擴(kuò)散硅壓阻式壓力傳感器的壓力測(cè)量試驗(yàn)

實(shí)驗(yàn)十一擴(kuò)散硅壓阻式壓力傳感器的壓力測(cè)量實(shí)驗(yàn)一、實(shí)驗(yàn)?zāi)康模毫私鈹U(kuò)散硅壓阻式壓力傳感器測(cè)量壓力的原理與方法。二、實(shí)驗(yàn)儀器壓力傳感器模塊、溫度傳感器模塊、數(shù)顯單元、直流穩(wěn)壓源+5丫、±15丫。三、實(shí)驗(yàn)原理在具有壓阻效應(yīng)的半導(dǎo)體材料上用擴(kuò)散或離子注入法，摩托羅拉公司設(shè)計(jì)出X形硅壓力傳感器如下圖所示：在單晶硅膜片表面形成4個(gè)阻值相等的電阻條。并將它們連接成惠斯通電橋，電橋電源端和輸出端引出，用制造集成電路的方法封裝起來(lái)，制成擴(kuò)散硅壓阻式壓力傳感器。擴(kuò)散硅壓力傳感器的工作原理：在X形硅壓力傳感器的一個(gè)方向上加偏置電壓形成電流，，當(dāng)敏感芯片沒有外加壓力作用，內(nèi)部電橋處于平衡狀態(tài)，當(dāng)有剪切力作用時(shí)，在垂直電流方向?qū)?huì)產(chǎn)生電場(chǎng)變化E=Ap?1?，該電場(chǎng)的變化引起電位變化，則在端可得到被與電流垂直方向的兩測(cè)壓力引起的輸出電壓Uo。%=d-E=d-Ap-1 （11.1）式中d為元件兩端距離。實(shí)驗(yàn)接線圖如圖11-2所示，MPX10有4個(gè)引出腳，1腳接地、2腳為Uo+、3腳接+5V電源、4腳為Uo-；當(dāng)P1>P2時(shí)，輸出為正；P1<P2時(shí)，輸出為負(fù)。圖11-1擴(kuò)散硅壓力傳感器原理圖四、實(shí)驗(yàn)內(nèi)容與步驟1.接入+5丫、±15丫直流穩(wěn)壓電源，模塊輸出端Vo2接控制臺(tái)上數(shù)顯直流電壓表，選擇20V檔，打開實(shí)驗(yàn)臺(tái)總電源。.調(diào)節(jié)Rw2到適當(dāng)位置并保持不動(dòng)，用導(dǎo)線將差動(dòng)放大器的輸入端Ui短路，然后調(diào)節(jié)Rw3使直流電壓表200mV檔顯示為零，取下短路導(dǎo)線。.氣室1、2的兩個(gè)活塞退回到刻度“17”的小孔后，使兩個(gè)氣室的壓力相對(duì)大氣壓均為0,氣壓計(jì)指在“零”刻度處，將MPX10的輸出接到差動(dòng)放大器的輸入端Ui,調(diào)節(jié)Rw1使直流電壓表200mv檔顯示為零。.保持負(fù)壓力輸入P2壓力零不變，增大正壓力輸入P1的壓力到0.01MPa,每隔0.005Mpa記下模塊輸出Uo2的電壓值。直到P1的壓力達(dá)到0.095Mpa；填入下表1。.保持正壓力輸入P1壓力0.095Mpa不變，增大負(fù)壓力輸入P2的壓力，從0.01MPa每隔0.005Mpa記下模塊輸出Uo2的電壓值。直到P2的壓力達(dá)到0.095Mpa；填入下表2。.保持負(fù)壓力輸入P2壓力0.095Mpa不變，減小正壓力輸入P1的壓力，每隔0.005Mpa

記下模塊輸出Uo2的電壓值。直到P1的壓力為0.005Mpa；填入下表3。9.保持負(fù)壓力輸入P1壓力0Mpa不變，減小正壓力輸入P2的壓力，每隔0.005Mpa記下模塊輸出Uo2的電壓值。直到P2的壓力為0.005Mpa；填入下表。10.實(shí)驗(yàn)結(jié)束后，關(guān)閉實(shí)驗(yàn)臺(tái)電源，整理好實(shí)驗(yàn)設(shè)備。五、實(shí)驗(yàn)數(shù)據(jù)記錄與分析1、保持負(fù)壓力輸入P2壓力零不變，增大正壓力輸入P1的壓力P1(kP)0.010.0150.0200.0250.0300.0350.0400.0450.050Uo2(V)1.6082.283.023.544.224.895.456.076.63P1(kP)0.0550.0600.0650.0700.0750.0800.0850.0900.095Uo2(V)7.057.798.218.739.259.6210.0310.5811.04（表1）用matlab繪制P1-Uo2曲線圖如下圖所示P1-U唾線圖121100.0050.010.0150.020.0250.030.0350.040.0450.050.0550.060.0650.070.0750.080.0850.090.0950.1P1/kP2、保持正壓力輸入P1壓力0.095Mpa不變，增大負(fù)壓力輸入P2的壓力P2(kP)0.010.0150.0200.0250.0300.0350.0400.0450.050Uo2(V)10.149.769.328.828.367.857.416.796.23P2(kP)0.0550.0600.0650.0700.0750.0800.0850.0900.095Uo2(V)5.725.084.593.853.262.561.681.120.32（表2）用matlab繪制P2-Uo2曲線圖如下圖所示

