumed by a circuit element (e.g., a resistor) is given P IV Recall that power is the rate at which energy is converted from one form to another. For example, a 1 Hp motor converts electrical energy to kinetic energy at a rate of 746 J/sec., or 746 Watts (1 Hp = 746 W). In other words, if K represents energy and t = time, K P I + t + Current is the rate of flow of charge, so if E R VR Q = charge and I = current, - Q - I t unit charge: and Voltage is energy per K V Q Power Rule So we can prove P = IV, the power rule: Q K K P IV P t Q T Resistive elements consume electrical energy by converting it to heat. This is called dissipation, because as heat is produced it is radiated away to the surrounding. The faster heat is consumed (that is, the faster electrical energy is converted to heat, which means greater power), the faster it must be radiated away. The only way (aside from increasing the surface area of I the component) to make it radiate heat faster is + + for the temperature to rise, but if the temperature E R VR rises too much, damage can result. All - - components have a maximum power dissipation rating, which must not be exceeded. Power Rule Consider our simple LED example. We found previously that VLED 2.5 V I 20 mA. PLED 20 mA. 2.5 V 50 mW So The LED’s power dissipation rating had better be at least 50 mW. We also found that VR = 9.5 V, so the VR + - resistor dissipates R I PR 20 mA. 9.5 V 190 mW 12V + + And the power delivered by the battery is VLED - - Pbattery 190 mW 50 mW 240 mW which can also be found like this: Pbattery 12 V 20 mA 240 mW Power Rule We’ve already seen that the power consumed by a circuit element (e.g., a resistor) is given P IV In other words, the power consumed by an element is equal to the product of the voltage drop across it and the current flowing through it. For the resistor shown below, PR IVR Ohm’s law tells us that I VR IR + + Substituting, E PR I IR I 2 R R VR - - Now we have another way of calculating the power consumed by a resistor: It’s the product of the resistance and the square of the current through the resistor. Power Rule Another form of the power rule is found as follows: Start with the basic power rule: P IV R R If we know VR and R, we can solve Ohm’s law for I: V I R R Substituting, VR2 PR VR VR I R R + + E R We now have yet another way of calculating VR - the power consumed by a resistor: It’s the - quotient of the square of the voltage across the resistor divided by the resistance. Power Ratings Through-hole-mount resistors, the kind we use in the lab, are available with power ratings of 1/8 W, ¼ W, ½ W, 1 W, and 2 W. Suppose we have a 470 resistor (a standard value) rated at ¼ W. What is the maximum voltage it can drop? V2 P V 2 Pm axR R Vm ax PR 0.25 W 470 I Vm ax 10.84 V + + E R VR - - A BJT Circuit Ohm’s law tells us Let’s look at the 3V + example in the book, RC 300 with a few 10 mA 3V RC - differences. We see + 10 mA that IC = 10 mA, and 2V LED IB = 430 mA. KCL - gives us 430 mA I E I C I B 10 .47 mA VRB + + - + The definition of 4V 9V + b gives us RB - + 0.7 V - - 10K I C bI B 5V I C 10 mA - b 23.3 I B 430 mA A BJT Circuit Let’s let b increase + to 40. The base- emitter loop is 6V RC unchanged, so IB is - + 20 mA still 430 mA. 2V LED I C bI B 40430 mA - 430 mA 20 mA. VRB + The voltage across + - + 4V 9V RC is now + VRC I C RC 20 mA.300 RB - + 0.7 V - - 10K 5V 6V - A BJT Circuit a. The base current + is unchanged, so the power generated by 6V RC the 5 V supply is still - + 20 mA 2.15 mW 2V LED - b. and c. Similarly, 430 mA the power dissipated VRB + by RB is still 1.85 + - + 4V 9V mW, and the power + dissipated by the RB - + 0.7 V - - base-emitter junction 10K 5V is still 301 mW. - d. The power delivered by th 9 V supply is doubled, to 180 mW A BJT Circuit e. The power dissipated by RC is quadrupled, + to 120 mW. 6V RC - f. power dissipated by + 20 mA the LED is doubled, to 2V LED 40 mW. - 430 mA VRB + + - + 4V 9V + RB - + 0.7 V - - 10K 5V - A BJT Circuit g. KVL gives us the + collector-emitter voltage, VCE 6V RC - 9 V VRC 2 V VCE + 20 mA VCE 9 V VRC 2 V 2V LED - 9 V 6 V 2 V 1V 430 mA The power VRB + dissipated by the + - + + 4V 9V collector-emitter RB - output is now + 0.7 V - - 10K VBJTce I CVCE 20 mA.1V 5V 20 mW - A MOSFET Circuit Here’s the circuit used by Herrick in his example. Let’s work that example, except that the load voltage VLoad is doubled to 6 V. If the load voltage is doubled, the source current must also be doubled, to 2 A. Examination of the MOSFET characteristic reveals that VGS 4.2 A Applying KVL around the