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MP21051MHz, 800mA Synchronous Step-Down ConverterThe Future of Analog IC TechnologyTMTMDESCRIPTIONThe MP2105 is a 1MHz constant frequency,current mode, PWM step-down converter. The device integrates a main switch and a synchronous rectifier for high efficiency without an external Schottky diode. It is ideal for powering portable equipment that runs from a single cell Lithium-Ion (Li+) battery. The MP2105 can supply 800mA of load current from a 2.5V to 6V input voltage. The output voltage can be regulated as low as 0.6V. The MP2105 can also run at 100% duty cycle for low dropout applications.The MP2105 is available in a low profile (1mm) 5-pin, TSOT package.EVALUATION BOARD REFERENCEBoard Number Dimensions EV2105DJ-00A2.0”X x 2.0”Y x 0.5”ZFEATURES• High Efficiency: Up to 95%• 1MHz Constant Switching Frequency • 800mA Available Load Current • 2.5V to 6V Input Voltage Range • Output Voltage as Low as 0.6V • 100% Duty Cycle in Dropout • Current Mode Control • Short Circuit Protection • Thermal Fault Protection • <0.1µA Shutdown Current•Space Saving 5-Pin TSOT23 PackageAPPLICATIONS• Cellular and Smart Phones• Microprocessors and DSP Core Supplies • PDAs • MP3 Players• Digital Still and Video Cameras • Portable Instruments“MPS” and “The Future of Analog IC Technology” are Trademarks of Monolithic Power Systems, Inc.TYPICAL APPLICATIONE F F I C I E N C Y (%)101001000LOAD CURRENT (mA)MP2105-EC01Efficiency vs Load CurrentMP2105 – 1MHz 800mA SYNCHRONOUS STEP-DOWN CONVERTER TMPACKAGE REFERENCE* For Tape & Reel, add suffix –Z (eg. MP2105DJ–Z)For Lead Free, add suffix –LF (eg. MP2105DJ–LF–Z) ABSOLUTE MAXIMUM RATINGS (1) V IN to GND..................................–0.3V to +6.5V V SW to GND...........................–0.3V to V IN +0.3V V FB, V EN to GND..........................–0.3V to +6.5V Junction Temperature.............................+150°C Lead Temperature..................................+260°C Storage Temperature .............–65°C to +150°C Recommended Operating Conditions (2) Supply Voltage V IN.............................2.5V to 6V Output Voltage V OUT..........................0.6V to 6V Operating Temperature.............–40°C to +85°C Thermal Resistance (3)θJA θJCTSOT23-5..............................220....110..°C/W Notes:1) Exceeding these ratings may damage the device.2) The device is not guaranteed to function outside of itsoperating conditions.3) Measured on approximately 1” square of 1 oz copper.ELECTRICAL CHARACTERISTICS (4)V IN = V EN = 3.6V, T A = +25°C, unless otherwise noted.Parameter SymbolCondition MinTypMaxUnits Supply Current V EN = V IN, V FB = 0.65V 440 600 µAShutdown Current V EN = 0V, V IN = 6V 0.10 1 µAIN Undervoltage Lockout Threshold RisingEdge 2.15 2.30 2.40 VIN Undervoltage LockoutHysteresis55 mVT A = +25°C 0.588 0.600 0.612Regulated FB Voltage V FB–40°C ≤ T A≤ +85°C 0.582 0.600 0.618VFB Input Bias Current V FB = 0.65V –50 0.5 +50 nA PFET On Resistance I SW100mA 0.42 ΩNFET On Resistance I SW–100mA 0.26 ΩSW Leakage Current V EN = 0V, V IN = 6V,V SW = 0V or 6V–1 +1 µAPFET Current Limit Duty Cycle = 100%,Current Pulse Width < 1ms1.2 1.62.1 AOscillator Frequency f OSC0.85 1.05 1.25 MHz Thermal Shutdown TripThreshold145 °CEN Trip Threshold –40°C ≤ T A≤+85°C 0.3 0.96 1.5 V EN Input Current V EN = 0V to 6V –1 +1 µA Notes:4) 100% production test at +25°C. Specifications over the temperature range are guaranteed by design and characterization.MP2105 – 1MHz 800mA SYNCHRONOUS STEP-DOWN CONVERTERTMTYPCIAL PERFORMANCE CHARACTERISTICSV IN = 3.3V, V OUT = 1.8V, L1 = 4.7µH, C1 = 4.7µF, C3 = 10µF, T A = +25°C, unless otherwise noted.MP2105-TPC03Load