減/」P2-Uc曲線10.0050.01 0.0150.020.0250.030.0350.040.0450.050.0550.060.0650.070.0750.080.0850.09P2/kPOV2OU10.0050.01 0.0150.020.0250.030.0350.040.0450.050.0550.060.0650.070.0750.080.0850.09P2/kPOV2OU234567890--------3、保持負(fù)壓力輸入P2壓力0.095Mpa不變，減小正壓力輸入P1的壓力P1(kP)0.0900.0850.0800.0750.0700.0650.0600.0550.050Uo2(V)-0.67-1.33-1.95-2.57-3.13-3.67-3.95-4.51-5.13P1(kP)0.0450.0400.0350.0300.0250.0200.0150.0100.005Uo2(V)-5.57-6.01-6.27-6.70-7.18-7.52-7.90-8.22-8.55（表3）用matlab繪制P1-Uo2曲線圖如下圖所示減小?1-U（曲線圖0.0050.010.0150.020.0250.030.0350.040.0450.050.0550.060.0650.070.0750.080.0850.09P1/kPO4-0.0050.010.0150.020.0250.030.0350.040.0450.050.0550.060.0650.070.0750.080.0850.09P1/kPO4-V2OU4、保持負(fù)壓力輸入P1壓力0Mpa不變，減小正壓力輸入P2的壓力P2(kP)0.0900.0850.0800.0750.0700.0650.0600.0550.050Uo2(V)-8.05-7.71-7.30-6.92-6.50-6.11-5.70-5.34-4.89P2(kP)0.0450.0400.0350.0300.0250.0200.0150.0100.005Uo2(V)-4.39-3.92-3.42-2.86-2.30-1.65-1.07-0.400.29（表4）用matlab繪制的P2-Uo2曲線圖如下圖所示減小?2-Uof曲線0.0050.01 0.0150.020.0250.03 0.0350.04 0.0450.05 0.0550.06 0.0650.070.0750.080.0850.09P2/kPro0.0050.01 0.0150.020.0250.03 0.0350.04 0.0450.05 0.0550.06 0.0650.070.0750.080.0850.09P2/kPro9-5、靈敏度及非線性誤差計(jì)算線性擬合直線的斜率即為所求的靈敏度，故對(duì)上面的曲線圖依次進(jìn)行線性擬合，如下圖所示（紅色為擬合直線），并由此算出其靈敏度和非線性誤差。靈敏度L=AU/AP=擬合直線斜率 非線性誤差3f=Am/yFSX100%（Am為輸出值與擬合直線的最大偏差，yF.S為滿量程輸出平均值）（1） P1-U（擬合直線0.0050.010.0150.020.0250.030.0350.040.0450.050.0550.060.0650.070.0750.080.0850.090.0950.1P1/kP2O864（1） P1-U（擬合直線0.0050.010.0150.020.0250.030.0350.040.0450.050.0550.060.0650.070.0750.080.0850.090.0950.1P1/kP2O864V-CMCOy=p1*x+p2pl=110.3p2=0.87657故靈敏度S=L=AU/AP=p1=110.3v/kpy=p1*x+p2pl=110.3p2=0.87657故靈敏度S=L=AU/AP=p1=110.3v/kp非線性誤差3f=(0.392/11.4)X100%=3.5%(2)由圖讀出Am=0.3920.0050.01 0.0150.020.025 0.030.035 0.040.045 0.050.055 0.060.065 0.070.075 0.080.0850.09 0.095 0.1氣壓P2/kPy=p1*x+p2 p1=-115.24 p2=11.765 由圖讀出Am=0.65842V-QJe故 靈敏度L=AU/AP=p1=-115.24v/kp非線性誤差3f=(0.65/10,72)X100%=6.1%減小P1-Uo2曲線圖減小P1-Uo2曲線圖0V0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08 0.085 0.09P1/kPy=p1*x+p2p1=91.498 p2=-9.3923 由圖讀出Am=0.45故靈敏度L=AU/AP=p1=91.498v/kp非線性誤差3f=(0.45/8.87)X100%=5.1%減小P2-Uo曲線0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08 0.085 0.09P2/kPy=p1*x減小P2-Uo曲線0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08 0.085