555 timer as square wave generator

I have seen some posts here about it but they all are talking about 5v supply voltage, I want to generate a 1 Hz square wave but my supply voltage is 3.3v , can someone guide me on how ?

thanks

CMOS versions of the NE555 can run on lower voltages. E. g. the TI LMC555 specifies "1.5V Supply Operating Voltage Ensured".

You can use 7555 for lower voltages(2V to 18V).

udhay_cit said:

You can use 7555 for lower voltages(2V to 18V).

Click to expand...

The ICM7555 is a Cmos 555 made only by Intersil but the LMC555 is the same and is made by National Semi recently bought by Texas Instruments and the TLC555 is also the same and is also made by Texas Instruments. All three of them are sold today but the TLC555 costs less than the others.

so what would be the circuit to generate a 1Hz square wave from this 555 timer?

thanks

Please check this:
https://mintelectronics.wordpress.com/2012/06/17/squarewave-generator-using-555/

The simple 555 oscillator does not produce a squarewave. Instead its output is a rectangular waveform because its timing capacitor charges through two resistors in series and discharges through only one of them. A diode can be added and another resistor to make the output more square.
If you need a perfect squarewave then use a CD4047 IC.

The link shows a 1Hz oscillator using a 10uF capacitor and a 68k resistor. If the capacitor is electrolytic then the accuracy of the frequency is poor because the capacitor value is not accurate. I would use a 1uF 5% film capacitor and a 680k capacitor for a much better frequency accuracy.

the link that was posted says voltages 4.5 -9v but I would be using 3.0- 3.3v , can i still use the same circuit with same components ?

Operating voltage range you can check in IC data sheet. You can use the same circuit. Higher operating voltage give you higher amplitude.

aliyesami, here you have a handy calculator (it is quoted in yadavvlsi's link):
http://www.ohmslawcalculator.com/555_astable.php
The bipolar and CMOS versions of the 555 generate a frequency and duty cycle that are quite independent from the supply voltage. So dont' worry porting a 9V schematic down to 3V. Only that you can't use the original bipolar 555. Just calculate, build, and if necessary you can tweak a little the resistor from pin 6 to 7 to get the exact frequency.
By the way, do you need a strictly square wave (50% high, 50% low)? The classic circuit as shown in ohmslawcalculator can't do that, to even approach 50% you have to choose values for the R from pin 7 to Vcc that may result too low, and too high current consumption into pin 7 (the Discharge transistor). My favorite topology is not using Discharge at all, nor its R to Vdd. The resistor in pin 7 will go to pin 3 (Output) instead of Discharge. This will give a nearly 50% duty cycle, provided the Output levels are nearly 0V and nearly Vdd. This will be true (for the CMOS versions) if you don't extract current from Output, e.g. adding a CMOS buffer there.

We do not know what your 1Hz oscillator is driving. The Cmos 555 has output levels that are rail-to-rail (+3.3V and 0V if the supply is 3.3V) ONLY if the load resistance is extremely high.

thanks all for your feeback. I dont need a exact square wave , not the 50% duty cycle , as long as I am getting a falling edge and 1Hz freq i am fine.
it will be used as the external interrupt to the AVRmega procecessor to generate the heartbeat.

regards

Then a Cmos 555 will be fine for your circuit.