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AD8534AR データシート(PDF) 9 Page - Analog Devices |
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AD8534AR データシート(HTML) 9 Page - Analog Devices |
9 / 16 page AD8531/AD8532/AD8534 REV. D –9– Power Dissipation Although the AD8531/AD8532/AD8534 is capable of providing load currents to 250 mA, the usable output load current drive capability will be limited to the maximum power dissipation allowed by the device package used. In any application, the absolute maximum junction temperature for the AD8531/ AD8532/AD8534 is 150 ∞C, and should never be exceeded for the device could suffer premature failure. Accurately measuring power dissipation of an integrated circuit is not always a straightforward exercise, so Figure 34 has been provided as a design aid for either setting a safe output current drive level or in selecting a heatsink for the package options available on the AD8531/AD8532/AD8534. TEMPERATURE – C 1.5 1 0 0100 25 50 75 0.5 85 TJ MAX = 150 C FREE AIR NO HEATSINK SOIC JA = 158 C/W TSSOP JA = 240 C/W SOT-23 JA = 236 C/W SC70 JA = 376 C/W Figure 34. Maximum Power Dissipation vs. Ambient Temperature These 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 ¥ qJA + TA where TJ = junction temperature; PDISS = power dissipation; qJA = package thermal resistance, junction-to-case; and TA = 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 output load current; VS = is supply voltage; and VOUT = is 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 1 illustrates the AD8531/ AD8532/AD8534 output voltage 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 might seem. In many cases power cannot be directly measured. This 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 pack- age temperature and the board temperature. The other is to directly measure the circuit’s supply current. Calculating Power by Measuring Ambient and Case Temperature Given the two equations for calculating junction temperature: TJ = TA + P qJA where TJ is junction temperature, and TA is ambient tempera- ture. qJA is the junction to ambient thermal resistance. TJ = TC + P qJC where TC is case temperature and qJA and qJC are given in the data sheet. The two equations can be solved for P (power): TA + P qJA = TC + P qJC P = (TA – TC )/ ( qJC – qJA) Once power has been determined it is necessary to go back and calculate the junction temperature to assure that it has not been exceeded. The temperature measurements should be directly on the pack- age and on a spot on the board that is near the package but definitely not touching it. Measuring the package could be diffi- cult. A very small bimetallic junction glued to the package could be used or it could 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. This could make rms current very difficult to calculate. It can be overcome by lifting the sup- ply pin and inserting an rms current meter into the circuit. For this to work you must be sure all of the current is being deliv- ered by the supply pin you are measuring. 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 device’s input overvoltage characteristic must be considered. When an overvoltage occurs, the amplifier could 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 (Figure 32), the AD8531/ AD8532/AD8534 does not have any internal current limiting resistors, so fault currents can quickly rise to damaging levels. |
同様の部品番号 - AD8534AR |
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同様の説明 - AD8534AR |
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