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Step 7: 555 Timer: Astable Mode Duty Cycle

The duty cycle of a pulse wave is the ratio of the time it spends high to the total duration of the high and low state. We calculated these durations in step 5, and we can combine them to calculate the duty cycle of the 555:

duty cycle = (time spent high) / (total time duration of high and low states)
substitute the equations from step 5 to get:
duty cycle = (0.7*(RA + RB)*C) / (0.7*(RA+2*RB)*C)

this simplifies to:
duty cycle = (RA+ RB) / (RA + 2*RB)

In the equation above, when RA is much larger than RB (you can ignore the RB terms) you end up with a duty cycle ~= 1 and when RB is much larger than RA (you can ignore the RA terms) you get a duty cycle =~ 1/2. So the limits of the duty cycle with the circuit shown in fig 2 are 50% to 100%. If you wanted to get a duty cycle that was less than 50%, you have to use a circuit like the one shown in fig 1. In this circuit a diode bypasses RB during the charging phase of the 555 (while the output is held high). So how does this affect the durations of the high and low phases of the output?

The duration of low output remains:
t = 0.7*RB*C seconds
this happens when the capacitor is discharging, so current is flowing from the capacitor, through RB (in the upwards direction in fig 1), and into the discharge pin of the 555. This is the opposite direction of current flow that the diode will accept, so no current flows through the diode. During this time, the circuit in figure 1 is functionally equivalent to the circuit in fig 2.

The duration of high output does change, most notably the RB contribution goes away because it is being bypassed by the diode. In this case the capacitor is being charged so current is flowing from the power supply Vcc, through RA (in the downward direction in the schematic), and through the diode to the capacitor. Current will not flow through RB because the path through the diode is the path of lease resistance; the diode is essentially acting as a wire across RB.

Previously the duration of high output was:
t = 0.7*(RA+RB)*C seconds
we cannot simply remove RB from the equation because we need to account for a slight voltage drop (about 0.7V for silicon diodes) across the diode. We calculated the general form of the duration of high output in step 5. I've reproduced it below:
t = -(RA+RB)*C*ln[1/3*Vcc/(Vcc - V0)]
we should subtract the voltage drop from the diode (Vd) from both instances of Vcc in this equation and remove the contribution from RB
t = -RA*C*ln[(1/3*Vcc-Vd)/(Vcc - Vd - V0)]
as in step 4, the initial voltage V0 equals 1/3Vcc
t = -RA*C*ln[(1/3*Vcc - Vd)/(Vcc - Vd- 1/3Vcc)]
t = -RA*C*ln[(1/3*Vcc - Vd)/(2/3*Vcc - Vd)]
t = -RA*C*ln[(Vcc - 3*Vd)/(2*Vcc - 3*Vd)]

so the the duration of the high output is now
t = RA*C*ln[(2*Vcc - 3*Vd)/(Vcc - 3*Vd)]
Notice how there is no RB dependance. Also notice how the voltage drop across the diode and the supply voltage have an effect on the equation.

It's good to note here that you can also use monostable mode with an external trigger to create a PWM signal of duty cycles between 0 and 100%. I explained how to do this at the end of step 4. Even more info about PWM with the 555 timer can be found on the datasheet.

