555 timer circuit power consumption

Question

My goal is to have an LED flash (rather than stay on all of the time) in order to conserve battery life. So, I learned how to modify a 555 timer circuit so that the LED flashes for about 0.1Hz and remains off for ~10 seconds.

The circuit diagram is like this:

Source: http://www.electronics-tutorials.ws/waveforms/tim58.gif?81223b

Are there other components in a 555 time circuit (resistors, capacitors, 555 itself, etc) that are drawing energy and will significantly affect my battery life, even though I have greatly reduced the duty cycle?

Answer 1

As Nasha has already pointed out, using a CMOS 555 will greatly cut down on the quiescent current required from the battery, as well as eliminate the huge shoot-through current spikes inherent in the bipolar 555's output totem-pole every time it switches.

Another area of concern is the timing capacitor's chemistry and capacitance.

In order to minimize its own leakage, the cap should be something other than electrolytic, (plastic or ceramic dielectric, say,) and its capacitance should be selected to be as small as possible and, in the circuit you've linked to, the value of the resistors should be made as large as possible since every time the cap charges and discharges all that energy is just wasted.

Also, C2 can be eliminated.

Answer 2

The 555 «traditionally» consumes a milliamp-wise current, like 5-10mA. In your modified circuit, what the LED will consume is negligible compared to the 555 itself. The LMC555 is its CMOS version, which brings the supply current down to the micro-amp domain, i.e. like 250µA max @ 5V. So the CMOS version will greatly increase your battery life.