09-power measurements in (wpt) systems -朱博
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Power measurements in (WPT) systems
Qi Developer Conference by Laurens S, JamesAugust 2017 - WPC1703 - Taipei
Laurens Swaans28 August 2017
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About nok9 ABnok9¨ >30 years all-in-one test & measurement
equipment¤ Quality control¤ Production¤ Conformance
¨ Calibration services¤ Optical media discs¤ AC Power measurements
¨ Consultancy¤ Security algorithm for Xbox¤ Wireless charging solutions
¨ Quality Certifications¤ ISO17025¤ ISO9001 / ISO14001
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Agenda¨ What is power?¨ Why is power important?¨ Power requirements?¨ Where do we measure power?¨ How to measure AC power?¨ What models to use?¨ Fortunately there’s help!¨ Tune your Rx design
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What is power?
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¨ Power: ¤ Indication of the energy usage per time unit (rate of work)¤ Expressed in Joule per second (SI), or Watt (more common)¤ Electrical power: 𝑃 𝑡 = 𝑣 𝑡 % 𝑖 𝑡
¨ DC power¤ 𝑃'( = 𝑉'( % 𝐼'(
¨ AC power¤ Instantaneous power: 𝑃 𝑡 = 𝑣 𝑡 % 𝑖 𝑡¤ RMS values power: 𝑃 𝑡 = 𝑉+,- % 𝐼+,- % cos𝜑
n RMS= Root Mean Square𝜑 = the phase shift between voltage and current
What is power?
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What is power?
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VS
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¨ Burning 1kg of coal vs 1kg of TNT¤ TNT will release all its energy in a very short but very powerful bang¤ The coal will burn and release heat for a very long time
¨ This exemplifies the difference between energy and power¤ Which has more energy: coal or TNT? Which has more power?
¨ Energy is power accumulated over time:¤ The battery status of your phone going from 0-100%
¨ Power is the amount of energy per time unit¤ The charging speed of your phone while charging (% per minute)
¨ So…Coal: 25MJ/kg vs TNT: 4.2MJ/kg¤ Power: TNT wins, Energy: coal wins
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What is power?
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¨ Foreign Object Detection (FOD)
Why is power important?
Receiver coil
Transmitter coil
Magnetic flux
Received Power 𝑃+2
Power loss < 250mW
𝑃3 = 𝑃42 − 𝑃+2
Power loss > 250mW
𝑃3 = 𝑃42 − 𝑃+2
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¨ Foreign Object Detection (FOD)¤ Studies done by FOD tiger team: July 2010 – July 2012
n >500mW of power loss into a typical everyday object can lead to dangerous temperatures
n Power loss budget of 500mW to be divided over Rx and Tx¤ Rx has to accurately report its power consumption to the Tx
n Not only the power into its load circuit (rectified power)…n …but also the power lost in the rectifier, modulator, coil, friendly
metals¤ Tx has to accurately determine its transmitted power
n All power dissipated outside the Tx product housing
Why is power important?
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Where do we measure power?
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¨ In Wireless Power Transfer (WPT) systems:¤ Transmitter input: 𝑇𝑥89¤ Transmitter output: 𝑇𝑥:;4¤ Receiver input: 𝑅𝑥89¤ Receiver output: 𝑅𝑥:;4
A simple model
V Power Transmitter
Power Receiver
Load
𝑇𝑥89 𝑇𝑥:;4 𝑅𝑥89 𝑅𝑥:;4
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¨ TxIN:¤ External DC power supply: 𝑃 = 𝑉'( % 𝐼'(¤ Mains connected transmitter: 𝑃 = 𝐴𝑉𝐺 𝑉 𝑡 % 𝐼 𝑡
¨ RxOUT:¤ Battery or other DC load: 𝑃 = 𝑉'( % 𝐼'(
¨ System efficiency: 𝜼𝑺𝒀𝑺 =𝑹𝒙𝑶𝑼𝑻𝑻𝒙𝑰𝑵
¤ Note the possibility to distinguish between energy efficiency and power efficiency
Conventional powers
V Power Transmitter
Power Receiver
Load
𝑇𝑥89 𝑇𝑥:;4 𝑅𝑥89 𝑅𝑥:;4
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¨ TxOUT:¤ AC power generated by the coil: 𝑃 = 𝐴𝑉𝐺 𝑉 𝑡 % 𝐼 𝑡
¨ RxIN:¤ AC power generated by the coil: 𝑃 = 𝐴𝑉𝐺 𝑉 𝑡 % 𝐼 𝑡
¨ Efficiencies:¤ Antenna: 𝜼𝑨𝑵𝑻 =
𝑹𝒙𝑰𝑵𝑻𝒙𝑶𝑼𝑻
¤ Transmitter: 𝜼𝑻𝒙 =𝑻𝒙𝑶𝑼𝑻𝑻𝒙𝑰𝑵
¤ Receiver: 𝜼𝑹𝒙 =𝑹𝒙𝑶𝑼𝑻𝑹𝒙𝑰𝑵
WPT characteristic powers
AC power measurements enables these
V Power Transmitter
Power Receiver
Load
𝑇𝑥89 𝑇𝑥:;4 𝑅𝑥89 𝑅𝑥:;4
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How to measure AC power?
