• Sinusoidal voltage: V = VPEAK sin(2πf t)
`o V = voltage (a time-varying voltage here). Units are volts
`o VPEAK = peak voltage of sinewave.
`o Also referred to as the amplitude. Units are volts.
`o f = frequency (in cycles per second or Hertz)
`o t = time in seconds
`
`1
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`
`o V = voltage (a time-varying voltage here). Units are volts
`o VPEAK = peak voltage of sinewave.
`o Also referred to as the amplitude. Units are volts.
`o f = frequency (in cycles per second or Hertz)
`o t = time in seconds
`
`1
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`
`
`If a source has a maximum output power
`of 10 dBm then that is the same as saying
`that the source has a maximum output
`power of 10 mW (milliwatts).
`
`2
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`
`
`• Period = T = 1/f
`o f = frequency (in cycles per second or Hertz)
`o T is the period of the sinewave in seconds
`o E.g. if f = 1 MHz then T = 1 microsecond = 1000 nanoseconds
`o E.g. if f = 1 GHz then T = 1 nanosecond
`
`3
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`
`
`• Power delivered to a resistor R with a
`sinusoidal voltage across it is
`o P = ½(VPEAK)2/R
`where VPEAK is the amplitude
`
`4
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`
`
`• Energy delivered to a resistor R in a time Δt:
`
`U = P * Δt
`
`where P = the power delivered to the
`resistor (shown on previous slide).
`
`5
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`
`
`Power delivered to a resistor R with a constant
`voltage V across it:
`
`P = V2/R
`
`6
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`
`
`Excerpt from Subramanian
`Reply Declaration at Par. 7.
`• Since E is often used as the voltage of a source, U is often used
`to denote energy, rather than E, when both are used.
`
`• Energy stored on a capacitor with a constant voltage V across it
`is
`o U = ½CV2
`
`7
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`
`
`• Replacing E by U, then the energy stored on a capacitor
`with a constant voltage V across it is:
`U = ½CV2
`o If the voltage across a capacitor is 0 volts, then no energy is
`stored on the capacitor.
`o If the voltage across the capacitor is 0.000 1 V and the
`capacitor C = 0.000001 farads then the energy stored on the
`capacitor is:
`U = ½CV2 = ½ (0.000001) (0.0001)2
`= ½ (0.000001) (0.0001)2
`= ½ (0.000001) (0.0000001)
`= ½ (0.000001) (0.000000000001)
`= 0.5 pJ (picojoules)
`
`8
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`
`
`The voltage across RL = ½ Vs, when RL = RS
`
`Source: Thomas L. Floyd,
`Principles of Electric Circuits, Fifth
`Edition, (Prentice-Hall, Inc.), 1997.
`Ex. 2024
`
`9
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`
`
`• Based on the Maximum Power Transfer Theorem,
`the maximum power that can transferred to a load
`RL, by a sinusoidal voltage source with internal
`resistance RS and a voltage amplitude of EPEAK
`occurs when RL = RS
`o The maximum power that can be delivered is also
`called the available power.
`o The available power is calculated when the load
`resistor RL equals RS.
`o Then the peak of the sinusoidal voltage across RL is:
`VPEAK = ½ EPEAK.
`o PMAX = ½(VPEAK)2/RS
`o PMAX = ½(½EPEAK)2/RS
`o PMAX = (EPEAK)2/(8RS)
`
`10
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`
`
`The maximum energy that can be delivered to a
`load (i.e. the available energy) during a time
`TON is:
`o UMAX = ((EPEAK)2/(8RS))TON
`
`11
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`
`
`• Resistance of two resistors R1 and R2 in
`series is
`o R = R1 + R2
` E.g. a 50 Ω resistor in series with a 50
`Ω resistor has a resistance of 100 Ω
`
`12
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`
`
`• Resistance of two resistors R1 and R2 in
`parallel
`o R = R1 // R2 = 1/(1/ R1 + 1/ R2)
` E.g. a 50 Ω resistor in parallel with a
`2000 Ω resistor has a resistance of
`48.78 Ω
`R = 50 Ω // 2000 Ω
`= 1/(1/ 50 + 1/ 2000) Ω
`= 1/(0.02 + 0.0005) Ω
`= 1/(0.02005)
`= 48.78 Ω
`
`13
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`
`
`Response of a simple RC circuit to a voltage step.
`
`14
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`
`
`• The time constant of an RC circuit is R times C.
`• The symbol typically used for time constant is τ.
`• E.g. for the circuit above τ = R times C = RC
`
`15
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`
`
`o Voltage, VC, across a capacitor charging for time TX
`through a resistor from a constant voltage E and
`starting from 0 V is
`
` VC = E (1 –𝑒𝑒(−𝑇𝑇𝑋𝑋τ))
` VC = E (1 –𝑒𝑒(−𝑇𝑇𝑋𝑋τ))
`
` E.g. TX = 100 ns and τ = 100 microseconds
` TX / τ = (0.0000001)/(0.0001) = 0.001
`
`= E(1−0.9990000) = 0.001 V
`
`16
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`
`
`o The time constant of a resistor-capacitor
`circuit (R and C in parallel or series) is
` τ = RC
`o If R is in ohms and C is in farads, the units of
`time constant is in seconds.
