Pressure Sensor Evaluation Dr. Lynn Fuller Lianna Dicke Webpage:

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ROCHESTER INSTITUTE OF TECHNOLOGY MICROELECTRONIC ENGINEERING Pressure Sensor Evaluation Lianna Dicke Webpage: http://people.rit.edu/lffeee 82 Lomb Memorial Drive Rochester, NY 14623-5604 Email: Lynn.Fuller@rit.edu Department webpage: http://www.microe.rit.edu 5-4-2016 PressureSensor.pptx Page 1

OUTLINE Introduction Theory Data Sheet Response Offset, Span, Linearity, Temperature, etc. Internal Electronics, Compensation Calibration Application Gas Mass Flow Measurements References Page 2

INTRODUCTION This document will evaluate a commercial pressure sensor and its application for gas flow measurements using Bernoulli s equation. The sensor is a single crystal diaphragm using piezoresistive sensors arranged in a bridge configuration. Pressure is applied to both top and bottom of the diaphragm creating a deflection proportional to the pressure difference. Thin silicon wafer Thickness = 10 µm Diameter 75 mm Page 3

SEM OF RIT PRESSURE SENSOR Front Back Page 4

CALCULATION OF EXPECTED OUTPUT VOLTAGE +5 Volts Vo1 R1 R3 R2 R4 Gnd Vo2 The equation for stress at the center edge of a square diaphragm (S.K. Clark and K.Wise, 1979) Stress = 0.3 P(L/H) 2 where P is pressure, L is length of diaphragm edge, H is diaphragm thickness For a 3000µm opening on the back of the wafer the diaphragm edge length L is 3000 2 (500/Tan 54.74 ) = 2290 µm Page 5

CALCULATION OF EXPECTED OUTPUT VOLTAGE (Cont.) Stress = 0.3 P (L/H) 2 If we apply vacuum to the back of the wafer that is equivalent to and applied pressure of 14.7 psi or 103 N/m 2 P = 103 N/m 2 L= 2290 µm H= 25 µm Stress = 2.49E8 N/m 2 Hooke s Law: Stress = E Strain where E is Young s Modulus s = E e Young s Modulus of silicon is 1.9E11 N/m 2 Thus the strain = 1.31E-3 or.131% Page 6

CALCULATION OF EXPECTED OUTPUT VOLTAGE (Cont.) The sheet resistance (Rhos) from 4 point probe is 61 ohms/sq The resistance is R = Rhos L/W For a resistor R3 of L=350 µm and W=50 µm we find: R3 = 61 (350/50) = 427.0 ohms R3 and R2 decrease as W increases due to the strain assume L is does not change, W becomes 50+50x0.131% W = 50.0655 µm R3 = Rhos L/W = 61 (350/50.0655) = 426.4 ohms R1 and R4 increase as L increases due to the strain assume W does not change, L becomes 350 + 350x0.131% R1 = Rhos L /W = 61 (350.459/50) = 427.6 ohms Page 7

CALCULATION OF EXPECTED OUTPUT VOLTAGE (Cont.) 5 Volts R1=427 R3=427 Vo1=2.5v R2=427 Gnd Vo2=2.5v R4=427 No stress Vo2-Vo1 = 0 With stress Vo2-Vo1 = 0.007v =7 mv R1=427.6 Vo1=2.4965v R2=426.4 5 Volts Gnd R3=426.4 Vo2=2.5035v R4=427.6 Page 8

IF RESISTORS ARE SINGLE CRYSTAL SILICON In addition to the effects of strain on the resistance if the resistor is made of single crystal silicon there is also a significant piezoresistive effect on the resistor value. Strain effects the mobility of holes and electrons in silicon. The resistors on the diaphragm of the pressure sensor drawn above have current flow longitudinal (R1 and R4) and transverse (R2 and R3) to the strain. The strain is tensile on the top surface of the diaphragm where the resistors are located if positive pressure is applied to the top of the diaphragm. The peizoresistive coefficient for R1 and R4 is 71.8 and for R2 and R3 is -66.3 E-11/Pa. The calculations above give the stress as 2.49E8 Pa thus the hole mobility will decrease in R1 and R4 (R increases in value) by 2.49E8 x 71.8e-11 = 17.9% while R2 and R3 (decrease in value) because the mobility increases by 2.49E8 x 66.3E-11 = 16.5%, thus the overall effect will be dominated by the piezoresistance rather than the effect of strain on the dimensions. Page 9

EXPRESSION FOR RESISTANCE R = Ro [ 1 + p L s xx + p T (s yy + s zz )] where Ro = (L/W)(1/(qµ(N,T) Dose)) p L is longitudinal piezoresistive coefficient p T is transverse piezoresistive coefficient s xx is the x directed stress, same direction as current s yy is the y directed stress, transverse to current flow s zz is the z directed stress, transverse to current flow In the <110> direction p L (E -11 /Pa) p T (E -11 /Pa) Electrons -31.6-17.6 holes 71.8-66.3 (100) wafer <110> directions Page 10

