Document Sample

ENGG1000 Engineering Design and Innovation Capacitors 1.0 The Capacitor Capacitors are a fundamental circuit element, which physically consist of two conducting plates very close to each other (separated by a dielectric material), as suggested by their circuit symbol, shown in figure 1. The plates are not connected (i.e. there is an open circuit between them), however if positive charges were to accumulate on one of the plates, then due to the proximity of the other plate, negative charges would be attracted to it, producing a current. The charge accumulation effect occurs when a voltage is applied to the capacitor, which creates an electric field between the plates, encouraging charge accumulation at the plates. If charges are stored at the plates and then the voltage is removed, the charges will flow away from the plates again. This behaviour leads us to think of the capacitor as a charge storage device. Figure 1. The relationship between charge stored q and the voltage across a capacitor V is given by q = CV , where C is the capacitance of the device, in Farads (F). Since the flow of dq charge is current I = , this leads us to the voltage-current relationship for the dt capacitor: dV (t ) I (t ) = C dt Where I(t) and V(t) are the capacitor current and voltage as a function of time t. What does this mean ? If the voltage across the capacitor is constant, then no current flows. If the voltage changes linearly, then a constant current flows. 2.0 First-order Capacitive Circuits An example of capacitive behaviour can be seen by analyzing the circuit of figure 2. R1 A B + + I Vin V R2 - - Figure 2. ENGG1000 Engineering Design and Innovation 1 Suppose initially the switch ( ) is in position B, so no current is flowing in the circuit and the capacitor voltage V = 0. Now the switch is moved to position A, and dV (t ) Vin = R1 I (t ) + V (t ) = R1C + V (t ) . dt − t This is a differential equation with solution V (t ) = Vin 1 − e R1C . Since t dV (t ) V − I (t ) = C , we find I (t ) = in e R1C . Let’s look at the capacitor voltage and dt R1 current at the time the switch closes (t = 0) and a long time afterwards (t → ∞): t = 0 : V(0) = 0 t → ∞ : V(∞) = Vin V t = 0 : I(0) = in t → ∞ : I(∞) = 0 R1 We find that initially, the voltage is zero, then it charges quickly and approaches Vin as t approaches infinity. Note that the capacitor voltage change is continuous, i.e. it does not change instantaneously. Initially, the capacitor current becomes quite large, in order to charge up the capacitor, but then drops as the capacitor voltage approaches Vin. At t = ∞, when V = Vin, the current is zero. Suppose we let a long time T pass, so that V = Vin, and flip the switch back to position V B again. At this instant, the only current flowing is in , through R2. Now R2 dV (t ) V (t ) = − R2 I (t ) = − R2C . This is another differential equation, with solution dt t −T t −T − dV (t ) Vin − R2C V (t ) = Vin e R2C . Since I (t ) = C , we find I (t ) = − e . Let’s look at the dt R2 capacitor voltage and current at the time the switch closes (t = T) and a long time afterwards (t → ∞): t = T : V(T) = Vin t → ∞ : V(∞) = 0 V t = T : I(T) = − in t → ∞ : I(∞) = 0 R2 Note that now the current flows in the opposite direction; the capacitor is acting like a voltage source supplying R2, in order to keep the capacitor voltage V constant when the switch opens again. The voltage simply drops continuously from Vin back to zero during the discharge. The voltage and current waveforms are shown in figure 3. ENGG1000 Engineering Design and Innovation 2 capacitor charging up capacitor discharging V(t) Vin T t Vin I(t) R1 T t Vin − R2 Figure 3. To summarise: The voltage across a capacitor cannot change instantaneously The following formula can be used to determine the capacitor voltage for circuits like figure 2 undergoing a change in voltage from Vinitial to Vfinal at time t = 0: t − V (t ) = V final − (V final − Vinitial )e RC Note that R is the charging/discharging resistance, which needs to be determined based on the circuit surrounding the capacitor. In our example, during charging R = R1, during discharging R = R2. Please note: These notes are intended to give the key basic knowledge required to get started in ENGG1000. They are certainly not comprehensive, and very likely you will need to consult other sources even just to complete your project in this course. ENGG1000 Engineering Design and Innovation 3

DOCUMENT INFO

Shared By:

Categories:

Tags:
Engineering Design, Mathematics 1B, Higher Mathematics, Electrical Engineering, the School, general education, Maths 1A, Year 1, Environmental Engineering, Civil Engineering

Stats:

views: | 13 |

posted: | 2/19/2011 |

language: | English |

pages: | 3 |

OTHER DOCS BY hjkuiw354

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.