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This a general question regarding what should we understand when we see the term "short circuit current" in a given device datasheet. Does it always exactly mean the output current when we short the output to the ground? Or should we look for further explanation in the datasheet?

Here is an example from the datasheet of "AD56x8" digital to analog converter below. AD5628/AD5648/AD5668

AD56x8 output characteristics

It says, the short-circuit current is 30 mA at Vdd = 5 V. I assume, it means that there is an output current protection on the device since it should be way above 30 mA considering the output impedance is 0.5 ohms. Is that a correct assumption? Does this also mean that the maximum current we can draw from the device is 30 mA (there is additional info about the current behaviour in this datasheet, what if there wasn't any)?

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    \$\begingroup\$ There are graphs of the source and sink capability in that datasheet, figures 32 & 33. \$\endgroup\$
    – Jeanne Pindar
    Commented Feb 11, 2020 at 21:11
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    \$\begingroup\$ This is under the heading of "OUTPUT CHARACTERISTICS", so it's safe to assume it means the behavior of the output when it is shorted. If you find it somewhere else in a datasheet you might have to think about what it means in that context. \$\endgroup\$
    – The Photon
    Commented Feb 11, 2020 at 21:12
  • \$\begingroup\$ I didn't notice the source/sink graph, I'll edit the question accordingly. \$\endgroup\$
    – packt
    Commented Feb 11, 2020 at 21:33
  • \$\begingroup\$ @packt When operating properly (say, \$\pm 5\:\text{mA}\$), it's about a \$50\:\Omega\$ output (easily seen on one of the charts in the later pages below.) But you want to minimize your loading to minimize the voltage drop at the output. If you short-cricuit the output, it's NOT operating properly. Keep in mind that there are lots of outputs here and temperature variations due to varying output loading is to be avoided, too. \$\endgroup\$
    – jonk
    Commented Feb 11, 2020 at 23:45

2 Answers 2

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Each buffer has active current limiting with a soft transition towards either rail. This is common practice for Op Amps to maintain low output impedance and low error load regulation , yet be short circuit protected.

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Note that this is a typical value. A transistor isn't a linear device. The 0.5 ohm DC impedance is probably the incremental resistance for Rl=2K to ground as seen at the top of the chart. For a short circuit, you'll be on the flat part of the curve, and the current will probably not change much from 0.5V (overloaded) to 0V (short circuit). It will typically be around 30mA.

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  • \$\begingroup\$ What do you mean by transistor? Do you mean the output stage of the dac? \$\endgroup\$
    – packt
    Commented Feb 11, 2020 at 21:39
  • \$\begingroup\$ Yes, if it's an active output, a transistor will be driving it (directly or indirectly). \$\endgroup\$
    – Cristobol Polychronopolis
    Commented Feb 12, 2020 at 14:26

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