CALCULATION OF EXPECTED OUTPUT VOLTAGE FOR SINGLE CRYSTAL RESISTORS 5 Volts R1=427 R3=427 Vo1=2.5v R2=427 Gnd Vo2=2.5v R4=427 No stress Vo2-Vo1 = 0 R1=503.4 Vo1=2.073v With stress Vo2-Vo1 = 0.854V = 854 mv R2=356.5 5 Volts Gnd R3=356.5 Vo2=2.9275v R4=503.4 Page 11

SENSOR AND INTERNAL ELECTRONICS One of 26,830 pressure sensors Page 12

MS4515 DIFFERENTIAL PRESSURE SENSOR Measurement Specialties www.meas-spec.com Page 13

MS4515 SPECIFICATIONS Measurement Specialties Specifications Pressure Range 0-2 inches H 2 O Amplified Ratiometric Analog Output Differential Pressure, Port 1 Port 2 Temperature Compensated 3.3V or 5V operation 1/8 barbed pressure ports Response time 1mS www.meas-spec.com Page 14

PRESSURE CONVERSION CHART 2 in-h 2 O x 1.8651 mmhg/in-h 2 O = 3.73 mmhg 2 in-h 2 O x 0.0361 psi/in-h 2 O = 0.0722 psi Page 15

MS4515 Page 16

TRANSFER FUNCTION measured Page 17

SENSOR AND INTERNAL ELECTRONICS The sensor is a single crystal diaphragm using piezoresistive sensors arranged in a bridge configuration. Page 18

INTRODUCTION The Analog Discovery module is used with your computer and the free WaveForms software to turn your computer into a two channel oscilloscope, curve tracer, 16 channel logic analyzer, arbitrary waveform generator, 16 channel digital pattern generator, power supplies and voltmeters, network analyzer, spectrum analyzer and more. The module connects to your USB port. Find more information on line at www.digilentinc.com Academic price ~$159 Page 19

ANALOG DISCOVERY PIN-OUT Page 20

TESTING Page 21

TESTING Apply Pressure to Port 1 0, 0.5 1.0, 1.5, 2.0 Page 22

CONVERSION OF PRESSURE MEASUREMENTS Depth in H20 Voltage Pressure, Pa Pressure, psi 0 2.50 0 0 0.5 2.90 124.8 0.018 1.0 3.30 249.6 0.036 1.5 3.70 374.4 0.054 2.0 4.10 499.2 0.072 Maximum Allowed Sensitivity 0.80V/in-H20 Sensitivity 0.00321 V/Pa Sensitivity 22.2 V/psi Page 23

MEASURING PRESSURE THROUGH 1um PIPETTE TIP ~1µm at tip Made at RIT by Olivia Scheibel from Epindorf M6-B3-AL Page 24

MEASURING PRESSURE THROUGH 1um PIPETTE TIP First attempt: It works but water goes into the tip by capillary action making subsequent measurements incorrect. The response time might be too slow. Needs improvement. Possibly treat the glass tip so it is hydrophobic or some other technique to keep the water out of the tip or fill the tip and tubing with liquid. Page 25

FLOW FROM DIFFERENTIAL PRESSURE Bernoulli s equation for steady-state flow of an incompressible, frictionless fluid in a channel of varying dimension, the pressure and the velocity of the flow along a streamline (path of a fluid molecule or particle in steady flow), and for level flow (no effect of gravity due to height differences. P 2 + ½ r v 2 2 = P 1 + ½ r v 1 2 where P 1 and P 2 are pressures at two points v 1 and v 2 are velocities at two points r is fluid density Static energy plus dynamic energy at point 1 and 2 are equal. Page 26

DIFFERENTIAL PRESSURE P1 DP P2 Flow pt1 pt2 Orifice At pt1 velocity is lower and pressure P1 is higher than at pt2 At pt2 velocity is higher and pressure P2 is lower than at pt1 Total kinetic plus potential energy is same at pt1 and pt2 Page 27

FLOW FROM DIFFERENTIAL PRESSURE Page 28

EXPERIMENTAL DP FLOW MEASUREMENTS Experimental verification of the basic concept. A flow channel is made from 1 ½ inch PVC pipe with a fan at one end to create air flow. An orifice is placed between the two sections and a sensor is used to measure the differential pressure across the orifice. The differential Microelectronic pressure Engineering is zero with no flow and finite with flow. Page 29

EXPERIMENTAL DP FLOW MEASUREMENTS fan pressure sensor Vout = 1.352 fan on, DP = 1E-3 N/m 2 1.378 fan off, DP = 0 Page 30