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¨ Power dissipation occurs in “resistive” loads¤ “resistive” means: voltage and current are in phase
AC power in resistive loads
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– Voltage: 100kHz, 1V amplitude– Current: 100kHz, 0.5A amplitude– Double frequency– Always positive– Power:
• 200kHz, 0.5W amplitude• Average P(t): 0.25W• RMS: 𝑉+,- % 𝐼+,- =
JK�% M.O
K�=0.25W
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¨ Power dissipation in WPT systems?¤ TxOUT and RxIN are not really “resistive”, not really “reactive”¤ Voltage and current are not exactly in phase
n Amount of phase shift depends on the impedance
¨ Power measurements when 𝛗 ≠ 𝟎¤ 𝑃 = 𝐴𝑉𝐺 𝑉 𝑡 % 𝐼 𝑡 still holds true¤ But also: 𝑃 = 𝑉+,- % 𝐼+,- % 𝑐𝑜𝑠 𝜑 (for sinusoidal signals)
AC power in reactive loads
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¨ AC power comparison
¨ Differences caused by sampling frequency (10MHz)
Comparison of calculations
ϕ AVG RMS Diff
0 0.249 0.249 0
18 0.237 0.237 7.6e-5
36 0.201 0.202 2.3e-4
54 0.146 0.147 3.2e-4
72 0.077 0.077 2.3e-4
90 3.5e-17 1.5e-17 2.0e-17
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¨ Power dissipation in WPT systems¤ TxOUT and RxIN are not really “resistive”, not really “reactive”¤ Voltage and current are not exactly in phase
n Amount of phase shift depends on the impedance¨ Power measurements when 𝛗 ≠ 𝟎
¤ 𝑃 = 𝐴𝑉𝐺 𝑉 𝑡 % 𝐼 𝑡 still holds true¤ But also: 𝑃 = 𝑉+,- % 𝐼+,- % 𝑐𝑜𝑠 𝜑 (for sinusoidal signals)
¨ So we look at the “resistive part” of a “complex load”¤ The actual amount of phase shift depends on the ratio between
resistance and reactance¤ How de we know “what to expect”?
AC power in reactive loads
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What model(s) to use?
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A simple model again
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V Power Transmitter
Power Receiver
Load𝑇𝑥89 𝑇𝑥:;4 𝑅𝑥89 𝑅𝑥:;4
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Typical WPT model¨ Transmitter parameters
¤ Vin: input voltage¤ Cp: Series resonance
capacitor on primary side¤ Lp: Primary inductance¤ Rp: ESR of Tx
¨ Receiver parameters¤ Cs: Series resonance
capacitor on secondary side¤ Ls: Secondary inductance¤ ZL: Receiver loading¤ Rs: ESR of Rx
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What signals to expect?
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¨ Invertor:¤ Transient: 10V, 160kHz¤ AC: 50kHz-250kHz
¨ Transmitter:¤ Qi TPT#2
¨ Receiver:¤ Qi TPR#1B
¨ Loading:¤ ZL=10Ω
¨ Coupling:¤ k=0.5
First simulation
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¨ Let the computer run through the complete load range
Now…analyze the system
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¨ Now add different couplings to the mix
Now…analyze the system
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¨ ..and different frequencies
Now…analyze the system
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¨ Simple visualization of the impact the circuit changes has on the Bode diagram¤ Example here shows the
Bode diagrams forn Primary Coil Current 𝐼Vn Secondary Voltage 𝑉+
¤ The red dot shows the steady-state operating point
¤ Stepping through different loads at a fixed coupling
Make visible what happens
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Fortunately there’s help
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¨ Lots of devices available…
Who can measure power?