Transient(I OUT =0mA to 500mA step)Light Load Operation(IOUT =0mA)F E E D B AC K V O L T A G E (V )-40+200-20+40+60+80+100-40+200-20+40+60+80+100TEMPERATURE (°C)MP2105-TPC01E F F I C I E N C Y (%)MP2105-TPC04TEMPERATURE (°C)MP2105-TPC02V OUT 100mV/div.I OUT 0.5A/div.V OUT 10mV/div.I L0.2A/div.SW 2V/div.MP2105 – 1MHz 800mA SYNCHRONOUS STEP-DOWN CONVERTERTMTYPCIAL PERFORMANCE CHARACTERISTICS (continued)V IN = 3.3V, V OUT = 1.8V, L1 = 4.7µH, C1 = 4.7µF, C3 = 10µF, T A = +25°C, unless otherwise noted.MP2105-TPC09Short Circuit Protection(No Load)MP2105-TPC07Heavy Load OperationI OUT= 800mAMP2105-TPC10Short Circuit Recovery (No Load)MP2105-TPC08Startup from ShutdownV OUT 10mV/div.I L0.2A/div.L =0SW 2V/div.V OUT 1V/div.I L0.5A/div.V OUT 1V/div.I L0.5A/div.V EN 2V/div.V OUT 1V/div.I L0.5A/div.PIN FUNCTIONSPin # Name DescriptionEN 1Regulator Enable Control Input. Drive EN above 1.5V to turn on the MP2105. Drive EN below0.3V to turn it off (shutdown current < 0.1µA).GND 2 Ground SW 3 Power Switch Output. Inductor connection to drains of the internal PFET and NFET switches. IN 4 Supply Input. Bypass to GND with a 2.2µF or greater ceramic capacitor.FB 5Feedback Input. Connect FB to the center point of the external resistor divider. The feedbackthreshold voltage is 0.6V.MP2105 – 1MHz 800mA SYNCHRONOUS STEP-DOWN CONVERTERTMOPERATIONThe MP2105 is a constant frequency current mode PWM step-down converter. The MP2105 is optimized for low voltage, Li-Ion battery powered applications where high efficiency and small size are critical. The MP2105 uses an external resistor divider to set the output voltage from 0.6V to 6V. The device integrates both a main switch and a synchronous rectifier, which provides high efficiency and eliminatesan external Schottky diode. The MP2105 can achieve 100% duty cycle. The duty cycle D of a step-down converter is defined as:%100V V %100f T D INOUTOSC ON ×≈××= where T ON is the main switch on time, and f OSC is the oscillator frequency (1MHz).INENFBGNDSWMP2105_BD01Figure 1—Functional Block DiagramMP2105 – 1MHz 800mA SYNCHRONOUS STEP-DOWN CONVERTERTMCurrent Mode PWM ControlSlope compensated current mode PWM control provides stable switching and cycle-by-cycle current limit for superior load and line response and protection of the internal main switch and synchronous rectifier. The MP2105 switches at a constant frequency (1MHz) and regulates the output voltage. During each cycle the PWM comparator modulates the power transferred to the load by changing the inductor peak current based on the feedback error voltage. During normal operation, the main switch is turned on for a certain time to ramp the inductor current at each rising edge of the internal oscillator, and switched off when the peak inductor current is above the error voltage. When the main switch is off, the synchronous rectifier will be turned on immediately and stay on until either the next cycle starts. Dropout OperationThe MP2105 allows the main switch to remain on for more than one switching cycle and increases the duty cycle while the input voltage is dropping close to the output voltage. When the duty cycle reaches 100%, the main switch is held on continuously to deliver current to the output up tothe PFET current limit. The output voltage then is the input voltage minus the voltage drop across the main switch and the inductor.Short Circuit ProtectionThe MP2105 has short circuit protection. When the output is shorted to ground, the oscillator frequency is reduced to prevent the inductor current from increasing beyond the PFET current limit. The PFET current limit is also reduced to lower the short circuit current. The frequency