8
\$\begingroup\$

So, I've been checking out the datasheet and I've gotten pretty confused... At the top it claims that 6mA drive at 5V but farther down in the absolute maximum ratings it says that the outputs can be up to 35mA.

So, my question is, how much current can each output pin of the chip take?

I'm using it for a controllable line of LEDs and I need to figure out what resistor value to use.

\$\endgroup\$
  • \$\begingroup\$ See Wouter's answer to another question here: electronics.stackexchange.com/questions/37648/… \$\endgroup\$ – The Photon Aug 3 '13 at 3:20
  • \$\begingroup\$ Also, notice note 1 under the absolute maximums table: "Stresses beyond those listed under absolute maximum ratings may cause permanent damage to the device. These are stress ratings only, and functional operation of the device at these or any other conditions beyond those indicated under recommended operating conditions is not implied." \$\endgroup\$ – The Photon Aug 3 '13 at 3:22
11
\$\begingroup\$

What it means by 6mA at 5v, is that at 6mA, the output voltage level of that output will be fairly close to 5v (or VCC). The more current you source out of an output, the lower that voltage will be. This is called Voltage Droop, or Sag. Same goes for sinking current, with the voltage level rising, so instead of it being 0.0 or 0.1, at 20mA it might be 1v.

enter image description here

Notice that at a VCC of 4.5V, Voh (Voltage Output High), at Ioh (Current output high) -20µA (or -0.02 milliamps, the - indicating sourcing current) the typical is 4.499 volts. While at -6 mA, the typical is 4.3? Your output just lost 0.2v with only 6 mA of current draw. Anything more, and that drop will increase an untested amount. Some datasheets actually have a graph for this. This one does not. So you can draw more than 6mA, up to the max Io output current of 35mA (plus or minus) on a single output, but the voltage level will change to the point where it might not be useful to you. Ex. At 35mA, you might not be able to light an led.

\$\endgroup\$
  • 2
    \$\begingroup\$ Wow, thank you! That's a very helpful answer. I learned a bunch in 2 paragraphs. \$\endgroup\$ – TheNickmaster21 Aug 3 '13 at 14:28
3
\$\begingroup\$

Under "Absolute Maximum Ratings" the datasheet gives 70 mA for the maximum continuous current through the Vcc or GND pins, so I would limit the individual output currents to less than 1/8 of that, or 8.75 mA if it is possible that all channels may be driving in the same direction at any time. Under "Electrical Characteristics" it gives the ouput voltage for 7.8 mA, with 6 V Vcc and 5.2 mA at 4.5 V Vcc, so I'd try to stay under ~6 mA with a 5 V Vcc.

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

Page 27 of SCLA007A, "HCMOS Design Considerations", "Drivers for LEDs and Relays", states that "[the recommended] operating source or sink current for HCMOS devices is 5.2 mA. Several SN74HC gates can be tied in parallel to drive LEDs and relays." Limiting current to that amount per gate will be safe.

\$\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.