Power measurements in (WPT) systems

Qi Developer Conference by Laurens S, JamesAugust 2017 - WPC1703 - Taipei

Laurens Swaans28 August 2017

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

¨ 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?

¨ 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?

¨ 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

¨ 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

¨ 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?

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

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

¨ 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

¨ 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

¨ 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

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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