and current limit will return to the normal values once the short circuit condition is removed and the feedback voltage reaches 0.6V.Maximum Load CurrentThe MP2105 can operate down to 2.5V input voltage; however the maximum load current decreases at lower input due to large IR drop on the main switch and synchronous rectifier. The slope compensation signal reduces the peak inductor current as a function of the duty cycle to prevent sub-harmonic oscillations at duty cycles greater than 50%. Conversely the current limit increases as the duty cycle decreases.APPLICATION INFORMATIONOutput Voltage SettingThe external resistor divider sets the output voltage (see Figure 3). The feedback resistor R1 also sets the feedback loop bandwidth with the internal compensation capacitor (see Figure 1). Choose R1 around 500k Ω for optimal transient response. R2 is then given by:1V6.0V 1R 2R OUT−=Table 1—Resistor Selection vs. OutputVoltage SettingV OUT R1 R2 1.2V 499k Ω (1%) 499k Ω (1%) 1.5V 499k Ω (1%) 332k Ω (1%) 1.8V 499k Ω (1%) 249k Ω (1%) 2.5V499k Ω (1%)158k Ω (1%)Inductor SelectionA 1µH to 10µH inductor with DC current rating at least 25% higher than the maximum load current is recommended for most applications. For best efficiency, the inductor DC resistance shall be <200m Ω. See Table 2 for recommended inductors and manufacturers. For most designs, the inductance value can be derived from the following equation:()OSCL IN OUT IN OUT f I V V V V L ×∆×−×=where ∆I L is Inductor Ripple Current. Choose inductor ripple current approximately 30% of the maximum load current, 800mA.The maximum inductor peak current is:2I I I LLOAD )MAX (L ∆+= Under light load conditions below 100mA, larger inductance is recommended for improved efficiency. Table 3 lists inductors recommended for this purpose.MP2105 – 1MHz 800mA SYNCHRONOUS STEP-DOWN CONVERTERTMTable 2—Suggested Surface Mount InductorsManufacturerPart NumberInductance (µH)Max DCR (Ω)Saturation Current (A)Dimensions LxWxH (mm 3)Coilcraft D01605T-472 4.70.150 1.20 5.4x4.2x1.8 Toko D52LC 4.7 0.087 1.14 5x5x2 Sumida CR43-4R7 4.7 0.109 1.15 4.3x4.8x3.5Table 3—Inductors for Improved Efficiency at 25mA, 50mA, under 100mA Load.ManufacturerPart NumberInductance (µH)Max DCR (Ω)Saturation Current (A)I RMS (A)Coilcraft DO1605T-103MX 100.3 1.0 0.9 Murata LQH4C100K04 10 0.2 1.2 0.8 Sumida CR32-100 10 0.2 1.0 0.7 Sumida CR54-100 100.1 1.21.4Input Capacitor SelectionThe input capacitor reduces the surge current drawn from the input and switching noise from the device. The input capacitor impedance at the switching frequency shall be less than input source impedance to prevent high frequency switching current passing to the input. Ceramic capacitors with X5R or X7R dielectrics are highly recommended because of their low ESR and small temperature coefficients. For most applications, a 4.7µF capacitor is sufficient. Output Capacitor SelectionThe output capacitor keeps output voltage ripple small and ensures regulation loop stable. The output capacitor impedance shall be low at the switching frequency. Ceramic capacitors with X5R or X7R dielectrics are recommended. The output ripple ∆VOUT is approximately:()⎟⎟⎠⎞⎜⎜⎝⎛××+×××−×≤∆3C f 81ESR L f V V V V V OSC OSC IN OUT IN OUT OUTPC Board LayoutThe high current paths (GND, IN and SW)should be placed very close to the device with short, direct and wide traces. Input capacitor C1 needs to be as close as possible to the IN and GND pins. The external feedback resistors shall be placed next to the FB pin. Keep the switching node SW short and away from the feedback network. Figure 2 illustrates an example of PCB layout and signal routing.Figure 2—MP2105 Suggested LayoutMP2105 – 1MHz 800mA SYNCHRONOUS STEP-DOWN CONVERTERNOTICE: The information in this document is subject to change without notice. Please contact MPS for current specifications. Users should warrant and guarantee that third party Intellectual Property rights are not infringed upon when integrating MPS products into any application. MPS will not assume any legal responsibility for any said applications.TMPACKAGE INFORMATIONTSOT23-5Gauge Plane0.25 BSC.