I'm thinking of using a SN74ABT244A octal buffer in a circuit that requires the buffer to sink 20mA of current on all eight outputs under worst-case input conditions. In the datasheet, it says that IOH is -32mA, which should be enough. However, I am confused as to whether or not this applies to each individual output, or all outputs on the entire chip, in which case I would be sinking 160mA of current and severely exceeding the current rating of this chip.
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If your application requires it to sink current, the rated maximum is IOL at 64mA max. This is a per output pin limit so it sounds like it will give you plenty of headroom in your application. |
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It's for each pin. Note that a datasheet may give maximum values for supply current (Icc) and ground current as well, usually mentioned under Absolute Maximum Ratings (AMR). The 74HC595 for instance gives an AMR of 35 mA per pin, but on top of that a supply current and ground current of both 70 mA AMR. The 74ABT244A doesn't mention these limitations, so should be OK with a ground current of 8 \$\times\$ 64 mA = 512 mA, which is a lot! Since this is a higher current device it will use thicker bonding wires than usual in TTL, at least for the Vcc and GND pins. |
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Each individual output. The current ratings on the chip are for outputs high or low, i.e. CMOS. If you are sinking that much current you won't have the output at the rail. |
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