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OBI CHIMA oc5118@ic.ac.uk 8704 6051 Energy Storage Elements A MedED LECTURESESSION STRUCTURE Intro to storage eleTransient DC Response ofC Response of capacitors & Inductorsacitors & Inductors Storage elements • Capacitors à Store Charge • Inductors à Store Magnetic Field • The current and voltage running through and across a capacitor is a function of timeSESSION STRUCTURE Intro to storage eleTransient DC Response ofC Response of capacitors & Inductorsacitors & Inductors Transient DC Response: Capacitors • As soon as the switch is closed, a current I begins to flow through the circuit & the capacitor acts like a short circuit (current flows through as if no component there) • Current flowing through the capacitor causes a potential differenceto be created across it • REMEMBER: Capacitors separate charge and Current is rate of flow of charge • The charge separated across a capacitor’s plates is related to the pd across the capacitor by the equation: NB: If the current stops flowing before Q = C *V C1(t) the capacitor is fully charged, the • Thus the current through a capacitor at any given time is: voltageC1 will remain at the voltage it was charged to at the time the current I = C* dV /dt was stopped (t) C1 Transient DC Response: Capacitors • Voltage across a capacitor at time, t, is defined by: V C1 =V 1 [1-et/R1C] • As capacitor charges, the voltage across it increases up until it reaches the supply voltage • As the voltage increases, the current through the capacitor decreases as the driving force for electron flow is reduced • Once fully charged, the capacitor acts like an open circuit and current is ZERO Transient DC Response: Capacitors • Voltage across a capacitor at time, t, is defined by: V =V * [1-e -t/R1C] C1 1 • At time t=0,V C1= 0 • There is no voltage drop across the capacitor so, functionsasashortcircuit -t/R1C1 • At time t = ∞, e à 0 ,V C1 =V1 • usually fully charges in 4-5ms • The capacitor is fully charged to the voltage of the source, so there is no ‘driving force’ for current to flow – it is essentially an open circuit -t/R1C1 -1 • At time t = R1C1, e = e ,V C1= O.632V 1 • R1C1 (product of the resistance and capacitance) is known as the charging constant ( ) – this is the time it voltageor a capacitor to charge to 63.2% of its chargingTransient DC Response: Capacitors Transient DC Response: Capacitors • Current through a capacitor at time, t, is defined by: -t/R1C1 IC1=V /R1* 1e ] • At time t=0, I =V /R C1 1 1 • The current has an initialvalue ofV/R (Ohm’s Law) as the capacitor acts as a short circuit • At time t = ∞, e -t/R1C1à 0 , IC1= 0 • The capacitor is fully charged to the voltage of the source, so there is no ‘driving force’ for current to flow – it is essentially an open circuit Discharging a Capacitor • To discharge a capacitor, remove the voltage source and short the circuit • capacitor acting as the powerged source but its voltage decreases over time until fully discharged (like a battery powering a device) • When discharging, the current The discharge equation is: flowsinthe opposite direction and decreases to zero over time -t/R1C1 V C1 =V *1[e ] Transient DC Response: Inductors • These store magnetic fields • Increasing the voltage across them causes an increase in the magnetic field – they only respond to changes in voltage • The current through an inductor at time, t, can be expressed as: IL1 =V /1 *1[1 - e -tR1/L] • At time t=0, I =L1 • Initially, the inductor behaves like an open circuit • At time t = ∞, e -tR1/Là 0 , I L1 /R 1 1 • When fully charged, t behaves like a short circuit • At time t = L1/R1, I = L1632V /R 1 1SESSION STRUCTURE Intro to storage eleTransient DC Response ofC Response of capacitors & Inductorsacitors & Inductors AC Response • With a DC signal we simply charge the capacitor/inductor, with anAC signal we are constantly charging and discharging the capacitor • Because capacitors and inductors have set time constants (tau) during which they charge and discharge, alternating the signal means (depending on the frequency) they may not be fully charged before they are discharged again • We’ve seen that as a capacitor charges, the current flowing through it decreases (the resistance increases) • In a low frequencyAC signal, the polarity will be switched later on during the charge, when there is already quite a large resistance to current flow – perpetuating the length of time during which resistance is high • In a high frequencyAC signal, the polarity will be switched very on in the charge, where there is little resistance to current flow -- perpetuating the length of time during which resistance is low AC Response of Capacitors • The relationship between frequency and reactance (frequency-dependent resistance) for a capacito: is X C 1/2πCf • X is reactance in Ohms • We can see that reactance is inversely proportional to frequency • As frequency increases (along x-axis) reactance decreases and therefore more current can flow through the capacitor • In this configuration, it acts as a high-pass filter, only allowing currents of high frequencies through the capacitor AC Response of Capacitors • The relationship between frequency and reactance (frequency-dependent resistance) for an inducto: is X C 2πLf • X is reactance in Ohms • We can see that reactance is directly proportional to frequency • As frequency increases (along x-axis) reactance increases and therefore less current can flow through the capacitor • In this configuration, it acts as a low-pass filter, only allowing currents of low frequencies through the capacitor AC Response of Capacitors • The relationship between frequency and reactance (frequency-dependent resistance) for an inducto: is X C 2πLf • X is reactance in Ohms • We can see that reactance is directly proportional to frequency • As frequency increases (along x-axis) reactance increases and therefore less current can flow through the capacitor • In this configuration, it acts as a low-pass filter, only allowing currents of low frequencies through the capacitor High and Low -pass filters • Often we want to remove certain frequencies of signals from another signal • Using a single resistor and a capacitor (or inductor) we can create both high and low-pass filters Low-pass High-pass Low -pass filters • The formula for the voltage dropped across the capacitor is: V =V * X /√(X 2+ R ) out 1 C C • Thecut-off frequency (f ) ic the frequency at which the original signal is attenuated by 70% • Vout 0.71 fc= 1/2πRC High -pass filters • The formula for the voltage dropped across the resistor is: 2 2 V out=V * 1/√(X C + R ) • Thecut-off frequency (fc) is the frequency at which the output (load) voltage equals70% of the input (source) voltage. • Vout 0.1V fc= 1/2πRCQuestionsQuestionsQuestionsTHANKYOU FOR COMING! 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