±0.10-+0.4000°0°4°(2 plcs)10°TYP.2.90 BSC(5PLCS)0.300(Min)LC TYP .0.950LC TYP .0.9502.80 B S C1.60 B S C330.500(Max)SEATING PLANE0.87±0.030.00-0.101.00M a x .(2 plcs)。

七年级下册数学《第九章不等式与不等式组》专题解一元一次不等式组（计算题共50题）1．（2022秋•越秀区校级期末）解不等式组：5−1＞4+2≥2−4．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：5−1＞4+2①≥2−4②，由①得：x＞3，由②得：x≤4，则不等式组的解集为3＜x≤4．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．2．（20231≤3+2．【分析】先求出不等式组中每一个不等式的解集，再求出它们的公共部分即可求解．1−≤3+2，由3K23＞1得x＞53，由4x﹣5≤3x+2得x≤7，故不等式组的解集为53＜x≤7．【点评】本题考查了解一元一次不等式组．解集的规律：同大取大；同小取小；大小小大中间找；大大小小找不到．3．（20233−1−2＜K56．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：解不等式5x≥3x﹣1得：x≥−12，解不等式r23−2＜K56得：x＜3，则不等式组的解集为−12≤x＜3．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．4．（20231≤−+1+23．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．1≤−+1①+23②，由①得：x≤23，由②得：x＞﹣1，则不等式组的解集为﹣1＜x≤23．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．5．（2023•陕西模拟）解不等式组：2+5≤3(+2)−1＜2．【分析】分别解两个不等式，然后根据大小小大中间找确定不等式组的解集．【解答】解：2+5≤3(+2)①−1＜2②，解不等式①得：x≥﹣1，解不等式②得：x＜3，∴不等式组的解集为：﹣1≤x＜3．【点评】本题考查了解一元一次不等式组：解一元一次不等式组时，一般先求出其中各不等式的解集，再求出这些解集的公共部分得到不等式组的解集．6．（2023•安徽模拟）解不等式组2+1≤4−−1＜32．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：2+1≤4−s−1＜32②，由①得x≤1，由②得：x＞﹣2，则不等式组的解集为﹣2＜x≤1．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．7．（2023≥+1≤．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：由3x﹣5≥x+1，得：x≥3，由3K42≤x，得：x≤4，则不等式组的解集为：3≤x≤4．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．8．（2023−3)≤−1＞0．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．−3)≤s−1＞0②，解不等式①得：x≥113，解不等式②得：x＞3，则不等式组的解集为x≥113．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．9．（2023−1)≤4−1．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：解不等式①得：≥−12，不等式②得：x＜4，∴不等式组的解集为：−12≤＜4．【点评】本题考查了解一元一次不等式组，正确掌握一元一次不等式解集确定方法是解题的关键．10．（20233≤13−2＜−1．【分析】先求出每个不等式的解集，再根据不等式的解集求出不等式组的解集即可．3≤13①−2＜−1②，由①得x≤2，由②得x＞﹣2，∴不等式组的解集为﹣2＜x≤2．【点评】本题主要考查解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．11．（2023+2)≥2+51＜K22并把它的解集在数轴上表示出来．【分析】分别求出每一个不等式的解集，将解集表示在数轴上，根据数轴求得不等式的解集即可求解．【解答】解：解不等式①得，x≥﹣1，解不等式②得，x＞0，所以不等式组的解集为x＞0．这个不等式组的解集在数轴上表示如图：【点评】本题考查了解一元一次不等式组，在数轴上表示不等式的解集，数形结合是解题的关键．12．（20232)＞8+9①2＞r23②．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：解不等式①，得：x＜32，解不等式②，得：x＞﹣5，则不等式组的解集为﹣5＜x＜32．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．13．（2023−7＜3(+1)−1≥7−32．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．−7＜3(+1)①−1≥7−32t，解不等式①得：x＜5，解不等式②得：x≥4，则不等式组的解集为4≤x＜5．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．14．（2023•碑林区校级三模）解不等式组：2(−2)≤3−1−2r13＞+1．【分析】首先解每个不等式，两个不等式的解集的公共部分就是不等式组的解集．【解答】解：2(−2)≤3−①1−2r13＞+1②，解①得：x≤73，解②得x＜−15．故不等式组的解集是：x＜−15．【点评】本题考查了一元一次不等式组的解法：解一元一次不等式组时，一般先求出其中各不等式的解集，再求出这些解集的公共部分，15．（2023−1)＜72≥．【分析】先解每个不等式，再求两个不等式解集的公共部分即可．−1)＜7①+2≥t，解不等式①得，x＜3，解不等式②得，x≤2，∴不等式组的解集为x≤2．【点评】本题考查了解一元一次不等式组，熟练掌握解一元一次不等式组的步骤是解题的关键．16．（2023•香洲区校级一模）解不等式组：4−2≤3(+1)①1−K12＜4②．【分析】分别求出不等式组中两不等式的解集，找出两解集的公共部分即可．【解答】解：由①得x≤5，由②得x＞2，故不等式组的解集为2＜x≤5．【点评】本题考查的是解一元一次不等式组，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．17．（20231＜−+21+23．【分析】分别将每个一元一次不等式求解，然后求出公共解集即可．【解答】解：解不等式2x﹣1＜﹣x+2，得x＜1，解不等式K12＜1+23，得x＞﹣5，故不等式组的解集是：﹣5＜x＜1．【点评】本题考查了解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．18．（20232≥4+1K32+1．【分析】分别解两个不等式，求解集的公共部分即可．