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| 型号： | AD8531ARZ-REEL |
| 厂商： | Analog Devices Inc |
| 文件页数： | 4/20页 |
| 文件大小： | 0K |
| 描述： | IC OPAMP GP R-R CMOS 3MHZ 8SOIC |
| 标准包装： | 2,500 |
| 放大器类型： | 通用 |
| 电路数： | 1 |
| 输出类型： | 满摆幅 |
| 转换速率： | 5 V/µs |
| 增益带宽积： | 3MHz |
| 电流 - 输入偏压： | 5pA |
| 电压 - 输入偏移： | 25000µV |
| 电流 - 电源： | 750µA |
| 电流 - 输出 / 通道： | 250mA |
| 电压 - 电源，单路/双路(±)： | 2.7 V ~ 6 V，±1.35 V ~ 3 V |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 8-SOIC（0.154"，3.90mm 宽） |
| 供应商设备封装： | 8-SO |
| 包装： | 带卷 (TR) |
AD8531/AD8532/AD8534
Rev. F | Page 12 of 20
The thermal resistance curves were determined using the
AD8531/AD8532/AD8534 thermal resistance data for each
package and a maximum junction temperature of 150°C. The
following formula can be used to calculate the internal junction
temperature of the AD8531/AD8532/AD8534 for any application:
TJ = PDISS × θJA + TA
where:
TJ is the junction temperature.
PDISS is the power dissipation.
θJA is the package thermal resistance, junction-to-case.
TA is the ambient temperature of the circuit.
To calculate the power dissipated by the AD8531/AD8532/
AD8534, the following equation can be used:
PDISS = ILOAD × (VS VOUT)
where:
ILOAD is the output load current.
VS is the supply voltage.
VOUT is the output voltage.
The quantity within the parentheses is the maximum voltage
developed across either output transistor. As an additional
design aid in calculating available load current from the
AD8531/AD8532/AD8534, Figure 5 illustrates the output
voltage of the AD8531/AD8532/AD8534 as a function of
load resistance.
POWER CALCULATIONS FOR VARYING OR
UNKNOWN LOADS
Often, calculating power dissipated by an integrated circuit to
determine if the device is being operated in a safe range is not
as simple as it may seem. In many cases, power cannot be directly
measured, which may be the result of irregular output waveforms
or varying loads; indirect methods of measuring power are
required.
There are two methods to calculate power dissipated by an
integrated circuit. The first can be done by measuring the
package temperature and the board temperature, and the
other is to directly measure the supply current of the circuit.
CALCULATING POWER BY MEASURING AMBIENT
AND CASE TEMPERATURE
Given the two equations for calculating junction temperature
TJ = TA + PDISS θJA
where:
TJ is the junction temperature.
TA is the ambient temperature.
θJA is the junction to ambient thermal resistance.
TJ = TC + PDISS θJA
where:
TC is the case temperature.
θJA and θJC are given in the data sheet.
The two equations can be solved for P (power)
TA + PDISS θJA = TC + PθJC
PDISS = (TA TC)/(θJC θJA)
Once power is determined, it is necessary to go back and calculate
the junction temperature to ensure that it has not been exceeded.
The temperature measurements should be directly on the package
and on a spot on the board that is near the package but not
touching it. Measuring the package could be difficult. A very
small bimetallic junction glued to the package can be used, or
measurement can be done using an infrared sensing device if
the spot size is small enough.
CALCULATING POWER BY MEASURING SUPPLY
CURRENT
Power can be calculated directly, knowing the supply voltage
and current. However, supply current may have a dc component
with a pulse into a capacitive load, which can make rms current
very difficult to calculate. It can be overcome by lifting the supply
pin and inserting an rms current meter into the circuit. For this
to work, be sure the current is being delivered by the supply pin
being measured. This is usually a good method in a single-supply
system; however, if the system uses dual supplies, both supplies
may need to be monitored.
INPUT OVERVOLTAGE PROTECTION
As with any semiconductor device, whenever the condition
exists for the input to exceed either supply voltage, the input
overvoltage characteristic of the device must be considered.
When an overvoltage occurs, the amplifier can be damaged,
depending on the magnitude of the applied voltage and the
magnitude of the fault current. Although not shown here, when
the input voltage exceeds either supply by more than 0.6 V, pn
junctions internal to the AD8531/AD8532/AD8534 energize,
allowing current to flow from the input to the supplies. As
illustrated in the simplified equivalent input circuit (see Figure 36),
the AD8531/AD8532/AD8534 do not have any internal current
limiting resistors; therefore, fault currents can quickly rise to
damaging levels.
This input current is not inherently damaging to the device, as
long as it is limited to 5 mA or less. For the AD8531/AD8532/
AD8534, once the input voltage exceeds the supply by more than
0.6 V, the input current quickly exceeds 5 mA. If this condition
continues to exist, an external series resistor should be added.
The size of the resistor is calculated by dividing the maximum
overvoltage by 5 mA. For example, if the input voltage could
reach 10 V, the external resistor should be (10 V/5 mA) = 2 kΩ.
This resistance should be placed in series with either or both
inputs if they are exposed to an overvoltage condition.
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相关代理商/技术参数 |
参数描述 |
|---|---|
| AD8532 | 制造商:AD 制造商全称:Analog Devices 功能描述:Low Cost, 250 mA Output Single-Supply Amplifiers |
| AD8532AN | 制造商:Analog Devices 功能描述:OP AMP DUAL HIGH O/P RRI/O 8532 |
| AD8532AR | 功能描述:IC OPAMP GP R-R CMOS 3MHZ 8SOIC RoHS:否 类别:集成电路 (IC) >> Linear - Amplifiers - Instrumentation 系列:- 标准包装:160 系列:- 放大器类型:通用 电路数:4 输出类型:满摆幅 转换速率:10 V/µs 增益带宽积:9MHz -3db带宽:- 电流 - 输入偏压:1pA 电压 - 输入偏移:250µV 电流 - 电源:730µA 电流 - 输出 / 通道:28mA 电压 - 电源，单路/双路(±):2.7 V ~ 5.5 V，±1.35 V ~ 2.75 V 工作温度:-40°C ~ 125°C 安装类型:表面贴装 封装/外壳:16-SOIC（0.154"，3.90mm 宽） 供应商设备封装:16-SOIC N 包装:管件 |
| AD8532ARM | 制造商:Analog Devices 功能描述: |
| AD8532ARM-R2 | 制造商:Analog Devices 功能描述:OP Amp Dual GP R-R I/O 6V 8-Pin MSOP T/R |
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