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So I am using this AND gate which has an open drain output. I am trying to determine how many loads that I can place on this output. I see in the datasheet, that the Iol is over 1mA, but I am unsure as to whether it can drive the loads when in high Z.

So if the input loads have a input current of .5uA, what will supply this current to these loads. Is it the open drain output?

Thanks!

circuit

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    \$\begingroup\$ Depends on your input requirements on the next gate. The open drain itself only care about the shorted current (limited by your resistor, 330 uA) and the voltage (3.3 V). If both are ok, the open drain does not care what you load it with downstream. \$\endgroup\$
    – winny
    Commented May 18, 2022 at 20:41
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    \$\begingroup\$ At the high state, the output current is delivered by the pull up resistor. So the input high current for all connected inputs should flow through the resistor while the voltage is at least the minimum high level potential. \$\endgroup\$
    – Uwe
    Commented May 18, 2022 at 20:49

2 Answers 2

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It is the resistor. In a static state you can drive more than 100 loads of this kind, but their input capacitance slows down the transition from low to high. So the real question is: What is the possible switching frequency at a given number of loads or vice versa. Another point is, that slow transitions on inputs without schmitt trigger internally draw a significant amount of additional supply current (see minimum input rise time in datasheet).

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The smaller the resistor, the more loads you can pull high (logic 1) - but watch total capacitive load, as previous poster. However, if you make the pull-up resistor too small, too much current will be drawn when the open-drain AND gate output is trying to drive to logic 0. Check your AND gate data sheet for how much is allowed (typically 1-4mA, but can be more, depending on the logic series used).

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