I was looking over a datasheet for the NE555 Timer and I was wondering if the max supply current given under the electrical characteristics page is the max current you can supply to the 555 without damaging the chip.

I ask because it seems like the supply current (for 15V input) is relatively low (10mA) compared to the output current (225mA).

Is the supply current the max current you can put on pin 8 without burning out the chip? If so, why is it so low compared to the output current? If not, how does one find the input current?

  • \$\begingroup\$ If you look at the supply current entry at the bottom of the page you mention, you may also notice the "No load" phase. This means exactly what it says. Also, you don't "supply" a specific current to the 555. Just a (usually stable) voltage difference between pin 8 and pin 1. \$\endgroup\$
    – jonk
    May 19, 2017 at 4:38
  • \$\begingroup\$ a 555 can handle hundreds of PSU amps, as can most ICs... \$\endgroup\$
    – dandavis
    May 19, 2017 at 6:01
  • 3
    \$\begingroup\$ @dandavis: better wording would be: A 555, like most ICs, can be powered by a high current power supply of suitable voltage, but will only draw as much current as it requires. \$\endgroup\$ May 19, 2017 at 6:09

3 Answers 3


With any electronic circuit, you make current available to it and the circuit draws it under the pressure of the applied voltage. The circuit allows a certain flow.

Your 555 requires a certain current to operate its internals. It can also deliver a current from its V+ supply pin to its OUTPUT pin. And it can conduct a current coming in on its OUTPUT pin down to its GND pin.

There are various terms for the current to just operate a circuit without its outputs connected to anything (usually). 'Operating current', 'supply current' and 'quiescent current' are some of the most common. The circuit needs this current at a minimum. The datasheet may state min(imum), typ(ical) and max(imum) values for it. Unless the values are significantly high or your power supply is significantly limited, your should allow for the specified maximum current in your power supply design and routing. But you can expect it to normally draw the specified typical current.

The OUTPUT pin driver is a totem-pole output, consisting of a transistor between V+ and OUTPUT and another transistor between OUTPUT and GND. Each can conduct up to 200 mA reliably, although the voltage dropped and lost across each transistor increases as the current through increases and so does the power dissipated in the device. If that power becomes too high, it can damage the chip or heat it up and affect its behaviour to some degree. It is good practice to only draw the minimum you need from the output of a device for the reliable operation of your load.

So if you connect a 1 K resistor between your 555's OUTPUT and GND and run it from a 15 V supply, you can expect the current drawn by the 555 through V+ to be around:

\$I_{total} = I_{q} + I_{load} = 10 mA + (15 - 0.2)/1000 = 24.8 mA\$

where \$I_{q}\$ is the quiescent (supply) current and \$I_{load}\$ is the V+ voltage minus the drop across the output transistor (roughly and from the datasheet) divided by the load resistance. I actually worked the load current out twice: once to get an idea of the load current then again as I now could introduce the OUTPUT drop at that load (0.2 V) into the calculation.

Going with these these values, you'd expect only the 10 mA to come out of the 555's GND pin because the 14.8 mA is going through OUTPUT and the 1 K resistor to the GND supply rail.

If you connect a 1 K resistor between your 555's OUTPUT and V+ and run it from a 15 V supply, you can expect the current drawn by the 555's V+ pin to be around:

\$I_{total} = I_{q} = 10 mA\$

This is because the 1 K resistor draws its current from the V+ supply rail separately, not through the 555's V+ pin. But all of the supply current Iq and the load current will flow down and out through the 555's GND pin to the ground rail.


Current is a quantity that you make available to the 555, and it takes what it needs. However, voltage is the quantity, which can destroy a part if you exceed the maximum rating. You can imagine voltage as pressure. If your town water system ran amok and increased your municipal water pressure to 1000 psi, your 100 psi copper plumbing would likely burst. Too much pressure.


You are surprised that the specified supply current (10mA) is lower that the specified output current (225mA). The absolute maximum ratings may help you to determine in some conditions if your component might be broken or not (for instance in case you had a short-circuit on your board and your 555 had to take more voltage and/or current that it normally does).

In fact, that supply current is specified for "No load", which implies that there is no output current at this supply current. It is the current that this 555 version may require for its own internal functions. There are many 555 versions, and the LMC555 has a supply current of maximum 400uA at 12V.

The extra output current may or may not come through the supply pin: when driving the load it will come from the supply pin, when pulling the load it will come from the load and flow to the GND pin.

The maximum current on the supply pin before it "hurts" the component is to be found in the "Absolute Maximum Ratings" that are different from the "Recommended Operating Conditions". We can see that the +/-225mA output current is specified in the Absolute Maximum Ratings while the Recommended Operating Conditions specify +/- 200mA. You really should not go above the Recommended Current for your designed current.

The maximum current is not determined only from the rated currents - it also depends on the thermal impédance of the package and the actual power consumption. At 15V, the high level output voltage is typically 12.5V (the worst case is not given) for -200mA, so there is a 2.5V drop in the device which corresponds to 0.5W. The die temperature should stay below 150°C. If the component's environment could get as hot as 55°C, this power has to dissipate in a 95°C drop through the casing. Your thermal resistance has to be better than 190°C/W for an air temperature of 55°C. All casings are better, so that seems fine. I write "seems" because we do not know the worst voltage drop - for the full temperature range and at only 100mA it is specified to be 12V, but it is not provided for 200mA.

In summary:

  • The supply current given in the specification does not include the load current;
  • The maximum load current should be taken from the recommended operating conditions, not the absolute maximum ratings;
  • The maximum current is also limited by the maximum power dissipation of the package that you must verify for your operating conditions.

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