`o E.g. if R = 100 Ω and C = 1 microfarad:
`τ = RC = 100(0.000001) =
`0.0001 = 100 microseconds = 100 µs
`
`17
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`
`
`
`• Given that the maximum power that can transferred from a source with
`sinusoidal generator having peak voltage and internal resistance RS is:
`PMAX = (EPEAK)2/(8RS)
`
`o Calculations:
`o Consider a source that can deliver a maximum power of 10 mW (0.01
`watts).
` If RS = 50 ohms, then:
`• PMAX = 0.01 W
`= (EPEAK)2 / (8RS)
`= (EPEAK)2 / (8 x 50)
`= (EPEAK)2 / (400).
`• Therefore (EPEAK)2 = 0.01(400) = 4 V2
`• So EPEAK = 2 V.
`
`18
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`
`
`
`Given that the energy available from the source during TX is:
`U = PMAXTX
`If PMAX = 0.01 watts, and TON = 100 nanoseconds, then the
`energy available is:
`U = PMAXTX = 0.01(100)(0.000000001)
`= (0.000000001) joules = 10−9 J = 1 nJ = 1000 pJ.
`
`19
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`
`
`Recall that when VC = 0.001 V is the
`voltage on a capacitor where C = 1 μF, the
`energy stored on the capacitor is:
`
`UC = ½C(VC)2
`= ½(1 μF)(0.001 V)2
`= ½(0.000001)(0.000001) J
`= ½(0.000000000001) J
`= 0.5 pJ
`
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`
`
`o Assume that the maximum available
`power from a source is 10 mW.
`o Therefore, the energy available in 25 ns
`is:
`UA = Power multiplied by time
`= (0.01 W)(0.000000025)
`= 0.00000000025 J
`= 250 pJ
`o Recall that the energy transferred to a 1
`μF capacitor in 25 ns is UC = 0.5 pJ
`o Therefore, the percentage of energy
`transferred to the capacitor in 25 ns
`relative to the energy available is:
`100 (UC / UA) = 100(0.5 pJ)/ (250 pJ)
`= 0.2%
`
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`
`
`If a capacitor starts off with 0 joules of energy stored on it,
`and that capacitor after charging has 1 pJ of energy stored on
`it, then 1 pJ of energy has been transferred to the capacitor.
`
`22
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`
`
`
`90°
`
`0°
`
`
`
`
`RF waveform
`
`(input signal)
`
`Average value of
`input signal
`
`
`
`270°
`
`—. Period T
`
`256. As shownby
`
`the annotation, the
`
`input signal f; that is received
`
`by
`
`the
`
`commutating
`
`switch 38 is shownin blue. The red annotation showsthe
`
`portion
`
`of
`
`the
`
`input signal f| (which
`
`are
`
`quarter cycles,
`
`each quarter cycle being 90 degrees
`
`out
`
`of 360 degrees
`
`of a full
`
`cycle)
`
`over which the
`
`voltages
`
`are
`
`averaged.
`
`The
`
`signal 125
`
`of
`
`Figure
`
`4 illustrates the waveform
`
`corresponding
`
`to the average value of the
`
`input
`
`signal f; and
`
`points 110, 105, 110, 115 illustrates the average voltages during
`
`each
`
`90 degree
`
`interval. /d. at 3:44-48.
`
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`IPR2020-01
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`
`
`11.
`
`Tayloe’s storage of
`
`non-negligible
`
`amounts of energy is further
`
`confirmed
`
`by Tayloe’s description
`
`of
`
`capacitors 72-78 as
`
`“integrating”
`
`the
`
`input
`
`signal.
`
`(Ex. 1004-Tayloe,
`
`2:42-44
`
`(“Each
`
`of the
`
`capacitors
`
`functions as a
`
`separate
`
`integrator,
`
`each
`
`integrating
`
`a
`
`separate quarter
`
`wave
`of the input signal’).)
`
`This
`
`language
`
`indicates an accumulation of energy
`
`on the
`
`capacitors,
`
`and it matches the
`
`°551 patent’s description
`
`of its storage element as
`
`“integrat[ing]” non-negligible
`
`amounts of energy. (Ex. 2007-’551, cl. 198-202 (“said storage module receives and
`
`integrates the non-negligible
`
`amounts
`
`ofenergy from the carrier
`
`signal’’).)
`
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`
`
`
`°551 patent’s description
`
`of its storage element as
`
`“integrat[ing]” non-negligible
`
`amounts of energy. (Ex. 2007-’551, cl. 198-202 (“said storage module receives and
`
`integrates the non-negligible
`
`amounts
`
`of energy from the carrier
`
`signal”’).)
`
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`
`
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