Download Sampling And Averaging Considerations For Measuring Input Power

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