does anybody know something about this parameter. I find no information in the datasheet but the AD8009 is short circiut protected. the AD8000 also?
Thanks and best greetings from germany
In most cases with high-current-output amplifiers the power dissipation limits the short duration, and the power calculations need to be made on a case-by-case basis. This is why the "Absolute Maximum Ratings" section of the AD8009 data sheet indicates "Observe Power Derating Curves" for the output short duration.
The short circuit output current of the AD8009 is 330 mA. In the case ot +/-5V supplies and a short to ground (mid-supply), one of the output transistors will supply the vast majority of output current The majority of the power consumed in the part due to the output short is therefore (5V)*(330 mA) = 1.65 Watts. The quiescent power is low compared to this, which in the worst case (presuming the part gets very hot, which it will) is (10V)*(18 mA) = 0.18 Watts. The total power is therefore about 1.83 mW.
The AD8009 comes in two packages -- SO-8 and SOT-23-5, with junction-to-air thermal resistances of 155C/W and 240C/W, respectively. The ambient-to junction rise in the SO-8 package is therefore 284C, and is 439C for the SOT-23-5. As you can see from this, the output short circuit duration before destruction for the AD8009 is extremely short. The best way to interpret this result is to say that a very brief short to ground, as may happen while bench testing in the lab, should be okay, but a short of any duration beyond this would destroy the part.
The AD8000 does not specify output short circuit current in its data sheet, but its linear output current, conditioned on HD < -50 dBc, is 100 mA. Since it can supply 100 mA reasonably linearly, the short circuit output current is likely somewhat greater than this. It's less than the AD8009, but is well over 100 mA. In light of this, I would conclude that the AD8000 should also not have its output shorted for any sustained period of time.
thanks for your response. In my circuit I have no chance to limit the output current in a failure case. So I want to meassure the supply current in the +6V and -6V supply for the op amp. I f the current rises higher 100mA I want to activate the shut down of the op amp. This need a certain time, I think nearly 100µs. Do you think this is sufficient to protect the op amp?
Best regards and a nice weekend
You should set the shut-down current at a safe operating point for the part. For instance, if you're using the AD8009 in the SO-8 package at a maximum ambient temperature of 50C, the maximum power consumption in the AD8009 should be less than about 0.6 W, which is what you have with 6V and 100 mA supply current. I would add some margin, however, since the part could be operating just under the trip point for an indefinite period of time. Do you need to set the shutdown current threshold so high?
the output voltage is max. 4V and the load resistance is 50 Ohm//150pF= 80 mA normal case. Ambient temperature is max. 50°C. Case is SOIC.
So I think to set the threshold to a level of 100mA is not critical but necessary. I know there is not much reserve. Another idea and cheap, is to place a fuse in series to the output.It is a SMD/100mA type verry fast.
Do you have other ideas or proposals?
With a ground-referenced load, the maximum power consumed by the part due to the load occurs when the output is at mid-supply. So, with your 6V supply, the maximum power consumed by the part due to the load will be 180 mW. This gives an ambient-to-junction rise of about 28C for the AD8009 in the SO-8 package, which leaves plenty of margin to the 150C junction temperature limit at a 50C ambient. You will be okay using your supply current measuring method as long as it responds to the shorted output within 100 us as you indicated earlier.
The difficulty that arises with short circuit (or very low load resistance) output current is that the output is at zero volts, or very low, due to the nonlinear behavior of the part, and not due to an actual 0V output level that is linearly derived from the input signal. In the short-circuit case, you have the amplifier trying to drive a very low resistance load with its maximum output current, so you have the part consuming power due to the load that equals the supply voltage times the short circuit current, which is very large. Under normal linear conditions, when the output voltage goes to 0V with a 0V referenced load, the power consumed by the part due to the load is zero since there is no current flowing from the supply to the load. A similar zero-power condition would also occur if the output voltage could be made equal to the supply voltage, since the voltage across the part (supply-to-load) would be zero, but op-amp outputs can never quite reach the supply voltage, though many with rail-to-rail outputs can come very close.
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