EXAMPLE CALCULATIONS Assume: B 4 = 0, C = 1 and e = 1, all are dimensionless Qm = (P d 2 /4) 2 Dp r1 caution: these assumptions not realistic!! Qm is mass flow rate in Kg/s Example: d is the orifice diameter in meters, m 0.02 m Dp is differential pressure in Pa 7.97E-3 Pa r1 is density of gas in Kg/m 3 1.0 Kg/m 3 Giving: Qm = 3.96E-5 Kg/s B, C and e are correction factors that depend on temperature, pressure, Rochester compressibility Institute of Technology and dimensions of the flow channel and orifice. Page 31

MASS & VOLUMN FLOW RATE and VELOCITY Mass flow rate Qm = 3.96E-5 Kgm/s Volume flow rate Qv = 3.96E-5 m3/s = 39 cm 3 /s Velocity V = 3.4 cm/sec 3.81cm V ~3 cm/s 3.4cm Page 32

VELOCITY MEASUREMENTS WITH ANEMOMETER sensor Anemometer Gives Velocity of 0.04 m/s = 4 cm/s This value compares well with the velocity derived from the differential flow measurements, 3.4 cm/s Page 33

Mass Flow, g/s DATA WHEN USING 0.025m DIAMETER ORIFICE Power Supply(V) Fan Voltage (V) Output Voltage (V) Differential of Output Voltage Convert to Pa Mass Flow (g/s) Mass Flow vs Fan Voltage, 0.025m Diameter Orifice 3 0 1.449 0 0 0 3 6 1.46 0.011 1.142264 0.821176 3 7 1.464 0.015 1.557632 0.958927 3 8 1.469 0.02 2.076843 1.107273 3 9 1.476 0.027 2.803738 1.286535 3 10 1.484 0.035 3.634476 1.464785 3 11 1.492 0.043 4.465213 1.623581 3 12 1.5 0.051 5.29595 1.768174 5 0 2.54 0 0 0 5 6 2.561 0.021 1.308411 0.878871 5 7 2.569 0.029 1.806854 1.032796 5 8 2.578 0.038 2.367601 1.182244 5 9 2.589 0.049 3.05296 1.342497 5 10 2.601 0.061 3.800623 1.497891 5 11 2.614 0.074 4.610592 1.6498 5 12 2.627 0.087 5.420561 1.788855 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 12, 1.788854628 12, 1.768173532 0 2 4 6 8 10 12 14 Fan Voltage, V q m = π 4 d2 2 p ρ 1 3 Volts 5 Volts ρ 1 = 1.225 kg m 3 p = pressure difference in Pa Page 34

Mass Flow, g/s DATA WHEN USING 0.02m DIAMETER ORIFICE Power Supply(V) Fan Voltage (V) Output Voltage (V) Differential of Output Voltage Convert to Pa Mass Flow (g/s) 3 0 1.449 0 0 0 3 6 1.466 0.017 1.765317 0.653347 3 7 1.474 0.025 2.596054 0.7923 3 8 1.482 0.033 3.426791 0.910284 3 9 1.492 0.043 4.465213 1.039092 3 10 1.503 0.054 5.607477 1.164439 3 11 1.514 0.065 6.74974 1.277546 3 12 1.529 0.08 8.307373 1.41731 5 0 2.539 0 0 0 5 6 2.569 0.03 1.869159 0.672289 5 7 2.583 0.044 2.741433 0.814182 5 8 2.598 0.059 3.676012 0.942804 5 9 2.615 0.076 4.735202 1.070045 5 10 2.633 0.094 5.856698 1.190034 5 11 2.652 0.113 7.040498 1.304772 5 12 2.675 0.136 8.47352 1.431413 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 Mass Flow vs Fan Voltage, 0.020m Diameter Orifice 6 7 9 10 11 12, 1.417309718 0 0 2 4 6 8 10 12 14 Fan Voltage, V q m = π 4 d2 2 p ρ 1 ρ 1 = 1.225 kg m 3 p = pressure difference in Pa 8 12, 1.431412649 3 Volts 5 Volts Page 35

SUMMARY The data shows that the mass flow is not dependent on the supplied power to the pressure sensor. It depends on the fan voltage and to a smaller extent on the orifice size. The bigger the orifice diameter, the larger the flow. A commercial pressure sensor and its application for gas flow measurements has been evaluated. An experimental air flow differential pressure measurement apparatus was constructed and used to evaluate gas mass flow measurements based on Bernoulli s equation. Page 36

REFERENCES 1. Measurement Specialties, 45738 Northport Loop West, Fremont, CA 94538, 1-800-767-1888, www.meas-spec.com 2. www.digikey.com 3. Digilent, 1300 Henley Court, Pullman, WA 99136, 509-334-6306, www.digilentinc.com Page 37

HOMEWORK: PRESSURE SENSOR EVALUATION 1. Go to Digikey and search for pressure sensors. Find: lowest price item, lowest pressure, ten manufacturer names, select a pressures sensor that would be good for measuring blood pressure and view the data sheet. 2. How might a differential pressure sensor be used to make a Spirometer? 3. Think of some other applications for pressure sensors. Page 38