2≥4+1①K32+1②解不等式①得：x≥﹣1，解不等式②得：x＜3．∴不等式组的解集为﹣1≤x＜3．【点评】本题考查解一元一次不等式组，解题关键是熟练掌握解一元一次不等式的步骤．19．（20233)＜41≤2r13．【分析】分别求出不等式组中两不等式的解集，找出两解集的公共部分即可．3)＜4s−1≤2r13②，由①得：x＞﹣3，由②得：x≤1，∴不等式组的解集为﹣3＜x≤1．【点评】此题考查了解一元一次不等式组，熟练掌握不等式组的解法是解本题的关键．20．（20231≤7−32K12+1．【分析】先解出每个不等式的解集，即可得到不等式组的解集，然后写出相应的整数解即可．1≤7−32①K12+1②解不等式①，得：x≤4，解不等式②，得：x＞﹣1，∴不等式组的解集是﹣1＜x≤4．【点评】本题考查解一元一次不等式组，熟练掌握解一元一次不等式的方法是解答本题的关键．1．（2023•河北区一模）解不等式组2＞−4①+3≤5②．请结合解题过程，完成本题的解答．（Ⅰ）解不等式①，得；（Ⅱ）解不等式②，得；（Ⅲ）把不等式①和②的解集在数轴上表示出来：（Ⅳ）原不等式组的解集为．【分析】根据解一元一次不等式组的方法，可以解答本题．【解答】解：2＞−4①+3≤5②，解不等式①，得x＞﹣2，解不等式②，得x≤2，把不等式①和②的解集在数轴上表示出来：故原不等式组的解集为﹣2＜x≤2．故答案为：x＞﹣2，x≤2，﹣2＜x≤2．【点评】本题考查了解一元一次不等式组、在数轴上表示不等式组的解集，掌握解一元一次不等式组的方法是关键．2．（2023•河西区模拟）解不等式组+5≥4，①4≥7−6．②请结合题意填空，完成本题的解答．（Ⅰ）解不等式①，得；（Ⅱ）解不等式②，得；（Ⅲ）把不等式①和②的解集在数轴上表示出来：（Ⅳ）原不等式组的解集为．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：+5≥4①4≥7−6②，解不等式①，得x≥﹣1，解不等式②，得x≤2，把不等式①和②的解集在数轴上表示出来：∴原不等式组的解集：﹣1≤x≤2．故答案为：x≥﹣1；x≤2；﹣1≤x≤2．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．3．（2023＜7①2≥+1②请按下列步骤完成解答．（1）解不等式①，得；（2）解不等式②，得；（3）把不等式①和②的解集在数轴上表示出来；（4）原不等式组的解集是．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：（1）解不等式①，得x＜4；（2）解不等式②，得x≥3；（3）把不等式①和②的解集在数轴上表示出来：（4）原不等式组的解集为3≤x＜4，故答案为：x＜4，x≥3，3≤x＜4．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．4．（2023•南昌模拟）解不等式组3＜92＞−3+5，并将解集在数轴上表示出来．【分析】先解出每个不等式的解集，即可得到不等式组的解集，然后在数轴上表示出其解集即可．【解答】解：解不等式3x＜9可得：x＜3；解不等式2x＞﹣3x+5可得：x＞1；故原不等式组的解集是1＜x＜3．其解集在数轴上表示如下所示：．【点评】本题考查解一元一次不等式组，解答本题的关键是明确解一元一次不等式的方法．5．（2023+3＞−K13≤1，并把它的解集在数轴上表示出来．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：由2x+3＞x得：x＞﹣3，由2−K13≤1得：x≤4，则不等式组的解集为﹣3＜x≤4，将解集表示在数轴上如下：【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．6．（2023春•东台市月考）解不等式组并将其解集在数轴上表示：3−2＜42(−1)≤3+1．【分析】分别求出不等式组中两不等式的解集，找出两解集的公共部分即可．【解答】解：3−2＜4①2(−1)≤3+1②，由①得：x＜2，由②得：x≥﹣3，则不等式组的解集为﹣3≤x＜2．．【点评】此题考查了解一元一次不等式组，以及在数轴上表示不等式的解集，熟练掌握不等式组的解法是解本题的关键．7．（20232＞3(−1)≤7−，并把解集在数轴上表示出来．【分析】先求出每个不等式的解集，再求出不等式组的解集，最后在数轴上表示出来即可．2＞3(−1)①≤7−t，解不等式①得：x＞−12，解不等式②得：x≤5，∴不等式组的解集为：−12＜x≤5，在数轴上表示不等式组的解集为：．【点评】本题考查了解一元一次不等式组，在数轴上表示不等式组的解集的应用，解此题的关键是能根据不等式的解集求出不等式组的解集．8．（2023•鼓楼区校级模拟）解不等式组，并把它的解集表示在数轴上：−1)≤3(1+p①−K12②．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：解不等式①得：x≤5，解不等式②得：x＞﹣1，则不等式组的解集为﹣1＜x≤5，将不等式组的解集表示在数轴上如下：【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．9．（2023＜6K12，并把它的解集在数轴上表示出来．【分析】分别求出各不等式的解集，再求出其公共解集并在数轴上表示出来即可．＜6①K12②，由①得，x＜1，由②得，x＞﹣1，故不等式组的解集为﹣1＜x＜1，在数轴上表示为：【点评】本题考查的是解一元一次不等式组及在数轴上表示不等式组的解集，熟知同大取大；同小取小；大小小大中间找；大大小小找不到的原则是解题的关键．10．（2023＞3(−1)．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解；解不等式5x+3＞3（x﹣1），得：x＞﹣3，解不等式8r29＞，得x＜2，则不等式组的解集为﹣3＜x＜2．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．11．（2023•蜀山区校级模拟）解不等式组：3−1≥+1+4＜4−2．并在数轴上表示它的解集．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：由3x﹣1≥x+1得：x≥1，由x+4＜4x﹣2得：x＞2，则不等式组的解集为x＞2，将不等式组的解集表示在数轴上如下：【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．12．（20234≥2−1，并将解集在数轴上表示出来．【分析】分别计算出方程组中两个不等式的解集，两个解集的公共部分就是不等式组的解集．4≥2−1①②解不等式①，得：x＜﹣1；解不等式②，得：x≤3；在数轴上表示为：∴这个不等式组的解集为x＜﹣1．【点评】此题考查一元一次不等式组的解集，在数轴上表示不等式的解集，解题关键在于掌握运算法则．13．（2023−3＜4s14≤r12②，并把它的解集在数轴上表示出来．【分析】先求出不等式组的解集，然后根据数轴上不等式组的解集表示出来即可．−3＜4①14≤r12②，解不等式①，得：x＜3，解不等式②，得：x≥﹣2，∴该不等式组的解集为：﹣2≤x＜3，把该不等式组的解集在数轴上表示为：【点评】本题考查了一元一次不等式组的解法以及数轴上表示不等式的解集，解题关键是熟练掌握确定不等式组解集的口诀：同大取大、同小取小、大小小大中间找、大大小小找不到．14．（2022−1＜3(−1)K22≥13，并把解集在数轴上表示出来．【分析】首先解每一个不等式，求得每一个不等式的解集，即可求得该不等式组的解集，再在数轴上表示出来即可．【解答】解：由5x﹣1＜3（x﹣1）得：5x﹣1＜3x﹣3，解得x＜﹣1，由23−K22≥13得：4x﹣3x+6≥2，解得x≥﹣4，故原不等式组的解集为﹣4≤x＜﹣1，把解集在数轴上表示出来，如下图：【点评】此题主要考查了解一元一次不等式组，关键是正确掌握解集的规律：同大取大；同小取小；大小小大中间找；大大小小找不到．在数轴上表示解集时，“≥”，“≤”要用实心圆点表示；“＜”，“＞”要用空心圆点表示．15．（20231)＜3−2①1≤r22②并将其解集在数轴上表示出来．【分析】先解出每个不等式的解集，即可得到不等式组的解集，然后在数轴上表示出其解集即可．1)＜3−2①−1≤r22②，解不等式①，得：x＜2，解不等式②，得：x≥﹣6，∴原不等式组的解集是﹣6≤x＜2，其解集在数轴上表示如下：．