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Sampling And Averaging Considerations For Measuring Input Power Bob White Staff Engineer Worldwide Technology Group Why Monitor Input Power? ● Improved Facility Utilization – Typical Draw Is 50-70% Of Nameplate Rating – Facilities Planned Based On Nameplate Rating – Can Add 30-80% More Server Capacity Without Additional Facility Capacity ● Important Metric For Virtualization Engines Needed Accuracy ● Revenue Accuracy: ±0.2% To ±0.5% – Not Economical For A Power Supply ● ±10% Not Good Enough – Significant Loss Of Increased Server Capacity – Economical Accuracy One Thing, ● ±5% Better Resolution Another! – Not Quite Good Enough ● OEMs Asking For ±3-4% – Requests For ±1% Have Been Reported Measurement Error: Ideal Measurement Location HVDC+ V LINE FILTER I HVDC– Measurement Error: Actual Measurement Location LINE FILTER HVDC+ V HVDC– V Lots Of Errors In Power Measurement ● Measuring Internally, Not At Power Input ● Component Tolerance Errors – Sense And Amplifier Resistors – Reference Voltages ● A/D Errors – Nonlinearities – Quantization Effects ● Sampling Related Errors Comprehensive Error Analysis Is A Big Job This Presentation Only About Sampling Related Errors Definition Of Power: Periodic Waveform 1 T P = ∫ v(t )i (t )dt T 0 Discrete Time Power Calculation V[5] V[4] V[6] V[3] V[7] V[2] V[8] V[1] V[9] V[0] V[10] TSAMPLE N −1 T/2 TSAMPLE T /2 = N 1 P = ∑ v[k ]i[ k ] N k =0 Discrete Time Power Calculation Discrete Time Power Calculation 7.00E+00 1400 6.00E+00 1200 Discrete Time Sum 1000 4.00E+00 800 3.00E+00 600 Continuous Time Integral 2.00E+00 1.00E+00 400 200 0.00E+00 0 0 1 2 3 4 5 6 7 8 9 10 -1.00E+00 -200 Sample/Time Difference Watts/Volts Joules 5.00E+00 Caveat Emptor ● ● ● ● ● Why Does This Work Out So Nicely? Perfect Sinusoid Fundamental Frequency Only Samples Uniformly Placed Nsamples × Tsample = Period Power Calculation Is Shift Invariant 0 1 2 3 4 5 6 7 8 9 10 7 1400 6 1200 5 1000 Joules 4 Discrete Time Sum 800 3 Continuous Time Integral 2 1 600 400 200 0 0 0 1 2 3 4 5 6 7 8 9 -1 10 -200 Sample/Time Difference Watts/Volts Power Calculation Is Shift Invariant Some Sampling Issues ● How Does The Number Of Samples In a Half Cycle Affect the Error? ● How Much Error Does A Time Difference Between The Voltage And Current Samples Introduce? ● How Much Error Does Not Being To Have An Exact Integer Number Of Samples In a Half Cycle Introduce? ● Does Sampling Over Many Half Cycles Reduce Or Increase The Error? Minimum Two Samples Per Half Cycle ● Two Sample Points Can Have Any Phase Shift And Arithmetic Still Works Minimum Two Samples Per Half Cycle ● Two Sample Points Can Have Any Phase Shift And Arithmetic Still Works ● Special Case: – Sample At Zero Crossing – Sample At The Peak ● Algorithm Is Now Simple ● Find The Peak And Divide By Two ● Just Have To Worry About Error Due To Waveform Distortion Average Power = ½ × Peak Power V And I Sampling Time Difference ● The Least Expense General Purpose Microcontrollers Have: – One A/D Converter – One Sample And Hold Circuit ● Cannot Sample Voltage And Current Simultaneously ● Time Difference Typically 20-40 µs ● Does This Affect Accuracy? ● Note: DSP/DSC Devices May Have More Than One Sample And Hold Circuit VOLTAGE V And I Sampling Time Difference START I SAMPLE CURRENT START V SAMPLE FINISH V SAMPLE FINISH I SAMPLE V And I Sampling Time Difference 0.20 0.1973% 0.10 Error (Joules) 0.00 0.1973% -0.10 -0.20 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 -0.30 0 µs -0.40 10 µs 20 µs 50 µs -0.50 100 µs Time (ms) 200 µs Tsample: 500 µs Line Frequency Variations ● Previous Examples Used Calculations Over Exactly One Half Or One Full Cycle Of The Line Frequency ● Consider Line Frequency Variation From 45 Hz To 65 Hz ● One Value Of TSAMPLE Will Not Permit Exact Alignment Over Line Frequency Range Line Frequency Variations: Constant N ● Keep Number Of Samples Per Half Cycle Constant ● Requires: – Detect Zero Crossings – Measuring Period – Adjusting TSAMPLE ● Detecting Zero Crossings Is Resource Intensive Line Frequency Variations: Constant TSAMPLE ● Vary Number Of Samples But Use Constant Sample Rate ● Don’t Have To Detect Zero Crossings Exactly ● Do Have To Figure Out When A Half Cycle Has Passed Line Frequency Variations: Constant TSAMPLE Voltage Past The Half Cycle Peak? Past The Zero Crossing Trough? Voltage Less Than Initial Value? Voltage GTE Initial Value? Line Frequency Variations: Constant TSAMPLE Error S Error As Function Of Line Frequency And Sample Time +12 W –16 W Question ●System Host Sends A READ_PIN Command To A Power Supply ●What Value Does The Power Supply Return? ●A: The Instantaneous V × I At That Moment ●B: The Power Based On The Last Half Cycle ●C: The Power Based On The Last Full Cycle ●D: The Power Averaged Over The Last Several Seconds Average Power Calculation ● Averaging Period (TAVERAGE) From 1 Second To 30 Seconds ● Number Of Half Cycle Values Needed For The Average: – 100 (1 Second, 50 Hz) – 3600 (30 Seconds, 60 Hz) How And When Is This Average Calculated? Possibility 1: Buffer & Calculate On Demand ● Create Circular Buffer This Value With Half CycleThis Power Only Calculations For Averaging Value Period 7 N N3 4 N- N- 5 Bu f Be fer f U p or e dat e fer f u B er Aft te a d p U 1 N- N- 4 5 7 N- 3 N- 2 N- N- N N- 6 2 N- N- Is Updated Too Much Time To Compute The Average 1 ● Update Buffer At Each Does New Half Cycle Not ● CalculateChange On Demand N- 6 N- Possibility 2: Keep A Running Sum Accumulator Always Has The Latest Information Possibility Three: Accumulate And Store ● Accumulator Adds Sample Values For The Averaging Time ● Average Is Calculated ● Average Is Put Into Storage Register, Overwriting Existing Value ● Advantages: – Minimal Hardware – Immediate Response ● Disadvantage: Data Can Be Up To One Averaging Period Old ACCUMULATE COMPUTE AVERAGE STORE IN REGISTER Averaging By Low Pass Filter ● A Low Pass Filter Is Sometimes Mentioned As A Way To Compute Average Power n Bit With MinimalWith Hardware Binary Integer ● A Low Pass Filter Generates A a < 1 Arithmetic: Weighted Average ak < 2-n => 0 a·a = a2 a·a2 = a3 Averaging By Low Pass Filter: Example ● P = 600 W – Constant In Averaging Time ● TAVERAGE = 10 seconds ● FAC = 50 Hz ● NSAMPLES = 1,000 – Summing Half Cycle Power Calculations ● 16 Bits Resolution (n = 16) Averaging By Low Pass Filter: Equations PLPF [n] = 1 N SAMPLES N SAMPLES P ∑ k =0 a k P[n − k ] N SAMPLES PLPF [n] = ∑ ak N SAMPLES Simple IIR k =0 LowNsamples Pass+1Filter a =0 Does Not Work Nsamples −n a = 2 ⇒ a = 0.988967896 PLPF = 54.386 W 600 W Summary ● Discrete Time Calculation Of Power Mostly Straightforward ● Optimizing Sampling Over Variations In Line Frequency Most Significant Challenge ● Averaging Requirements May Make For Expensive Hardware – Caution Is Needed Not To Overspecify