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¨ Lots of devices available…¨ Power Accuracy < 175mW?
Who can measure power?
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¨ Lots of devices available…¨ Power Accuracy < 175mW?¨ Frequency range > 100kHz?
Who can measure power?
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¨ Resolution: 1mW¨ Accuracy: ±112.5mW
¤ @7.5W range, 100-400kHz¤ Only active power (P)
¨ Resolution: 0.1mW¨ Accuracy: ±10mW
¤ @35W range, 100-500kHz¤ All waveforms (sine, square)¤ Active (P), reactive (Q) &
apparent power (S)
Who can measure power?
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¨ Research & Development from technology providers ¤ WPC seminars and online documentation¤ IEEE & other research papers
¨ Silicon providers (e.g. Integrated Device Technology, Texas Instruments, ON semi, Maxim, etc…)¤ Video tutorials, application notes, white papers, datasheets…
n Listen to these guys, they know what they’re talking about!¨ Test and measurement equipment
¤ ISO17025 accreditation
Help from the industry pioneers
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Tuning your design
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¨ Available parameters:¤ 𝑉:;4, 𝑉+W(4, 𝐼:;4(= 𝐼+W(4)¤ Power after rectifier
n 𝑃+W(4 = 𝑉+W(4 ⋅ 𝐼+W(4¨ But… 𝑃+W(W8[W' ≠ 𝑃+W(4
¤ 𝑃+W(W8[W' = 𝑃+W(4 + 𝑃3:--¨ Need to estimate losses in:
¤ Rectifier¤ ESR of coil and capacitors¤ Friendly metals
Tune your Rx
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Example circuit implementation
VOUTVRECT IOUT
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¨ 𝑃 = 𝑉 ⋅ 𝐼¤ 𝑃'J = 𝑉]^_ ⋅ 𝐼'J
¨ AC signal:¤ So two diodes conduct during
positive wave, 2 diodes conduct during negative wave
¨ Buffer capacitor:¤ Diodes only conduct to
recharge the capacitor¤ Short bursts when 𝑉 ( > 𝑉+W(4
Rectifier losses
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¨ But…average load current is quite stable¤ Average current is flowing
through the rectifier¨ Use average IOUT and find the
equivalent average VF
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¨ Communication window is excluded from power calculation
¨ Choose high quality resonance capacitors ¤ high temperature stability and
low ESR (NP0, C0G)¤ ESR is low enough to ignore
¨ Leaves inductor ESR¤ 𝑃3b = 𝐼+,-K ⋅ 𝐸𝑆𝑅3b¤ With 𝐼+,- ∝ 𝐼:;4
ESR losses
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¨ This one is tricky as it depends on the exposure of your Rx device (end product) to the magnetic field generated by the Tx¤ Amount of shielding¤ Materials used / exposed¤ Misalignment¤ Coil geometry¤ Coil current (with 𝐼+,- ∝ 𝐼:;4)
Friendly metals
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¨ Typically you define the “potential friendly metal losses” as:¤ 𝑃], = 𝑃42 − 𝑃3:`' − 𝑃3:--¤ 𝑃], = 𝑃42 − 𝐼:;4 ⋅ 𝑉+W(4 − 𝐼:;4 ⋅ 𝑉] − 𝐼:;4K ⋅ 𝐸𝑆𝑅¤ 𝑃], = 𝑃42 − 𝐼:;4 𝑉+W(4 + 𝑉] + 𝐼:;4 ⋅ 𝐸𝑆𝑅
¨ And then model them as a resistive loss:¤ 𝑃], = 𝐼:;4K ⋅ 𝑅],
¨ This allows to link all power values to the output current of your Rx
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1. Measure PTX, VRECT and IOUTover your entire operating range
2. Find appropriate values for VF, ESR and RFM
¤ Such that 𝑃42 = ∑𝑃+g�� in
every operating point3. Add an offset POS to ensure the
Rx always underestimates its power consumption¤ A device is not allowed to
overestimate its received powerTune your Rx
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