Say I am running my 74LVC125 at 2.5V. At the A input I have a logic 1 and at the nOE input, I have a logic 0, so the device should be driving a logic 1 onto the Y output.

With no load connected to the Y output, I should observe 2.5V at the output, obviously.

Now say I have a load along the lines of 1kohm to 5V. What happens? I choose 74LVC125 for my example because for tristate LVC logic, in the high-Z state (nOE high), applying higher voltage than Vcc on an output is allowed. But what about with the output enabled?

  • Will it sink something like (5.0V - 2.5V) / 1kOhm = 2.5mA?
  • Will very little current flow?
  • Will it sink more current than I would expect? The TVS diodes of the LVC device will not yet flow current, but will some kind of reverse bias damage happen to the rest of the LVC driver IC?

I want the first thing to happen, that is for the 'LVC125 to sink something like 2.5mA into 2.5V. Using a logic circuit like 74LVC125 is cheaper and than applying several MOSFETs for this purpose. Can it be done?

  • \$\begingroup\$ Most devices with open-drain outputs (e.g., '07) are designed to allow higher output voltages. \$\endgroup\$ – CL. Nov 16 '18 at 20:40

It is not allowed, according to the datasheet, to apply any more than Vcc + 0.5V (absolute maximum) to the output.

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However the built in clamp diodes can withstand a certain amount of current (not specified, but below GND -50mA is permitted.

So with the output high-Z and a pullup to +5 the output will be about at the Vcc rail + 0.7V (or ~5V, whichever is less). The Vcc rail may well not stay at 2.5V and may be pulled up to approaching 5V depending on what else is on the rail, since most voltage regulators will not sink current.

With the output enabled and high, the high-side MOSFET (p-channel) will be "on' and conducting (in reverse) and the above sentence also applies, however there will no longer be a diode drop involved.


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