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Good morning,

I am still learning, and often I see a lot of datasheet value that I dont understand. For this case specifically, what is the difference between output current and supply current?

The data sheet is here http://assets.nexperia.com/documents/data-sheet/74HC_HCT08.pdf

1)The ratings on page 3 list the supply current as 50ma, but the supply voltage is -0.5 to 7 volts. Does this mean I must limit the current to pin 14(in this case) to 7 volts, and the current no greater than 50ma?

2)Regardless of voltage, as long as it is within the allowed range(-0.5v - 7v), the IC will out put Vcc at +- 25ma at any of the output pins?

Thank you

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4 Answers 4

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It means you give it no more than 7Volts. You don't limit the current for Vcc. You regulate the voltage.

It can deliver upto 25mA on any output, but the 50mA will be divided among all the outputs. So, sum total of all outputs should not exceed 50mA.

If you try to draw more than total of 50mA through the outputs you will very likely damage the IC. You prevent that from happening by designing the loads on the outputs so that the total cannot ever be over the limit.

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    \$\begingroup\$ No it doesn't guarantee that voltage less than 7 V will draw less than 50 mA. It means you must not apply more than 7 V or you may damage the chip. There's no way they can promise this won't result in more than 50 mA current since you control the output loads, not them. \$\endgroup\$
    – The Photon
    Commented Mar 25, 2017 at 15:54
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You are looking at the "Limiting Values" table, also known as "Absolute Maximum Rating" The part may be damaged if you exceed these ratings. You should never operate the part at the ratings given in that table.

You should look at the "Recommended Operating Conditions" and other tables to find the normal supply voltage and other normal operating conditions.

Under Recommended Operating conditions, you will see that the 74HC08 will operate correctly with a supply voltage between 2 and 6 volts, while the 74HCT08 requires 4.5 to 5.5 volts, and the output current in either case should be limited to 5 mA or so. The actual power supply current for both types is 40 uA, plus any current delivered to the outputs.

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the supply voltage is -0.5 to 7 volts.

These values are "limiting values", which other vendors more often call "absolute maximum values". This doesn't tell you what voltage to operate the chip at, it tells you what limits to remain within to avoid damaging the chip.

To see what voltages you should operate the chip at, look at table 5, "recommended operating conditions." There it gives the limits of 2.0 to 5.0 V for regular operation. All the performance specs in the datasheet will assume you operate the chip between these limits, not the limiting values limits.

The supply current has a limiting value because the total supply current will be the static supply current (given in table 6) plus dynamic current that depends on how often the outputs are switching, plus whatever current is delivered by the output to the loads. This puts a limit on how much total load can be put on all the outputs.

what is the difference between output current and supply current?

The output current limit is for any one individual output.

The supply current will be the static current, plus switching current, plus (the sum of all the) load currents delivered to the loads through the outputs.

So if you have more than two outputs delivering the limit of 25 mA per output, you will exceed the 50 mA total supply current limit.

The ratings on page 3 list the supply current as 50ma, but the supply voltage is -0.5 to 7 volts. Does this mean I must limit the current to pin 14(in this case) to 7 volts, and the current no greater than 50ma?

You should use a Vcc voltage according to the recommended operating conditions, not the limiting values.

If you do that, and don't connect too low an impedance to the outputs, the device will limit its current draw well below the absolute maximum suppply current limit.

Regardless of voltage, as long as it is within the allowed range(-0.5v - 7v), the IC will out put Vcc at +- 25ma at any of the output pins?

No. To make the chip operate normally, you must respect the recommended operating conditions limits, not the absolute maximum or limiting values limits.

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  • \$\begingroup\$ I find it curious that Max Power is < Max V x Max I. This makes me wonder whether Icc MAX is instantaneous or average. \$\endgroup\$
    – Trevor_G
    Commented Mar 25, 2017 at 16:44
  • \$\begingroup\$ HUH? I guess they use superconductors in those output switches then. \$\endgroup\$
    – Trevor_G
    Commented Mar 25, 2017 at 17:56
  • \$\begingroup\$ Yes I know, I was just noting that the max power rating stated if HIGHER than you can calculate from the other MAX values. Which makes me wonder which ones are "integrated" numbers. \$\endgroup\$
    – Trevor_G
    Commented Mar 25, 2017 at 18:11
  • \$\begingroup\$ EEK.. you are right... it should read >. sigh... \$\endgroup\$
    – Trevor_G
    Commented Mar 25, 2017 at 18:13
  • \$\begingroup\$ CORRECTION : I find it curious that Max Power is > Max V x Max I. This makes me wonder whether Icc MAX is instantaneous or average. \$\endgroup\$
    – Trevor_G
    Commented Mar 25, 2017 at 18:16
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The absolute maximum voltage that you can apply to the IC is 7 volts. This will make the IC draw up to 50 mA (depending on your load). And as The Photon mention in the comments, It is the absolute maximum rating. The recommended operating conditions is 5.5V or 6V according to the datasheet. The datasheet mention the supply current so that you're able to choose the capacity of your supply correctly.

Example: If you're using one supply to feed ten IC's, The supply must able to produce 500mA. I'm assuming the worst case for safety.

The maximum output current that the IC can provide is 25 mA.

Example:

let's assume that you want to connect a resistor as your load. If the output is 7 volts. The minimum resistor that you can use is = 7 / 25mA = 280 ohm.

If you used a 100 ohm resistor, The output current will be = 7 / 100 = 70mA. In this case the limit is exceeded and The IC will get hot and a damage is likely to occur.

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    \$\begingroup\$ Note the 7 V and 50 mA values are abs max values, not recommended operating conditions or performance specs. \$\endgroup\$
    – The Photon
    Commented Mar 25, 2017 at 15:45
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    \$\begingroup\$ Also, the 50 mA value has to do with how much load is on the outputs. There's no reason the static current should be 50 mA when 7 V is applied to Vdd. \$\endgroup\$
    – The Photon
    Commented Mar 25, 2017 at 15:46
  • \$\begingroup\$ @ThePhoton Thank you very much for your comment :) I edited my answer \$\endgroup\$ Commented Mar 25, 2017 at 15:54

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