As per the Datasheet of LMC555, the maximum supply current is 250 μA @ 5 volt supply voltage.

How come the output current reach 2 mA, I mean where did the IC get the extra (2*1000-250) μA?

Maybe I miss something or misunderstand these attributes at the datasheet, so it will be helpful if anyone clarify this.

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I was thinking about using the CMOS version of the 555 timer to create a variable duty cycle 150 kHz square wave generator to drive a MOSFET in order to learn designing flyback converter and SMPS in general.

But the issue is that I need very low current 555 timer, so that I can drive it using simple resistor or couple or resistors in parallel from the mains after rectifying and smoothing the current. I needed the CMOS low current version so that the resistance can handle the power (current and voltage across it) without burning.

I didn't like to use the offline switching IC because it can be blown easily while learning and due to its higher cost. So how can I know the output current at 5 volts exactly from the datasheet at 5 V, and how can I solve the contradiction I felt and highlighted in yellow?

  • \$\begingroup\$ All switches open, it says. \$\endgroup\$ Commented Jun 18 at 11:37
  • \$\begingroup\$ Note that many other devices are spec'd in a similar fashion. '485 & LVDS drivers, for example, usually don't include the load current (through a termination resistor) in their supply specification. \$\endgroup\$
    – SteveSh
    Commented Jun 18 at 12:46

2 Answers 2


The 250 μA is the maximum at 5 V what the chip itself consumes with no load. (Note also that the 555 timer chip can produce some output current on other pins – even the other pins are unloaded in this test.)

When the chip is driving 2 mA out into a load, it will take max 2.25 mA in and uses at most 0.25 mA for itself and drives the 2 mA out.


The "maximum supply current" is only the current used internally by the timer chip. Any output current gets added to this.

  • \$\begingroup\$ Thanks dave , i have accepted other answer with same meaning of your answer since it was posted first according to time stamp .. but thanks \$\endgroup\$ Commented Jun 18 at 11:55
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    \$\begingroup\$ It wasn't, but that's okay. \$\endgroup\$
    – Dave Tweed
    Commented Jun 18 at 12:29
  • 1
    \$\begingroup\$ Thanks any way .. sorry for that \$\endgroup\$ Commented Jun 18 at 16:08

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