【点评】本题考查解一元一次不等式组，解答本题的关键是明确解一元一次不等式的方法．1．（20233)≤−4在数轴上表示出它的解集，并求出它的整数解．【分析】分别求出不等式组中两不等式的解集，找出两解集的公共部分求出不等式组的解集，进而求出整数解即可．3)≤−4①t ，由①得：x ≤2，由②得：x ＞﹣2，∴不等式组的解集为﹣2＜x ≤2，解集表示在数轴上，如图所示：则不等式组的整数解为﹣1，0，1，2．【点评】此题考查了解一元一次不等式组，以及在数轴上表示不等式的解集，熟练掌握不等式组的解法是解本题的关键．2．（2023•鼓楼区一模）解不等式组4(−1)＞3−22−3≤5，并写出该不等式组的整数解．【分析】首先解每个不等式，两个不等式的解集的公共部分就是不等式组的解集，然后确定整数解即可．【解答】解：4(−1)＞3−2①2−3≤5②，解①得x ＞2，解②得x ≤4．则不等式组的解集是：2＜x ≤4．则整数解是：3，4．【点评】本题考查了一元一次不等式组的解法：解一元一次不等式组时，一般先求出其中各不等式的解集，再求出这些解集的公共部分，解集的规律：同大取大；同小取小；大小小大中间找；大大小小找不到．3．（2022秋•道县期末）解不等式组3−2＜4①2(−1)≤3+1②，并求出它的非负整数解．【分析】【先分别解不等式，求出不等式组的解集，然后找出负整数解．【解答】解：解①得：x＜2，解②得：x≥﹣3，∴不等式组的解集为﹣3≤x＜2，∴不等式组的非负整数解为0，1．【点评】本题考查了解一元一次不等式组，解题关键是求不等式的公共解，要遵循以下原则：同大取较大，同小取较小，小大大小中间找，大大小小无解了．4．（2022≤3(+1)≥−1的最大整数解．【分析】先求出不等式组的解集，再求出最大整数解即可．【解答】解：由5x﹣1≤3（x+1），得：x≤2；由1+23≥−1，得：x≤4；∴不等式组的解集为：x≤2，∴不等式组的最大整数解为：2．【点评】本题考查求不等式组的整数解．正确的求出不等式组的解集，是解题的关键．5．（2022秋•湘潭县期末）求不等式组4−7＜5(−1)2≤18−3+7的正整数解．【分析】先求出不等式组的解集，再求出正整数解即可．【解答】解：4−7＜5(−1)①2≤18−3+7②，解不等式①得：x＞﹣2，解不等式②得：x≤5，∴不等式组的解集为：﹣2＜x≤5，其中正整数解是1，2，3，4，5．【点评】本题考查了解不等式组及不等式组的解集，熟练掌握不等式组的解法是解决问题的关键．6．（2023•长清区校级开学）解不等式组：2+＞7−4＜4+2，并求出所有整数解的和．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：由2+x＞7﹣4x，得：x＞1，由x＜4+2，得：x＜4，则不等式组的解集为1＜x＜4，所有整数解的和为2+3＝5．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．7．（2023−1)≥1−1，并写出它的所有非负整数解．【分析】分别求出两个不等式的解集，然后求出两个解集的公共部分，再写出范围内的非负整数解即可．−1)≥1①−1②，解不等式①得，x≤1，解不等式②得，x＞﹣3，所以不等式组的解集是﹣3＜x≤1，所以不等式组的非负整数解是0、1．故答案为：0、1．【点评】本题主要考查了一元一次不等式组解集的求法，其简便求法就是用口诀求解．求不等式组解集的口诀：同大取大，同小取小，大小小大中间找，大大小小找不到（无解）．8．（2022秋•鄞州区期末）解不等式组：−4＜2+3−2≤1，并求出所有满足条件的整数之和．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：由x﹣4＜2x，得x＞﹣4，由x+3−2≤1，得：x≤﹣1，则不等式组的解集为﹣4＜x≤﹣1，不等式组的整数解的和为﹣3﹣2﹣1＝﹣6．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．9．（2023−2)＞4≥3r26−1并写出该不等式组的最小整数解．【分析】分别求出每一个不等式的解集，根据口诀：同大取大、同小取小、大小小大中间找、大大小小找不到确定不等式组的解集．【解答】解：由x﹣3（x﹣2）＞4，得：x＜1，由2K13≥3r26−1，得：x≥﹣2，则不等式组的解集为﹣2≤x＜1，∴该不等式组的最小整数解为﹣2．【点评】本题考查的是解一元一次不等式组，正确求出每一个不等式解集是基础，熟知“同大取大；同小取小；大小小大中间找；大大小小找不到”的原则是解答此题的关键．10．（2023−1)≥1−5r12＜1，并写出它的整数解．【分析】分别求出不等式组中两不等式的解集，找出两解集的公共部分确定出不等式组的解集，进而求出整数解即可．−1)≥1①−5r12＜1②，由①得：x≤1，由②得：x＞﹣1，∴不等式组的解集为﹣1＜x≤1，则不等式组的整数解为0，1．【点评】此题考查了一元一次不等式组的整数解，以及解一元一次不等式组，熟练掌握运算法则是解本题的关键．11．（2022+22r15，并直接写出这个不等式组的所有负整数解．【分析】先解出每个不等式的解集，即可得到不等式组的解集，然后即可写出这个不等式组的所有负整数解．+2①2r15②，解不等式①，得：x＜1，解不等式②，得：x＞﹣3，∴该不等式组的解集为﹣3＜x＜1，∴这个不等式组的所有负整数解是﹣2，﹣1．【点评】本题考查解一元一次不等式组、一元一次不等式组的整数解，解答本题的关键是明确解一元一次不等式的方法．12．（2022春•大兴区校级期中）解不等式组4(+1)≤7+10−5＜K83，并求出这个不等式组的所有的正整数解．【分析】求出每个不等式的解集，根据找不等式组解集的规律找出即可．【解答】解：4(+1)≤7+10①−5＜K83②，解不等式①得：x≥﹣2，解不等式②得：x＜72，所以不等式组的解集为：−2≤＜72，所以不等式组的所有正整数解为：1，2，3．【点评】本题考查了一元一次不等式组的整数解的应用，关键是能根据不等式的解集找出不等式组的解集．13．（2023−5r12≤1＜3(+1)，在数轴上表示它的解集，并写出它的最大整数解和最小整数解．【分析】先求出每个不等式的解集，再求出不等式组的解集即可．−5r12≤1①＜3(+1)②，∵解不等式①得：x≥﹣1，解不等式②得：x＜2，∴不等式组的解集为：﹣1≤x＜2，在数轴上表示不等式组的解集为：，∴不等式组的最大整数解为：1，最小整数解为：﹣1．【点评】本题考查了解一元一次不等式组，在数轴上表示不等式组的解集的应用，解题的关键是掌握不等式组的解法．14．（2022•会东县校级模拟）解不等式组3(−1)＜5+1(−1)≥2−4并求它的所有的非负整数解．【分析】首先解每个不等式，两个不等式的解集的公共部分就是不等式组的解集，然后确定非负整数解即可．【解答】解：3(−1)＜5+1①(−1)≥2−4②，解①得x＞﹣2，解②得x≤3．则不等式组的解集是：﹣2＜x≤3．则非负整数解是：0，1、2、3．【点评】本题考查了一元一次不等式组的解法：解一元一次不等式组时，一般先求出其中各不等式的解集，再求出这些解集的公共部分，解集的规律：同大取大；同小取小；大小小大中间找；大大小小找不到．15．（2023•鼓楼区模拟）解关于x的不等式组：4(+1)≤7+102−3＜K12，并求出它所有整数解的和．【分析】先求出两个不等式的解集，再求其公共解，然后写出范围内的整数求其和即可．【解答】解：4(+1)≤7+10①2−3＜K12②，解不等式①得，x≥﹣2，解不等式②得，x＜53，所以不等式组的解集为﹣2≤x＜53，所以原不等式组的整数解是﹣2、﹣1、0、1，所以所有整数解的和为﹣2．【点评】本题主要考查了一元一次不等式组解集的求法，其简便求法就是用口诀求解．求不等式组解集的口诀：同大取大，同小取小，大小小大中间找，大大小小找不到（无解）．。

