How to calculate the values of resistors and capacitor for the 555 timer. Please provide the working also.?

Question

I am planning to use 10e-7 F capacitor. And I want t high and t low of 1 sec each. That means a duty cycle of 50%. I check many websites. They only provide the equations to calculate. They never show the working. Since R1 << R2, when I calculate I got R1 to be 0. It is wrong. Please provide the working also. Thanks

Answer 1

You CAN get 50% duty cycle from 555 timer by bypassing one of the resistors during charging by a diode...since the calculations won't be accurate as it is tough to get the exact model for whichever diode you will use, I suggest that you use a potentiometer in series with one of the resistors for fine-tuning the duty ratio; if you want to see a sample circuit, visit this:
http://www.williamson-labs.com/555-circuits.htm

Moreover, the best tutorial on 555 ever (& a lot of other topics too!) is by Tony van Roon, and incidentally this would be the first result on google if you type "555 tutorial"; the link is:
http://www.uoguelph.ca/~antoon/gadgets/555/555.html

Answer 2

There are 2 modes of 555 operation, 1) astable, where the 555 continuously outputs a pulse train. Pin 1 of the 555 is connected to ground, Pin 3 is the output, 8 is the power pin, it is connected to Vcc, pin 4 the reset pin is also tied to Vcc unless you want to reset the 555 timer, this is done by grounding the pin A 0.1 uF capacitor is tied from pin 5 to ground. You connect 2 resistors and a capacitor to the 555 timer. The 1st resistor call it R1 is tied from +Vcc to pin 7 of the 555. The second resistor R2 is also tied to pin 7 of the 555 and to pins 2 and 6 of the 555. The capacitor C is also tied to pins 2 and 6 of the 555 and the other end to ground. Pin 3 is the output pin The output pulse time T1 = 0.693(R1 + R2)C, where R1 and R2 are in ohms and C is in farads, T1 is seconds. The 555 output is low for time T2 which is T2 = 0.693(R2)C. The duty cycle is R1/(R1 + R2) The frequency is 1/(T1 +T2) 2) One shot ( Monostable vibrator) where the 555 is triggered by a low going pulse and outputs a pulse T seconds long when triggered. again pin 1 is ground, 3 is the output pin, 4 is the reset pin, 5 is where a 0.1uF capacitor is connected to ground and 8 is the power pin Pin 2 is the trigger pin a low voltage triggers the 555 timer. The timing resistor R is tied from Vcc to pin 6, and the timing capacitor C is tied from 6 to ground The pulse duration is T where T = 1.1(R)(C) again R is in ohms, C in farads, and T in seconds.

Answer 3

"They only provide the equations to calculate"

Do the calculations and use the parts they suggest. All I can do is go look at the datasheet for the 555 and tell you the equations it has listed. One of the main things to look for is that when you do the calculations, use reasonable values. For example R1 << R2, then don't pick R2 to be 50 ohms, meaning you have to get a really small R1 (also consider the current draw when you are using such small resistors).

A more reasonable case would be R1 = 1k and R2 = 100k or something of the sort.

Answer 4

555 Timer Circuit Calculator

Answer 5

You're not going to get a 50% duty cycle out of a 'triple nickle' timer. About he best you'll get is around 35% to 40%. (Go to Nationals website, D/L a copy of the 555 Timer .pdf, and look at the equations for 'on' and 'off' times and you'll see why) If you really have to have a 50% duty cycle, I'd suggest you run it at 1 Hz and use that to drive half of a 7474 type-D flip flop. Tie the ~Q output to the D input, tie the preset and clear inputs to Vcc, drive the clock input with the output of the '555, and take the output from the Q side of the flip-flop. That will give you a .5 Hz square wave with a 50% duty cycle.

Doug