0
\$\begingroup\$

How much current can this 74HC245 handle on each output pin? Is it 20mA, 35mA, or 70mA?

I don't understand what the difference between output current, output clamping current, and supply current is.

http://www.nxp.com/documents/data_sheet/74HC_HCT244.pdf

\$\endgroup\$
0
2
\$\begingroup\$

Icc is the max current to the Vcc pin. Ignd is the max current to the ground pin. Both of these are listed as 70 mA. The clamp values are the max current through the ESD protection/clamping diodes that connect each pin to Vcc and GND. Current will only flow through these diodes if you apply a voltage to an input or output pin which falls outside the supply rails (greater than Vcc or less than GND). Both of these are specified at 20 mA. The output current is how much current the driver on the pin can actually supply, which is specified as 35 mA. Note that you cannot draw 35 mA from more than two pins at the same time as this would exceed the Icc/Ignd limits.

The interesting thing about these chips is they actually have a spec for the current through the ESD protection diodes. This means that it is actually perfectly reasonable to connect the inputs to a relatively high voltage source (say, 12V) and only use a series current limiting resistor for the level translation.

\$\endgroup\$
2
\$\begingroup\$

The clamping currents are currents that will flow if you apply a voltage to the input or output voltage below ground or above the supply voltage, see the conditions: Vi < -0.5V or Vi > Vcc + 0.5V

There are ESD protection diodes on all inputs and outputs, on each pin there is a diode to GND and one to Vcc. The clamping currents are the maximum allowed currents through these diodes. Normally you would/should not apply voltages above or below the supply and GND rails. But if you do, do not exeed these limits.

The maximum current flowing in to / out of an output pin may not exeed 35 mA. The total supply current (which is also the sum of all currents supplied by outputs) should not be more than 70 mA.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.