小学五年级上册英语第3单元期末试卷(含答案含解析)英语试题一、综合题(本题有50小题，每小题2分，共100分.每小题不选、错误，均不给分)1.My brother _______ play football every day.2.What is the opposite of "sad"?A. HappyB. AngryC. ExcitedD. Sleepy3.Which of these is a number?A. SevenB. ChairC. CarD. Spoon4.What time of day do we usually eat breakfast?A. MorningB. AfternoonC. EveningD. Night5.Which one is a day of the week?A. BananaB. MondayC. RedD. Carrot6.Which one is a vegetable?A. BananaB. TomatoC. CarrotD. Orange7.Tom and his sister Lily are going to the library. Tom wants to borrow a book about __________, while Lily is looking for a __________ book. They both love reading and spend time picking out the books they want to borrow. Afterward, they go to the__________ to check out their books.8.How many days are there in a week?A. FiveB. SixC. SevenD. Eight9.Which of these is a color?A. CarB. GreenC. HouseD. Tree10.Tom ______ (study) hard for his English test. Yesterday, he ______ (spend) three hours reviewing his notes. Today, he ______ (feel) confident about the test. He ______ (believe) he ______ (do) well.11.Which of these is a school subject?A. MathB. SpoonC. TableD. Car12.My parents __________ (visit) their friends next month.13.Which of these is a type of fruit?A. AppleB. CarrotC. PotatoD. Cucumber14.Which of these is a drink?A. MilkB. PlateC. ForkD. Shoe15.Which of these is a color?A. RedB. DogC. TableD. Shoe16.My mom _______ (love) gardening. She _______ (plant) flowers and vegetables in the garden every spring. This year, she _______ (grow) tomatoes and cucumbers. She always _______ (ask) me to help her water the plants.17.They _______ their homework in the classroom now.18.Which one is used for drinking?A. ForkB. CupC. KnifeD. Plate19.I __________ (1) my homework before playing video games. It __________ (2) a long time, but I __________ (3) it quickly today. After finishing, I __________ (4) a game with my friends.20.David is very excited because it is his birthday today. He invites all his friends to his house for a __________. There are many balloons and __________ hanging from the ceiling. David’s mom has baked a big chocolate __________, and everyone sings "Happy Birthday" to him.21.I ________ (be) in the library at the moment.22.We ________ (study) English every day.23.Which of these is a tool?A. WrenchB. PlateC. SpoonD. Chair24.I ________ (do) my homework every evening.25.What is the past tense of "run"?A. RunningB. RanC. RunedD. Runs26.Anna is learning to play the piano. She practices every __________ after school. Anna’s teacher, Mr. Lee, teaches her how to read __________ and play simple__________. Anna is very proud of herself because she can now play a __________ tune. She also enjoys practicing with her __________, who plays the __________. Together, they make beautiful music and hope to perform at the __________ concert.27.We __________ (1) a field trip to the museum last Friday. It __________ (2) a bit rainy in the morning, but by the time we __________ (3) the museum, the weather__________ (4) sunny again. We __________ (5) many interesting things about history and science. After the museum visit, we __________ (6) to a nearby park and__________ (7) some pictures. It __________ (8) a wonderful day.28.Which of these is a color?A. GreenB. TableC. ChairD. Book29.My grandparents __________ (live) in the countryside. Every summer, we__________ (visit) them and __________ (help) with the farm work. Last year, we__________ (pick) apples and __________ (make) apple pie.30.We ______ (visit) the museum last weekend. We ______ (see) many interesting things, including dinosaur bones.31.What do you wear on your feet?A. ShoesB. GlovesC. HatD. Scarf32.What do we use to write?A. SpoonB. PenC. ShoesD. Cup33.What is the opposite of "slow"?A. FastB. TallC. BigD. Heavy34.My brother _______ a big dog.35.What is the opposite of "hot"?A. WarmB. ColdC. CoolD. Sweet36.I ______ (have) a cat named Whiskers. He ______ (like) to sleep on my bed. Every morning, he ______ (jump) onto my