gate- + + source loop yields VDS 12 V + - VDD VIN VGS VLoad + VGS - - + 4.2 V 6 V VIN Rload 3 Vload 10.2 V - - A MOSFET Circuit The load power is now given by PLoad 8W 6 V 2 3 so the load power is more than doubled. The supply power is doubled, to 18 W. The MOSFET output power is PMOSFET PSUPPLY PLoad 18 W 8 W 10 W + + VDS 12 V + VDD - + VGS - - + VIN Rload 3 Vload - - An Opamp Circuit Here’s Herrick’s opamp example. Suppose we change VIN from 1 V to 2 V: The ideal op-amp assumptions say that Va = VIN, because the Voltage difference between the inverting and noninverting inputs must be zero. +Esupply Having found Va, we Ini see that the VRi = Va, so VIN + Iout Va 2V 2V VOUT I Ri 2 mA. - Ri 1000 Iinv KCL tells us that IRf = IRi, -Esupply so Ohm’s law says IRf 2V Rf=2.7K VRf I Rf R f I Ri R f a Ri - + Rload + VRf 1K Va Rf VRi 1K R f Vin - IRi Ri Ri An Opamp Circuit KVL tells us that VOut VRf VRi Rf Vin Vin Ri +Esupply R Ini 1 f Vin VIN + Iout Ri 2V VOUT - 2700 Iinv 2V 1 7.4 V -Esupply 1000 ILoad IRf so 2V Rf=2.7K VOUT 7.4 V a I Load 7.4 mA. Ri - VRf + Rload RLoad 1000 + 1K 1K VRi - IRi Energy Our lab exercises all require the use of a power supply. What does a power supply do? It supplies power! What does that mean? What’s Power? Power is the rate at which energy is converted from one form to another. If K represents the energy converted in a time interval Dt, the rate of conversion is K/Dt, and the power which must be supplied for conversion is K P Dt An electric motor converts electrical energy to mechanical (kinetic) energy. A motor which is 100% efficient (that is, it converts all of the electrical energy supplied to it to useful work, none is wasted as heat or other forms) and is rated at 1 horsepower (1 hp = 746 Watts) converts energy at the rate of 746 Joules P 746 Watts 1 second. Energy So a 1 hp. motor can convert 746 Joules of electrical energy to 746 Joules of work every second. Where does the electrical energy come from? A power supply! 120 V I + 1 horsepower M DC motor - Here the power supply is a 120 Volt battery. The motor converts 746 J/sec. of energy, so the battery must supply 746 J/sec., or 746 W. of power. remember that P = IV, so the current supplied must be P 746 W I 6.22 A V 120 V Energy Notice that in this example the current is flowing out of the battery’s + terminal, and into the more positive terminal of the motor. There is a voltage rise across the battery terminals, and a drop across the motor. 120 V I + + 1 horsepower VM M DC motor - - Since current flows out of the positive battery terminal, the battery is supplying energy to the motor. Energy Some batteries, like the one in your car, are rechargeable. That is, they can convert electrical energy to chemical energy for storage, then reverse the process to supply electrical energy. 120 V I + + DC generator VM M - - Here’s the same battery, connected to a DC generator (which may be identical to the DC motor, except that its shaft is driven by a gasoline or diesel engine). The generator converts mechanical energy to electrical energy. Current flows out of the generator’s positive terminal and into the battery’s positive terminal, so now the generator is supplying energy to the battery. Energy The battery now receives energy from the generator at the rate of Pch arg e 120 V I Ch arg e Joules per sec., or Watts 120 V ICharge + + DC generator VM M - - So the battery stores energy at that rate. If the generator supplies charging power to the battery at 746 Watts, then the battery must charge 1 minute for every minute the motor is to run at full power. Energy Of course, the wires connecting the battery to the generator have some resistance. Let’s say it’s 1 Ohm. The generator voltage must be slightly greater than the battery voltage, to force current to flow from the generator to the battery. To 1 charge at a 746 Watt rate, the current must be 6.22 A, so VM 6.22 A 1 120 V 126 V + + ICharge 120 V VM M DC generator - - Similarly, the voltage across the motor when it’s receiving power from the battery must be lower than the battery voltage, so current flows out of the battery. to run at full power, VM 120 V 6.22 A 1 120 V 113 .78 V Energy A DC motor does produce a voltage across its terminals while its running, called the back emf, or counter emf. That’s where VM comes from. 1 + ICharge + 120 V VM M DC Motor - -

DOCUMENT INFO

Shared By:

Categories:

Tags:

Stats:

views: | 2 |

posted: | 8/24/2012 |

language: | English |

pages: | 38 |

OTHER DOCS BY ert554898

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.