bed and ______ (purr) loudly. I ______ (give) him food, and he ______ (eat) it happily.37.I __________ (not/like) doing homework, but I __________ (do) it every day. My brother __________ (help) me with my math problems. After we __________ (finish) our homework, we __________ (play) games together. I __________ (always) enjoy that time.38.It __________ (be) my birthday yesterday, and I __________ (celebrate) it with my friends. We __________ (eat) cake, __________ (play) games, and __________ (have) a lot of fun.39.It ______ (be) a sunny day when we ______ (go) to the park. We ______ (bring) a ball and ______ (play) soccer. After we ______ (finish) playing, we ______ (sit) on the grass and ______ (talk) about our favorite games. My friend Sarah ______ (tell) me that she ______ (like) to play badminton, so we ______ (decide) to play it next time.40.Yesterday, my friends and I _______ (go) to the park. We _______ (play) soccer for an hour. It _______ (be) a very sunny day, so we _______ (take) off our jackets and _______ (run) around. After playing, we _______ (sit) on the grass and _______ (talk) about our favorite sports.41.Jake is getting ready for school. He puts on his __________ and __________ before leaving the house. Jake also packs his __________, including his books and lunchbox. Before he leaves, he checks to make sure he has his __________, so he can catch the bus.42.She _______ the bus to school every day.43.How many fingers do humans usually have?A. 8B. 10C. 12D. 1444.I ______ (study) English every day after school. Yesterday, I ______ (study) for my English test, and I ______ (feel) confident. After the test, I ______ (celebrate) with my friends at the café.45.This afternoon, I __________ (1) to the library with my friend Lisa. We__________ (2) some books about animals. I __________ (3) a book about dolphins, and Lisa __________ (4) a book about lions. After reading, we __________ (5) to the caféand __________ (6) some hot chocolate.46.Jack and his friends are going to the park to play basketball. They have divided into two teams. Jack is playing on the __________ team, and his friend Tom is on the__________ team. They are using a __________ to play the game. After a fun match, they will go to the __________ to get a cold drink and rest. Jack’s team is hoping to win, but they know it will be a __________ game.47.My brother ______ (like) to play basketball, but he ______ (not/play) it every day. Last week, he ______ (practice) with his friends after school and ______ (score) a lot of points.48.What do you wear to keep warm in winter?A. TshirtB. ShortsC. JacketD. Sandals49.Which of these is used to cut food?A. SpoonB. PlateC. KnifeD. Cup50.What do we wear on our head?A. HatB. GlovesC. ShoesD. Scarf(答案及解释)。

2361 呼吸阀结构及工作原理呼吸阀是储罐的一个重要安全附件，当罐内气体的压力超过呼吸阀的整定压力值时，压力阀顶开，真空阀仍处于关闭状态，罐内上部气体从罐内呼出，使罐内的压力不再继续增高；而当罐内气体的真空度超过储罐的设计真空度时，真空阀开，压力阀仍处于关闭状态，吸入新鲜惰性气体维持储罐内的压力平衡。

2 呼吸阀超压/真空工况计算原理2.1 超压/真空的原因在确定储罐超压或真空的可能原因时，应考虑以下因素：（1）由于液体从罐中最大流出速率而导致的正常吸入（液体转移效应）；（2）由于蒸汽空间温度的最大降低量（热效应）引起的蒸汽收缩或冷凝而导致的正常吸入；（3）因液体流入罐内最大速率而导致的正常呼气，以及由此产生的最大汽化量（液体转移效应）；（4）由于蒸汽空间温度的最大增加量（热效应）引起的膨胀和汽化而导致的正常呼气；（5）火灾暴露引起的紧急排放。

在确定总的正常吸气或呼气时，至少应考虑液体转移效应和热效应所导致的正常排气的组合。

2.2 进液和出液所需的流通能力（1）呼气在储罐蒸汽空间的实际压力和温度条件下，呼出的体积流量V op 应通过以下公式给出：V op =2×V pf （1）式中：V pf 是挥发性液体的最大体积填充率，单位为m 3/h。

（2）吸气吸气通风要求V ip （单位：m 3/h），应为储罐的最大规定液体排放量，应通过以下公式给出：V ip -V pe （2）式中：V pe 是液体排出的最大速率，单位为m 3/h。

3 因热呼气和热吸入所需的流通能力（1）热呼气计算热呼出量（即加热时的最大热流量）V ot ，用国际标准单位表示：m 3/h。

V ot =Y ×V 0.9tk ×R i （3）式中：Y 是纬度的一个因子（见表1）；V tk 为储罐容积，m 3；R i 是保温的降低因子（如果储罐无保温R i =1；如果储罐部分保温R i =R inp ；如果储罐全保温R i =R in ）；˗�������������������1����������1（4）式中：A TTS 是储罐总表面积（外壳和顶部），m 2；A inp 是储罐保温表面积，m 2。