thanks for this. will be using this for a class report. ?
<p>Thanks a lot!!!!. It was very much helpful. I wanted to understand 555 from basics and it helped me a lot. Thanks :-)</p>
https://youtu.be/7KxIrGgPh9g
hey visit my YouTube channel for interesting 555 timer projects <br>https://youtu.be/7KxIrGgPh9g
<p>Can any of the pin be used to give an output logic 1 unless I open the switch??</p>
<p>Can any of the pin be used to give an output logic 1 unless I open the switch??</p>
Can any one help me to calculate the resistance of PWM to make astable pulse 50% duty cycle time on/off 5 second.. <br>The formula T=0.693 x R x C or T= 1.1 x R x C ?
<p>so if say a triangular input is applied at the trigger input of monostable 555 timer is it still possible to get rectangular waveforms from the output of the timer ? and how does the period of trigger input affect the output? thank you</p>
<p>The output on pin 3 is always a switched ON/OFF output. Pin 7 is the same except that it is open collector (no internal or active pull-up). That is, the rise and fall time are fast. The output is from a comparitor which produces an instant ON and OFF. The trigger level is when the voltage on pin 2 and pin 6 reach their respective trigger voltages (1/3 and 2/3 of the voltage that is on pin 5). The dead-band or hysteresis is the voltage difference between 1/3 and 2/3 of the voltage on pin5 and the difference in voltage of pins 2 and 6. Most simple timers/oscillators have pins 2 and 6 tied together. </p><p>The 555 is a very versatile device and can be operated in many nonstandard ways. The 7555 is a little better equivalent because it is 100% CMOS and can operate far more symmetrically than the 555 at lower operating currents. </p><p>The input on pins 2 &amp; 6 can have any shape from square to sine depending on what you are designing.</p><p>good Luck </p><p>Pixeltamer</p>
<p>like in this picture ???</p>
<p>Hi,</p><p>Thanks for the detailed instructable!</p><p>I was wondering if anyone knows the minimum input pulse length in monostable mode (for catching a <em>very</em> short sensor input pulse in a ballistic chronograph, for example).</p><p>Thanks</p>
<p>For what I have found in the datasheets, the minimum triggertime should be about 25nS. So... quite fast I think ;-)</p>
Thanks!
<p>hi . If i'm gonna change the momentary switch to a electret mic to make it a clap switch circuit. is it possible? what should be change on the connections?</p>
<p>Johanna, you have to amplifie the output of that micro first. Then you can connect the amplified output to the trigger input .</p>
Thank you for this instructable! Question: if I set up the resistors and capacitors right, would I be able to create a 5 minute pulse? Or would I need a microcontroller? Probably a stupid question but I am very new to electronics
<p>You can do that, Brad. Although you have to take care about your capacitators. When using ELCO's, high values give high leakage-currents. So the pulses might not be stable. For very long timing, it is possible to use a 10x divider (or more than one).</p>
<p>Great information, you put a lot of work into this to help people learn useful info.</p>
<p>Very detailed instruction set, Thank You for the explanation, I was struggling with trying to figure out how to get a 10% duty cycle on a &quot;blinker&quot; and this worked great. Again Thank You very much</p>
<p>Thanks sir, very nice and very clear.</p>
<p>Hi!<br>Is there any way to mix the mono and bistable mode? I mean pressing 1 button to turn it on the led, and keep it on untill press the other button (or the same button, this is not a critical issue) then after a time, the led goes off?</p>
<p>In the diagram pin 7 is connected directly to the pot and then to Vcc, but in your breadboard the pot goes to pin 6. Can you explain this discrepancy?</p>
<p>Nice article. One mistake though. You use a non standard layout for your astable circuit diagram. This is very amateurish.<br><br>http://www.555-timer-circuits.com/common-mistakes.html</p>
<p>put the mic at the trigger but light will only stay on for 5 seconds lol, i suppose this circuit is to trigger something for only 5 seconds after that its an off circuit</p>
<p>Thank you so much for the information you provided and used simple words.Can you please tell what is the use of capacitor in this circuit? Can anyon please...............answer?</p>
<p>Well, you could specify a little bit better what circuit you mean, but in general the use of the capacitors in these examples is because they need a very stable Voltage. The capicator insures this, ( just because of how a capacitor is build and where it is placed). Hope this helps.</p>
<p>You have an error in one of your equations. In point 3 you have:<br><br><strong>2/3*Vcc/</strong><strong>(Vcc - V0)</strong><strong> = 1- e^(-t / [(RA+RB)*C])</strong></p><p><strong>1/3*</strong><strong>Vcc/(Vcc - V0)</strong><strong> = e^(-t / [(RA+RB)*C])<br><br></strong>But this is not correct, it just happens to work for V0 = 0. This becomes a much larger problem in point 5 when you reuse the result to determine the amount of time the output is HIGH. <br><br>The way to calculate the time the output is HIGH is to subtract the time it takes to repeat the calculation from point 3, except solve for 1/3*VCC. Then subtract this result from the solution to point 3.<br><br>Your answer coming out correct is actually quite surprising. </p>
<p>does anyone know how to configure the 555 timer as a monostable multivibrator with a delay of about 100-500ms?? thx</p>
<p>T = 1.1 RC select C= 1uf @T= .1 sec R= T/1.1xC,,Do the same with T= .5 sec</p>
<p>This is exactly what i was looking for! I have little to no base in electronics and i'm trying to conceive a led dimmer in order to control a rgb led color. Your article mad me understand how a 555 timer worked and your equations will come in very handy! Thank you!</p>
<p>I'm planning on creating an astable multivibrator with as close to 50% duty cycle as possible. Thank you for the equations!</p>
<p>I followed your Astable guide to make a 200 Hz tone using Ra=2700, Rb=2200, and C=1uF. Was going crazy at first because I had two broken 555 chips in a row and didn't know. I wanted to say that this instructable is such good documentation to have, it is both thorough and informative -- thank you!</p>
<p>thanks for the post... the bistable circuit was exactly what i needed to toggle 2 hall effect sensors (A3144E) when all i had laying around were some 555 chips.</p>
<p>Im working on a model train layout, what i want is for a kid to press a button the train goes for 2 <strong>min</strong> then stops. the child does this 4 more times. after the train stops for the 5th time it activates a cool down clock for 10 min. it would have to work with electricity from an out lit. Any ideas? </p>
<p>I want this circuit, but I have 741 opamps in stead of the momentary switches. Can I connect the 741 output to the bottom of the pull-up resistors to set or reset the 555? or is some sort of buffering or isolation necessary between the 741 and the 555?</p>
question looking a building a 555 monostable mode can I use a switch that when trigger high, it will delay off after x seconds and stay off without the trigger input going back to low?
<p>Nice! Well written.</p>
<p>Hey everyone, I need to run a motor for a 3 second pulse, but only once, not a continuous on-off situation. and i need it to run in both directions. Is it possible to use this circuit for that or will i need something else? </p>
<p>sounds like you will need a microcontroller</p>
<p>Hi.. I have tried to make the monostable mode of 555 timer exactly as it is shown here for 5.17s pulse width. But i m not getting the output.The pulse becomes high when the push button is on but it comes to low only when the push button is switched off. I have checked the circuit connections numerous times but there is no error. Can you please tell me what might be the problem?</p>
<p>i had the same problem on my first trial,try changing your breadboard,and see it is works. :)</p>
<p>Thank you :)</p>
hi,pls am new in d electronic world but understand the basis.how can i increase or amplifer dc 2v to 12v dc to charge 12v battery.pls state all d component require and the diagram to build dis circult,and i want the 12v dc to constant. <br>thanks
For astable mode,when I added a load to output, the frequency increase. But it seems like the frequency is independent of load. Anyway to overcome it?
I'm not sure why that's happening, try putting a buffer between the load and the 555, use an op amp or transistor wired up like this: <br />http://en.wikipedia.org/wiki/Buffer_amplifier <br />
Thanks a million :)
Is there a way to adjust the low pulse so it's not 0?
nope sorry, these are digital circuits so 0 and 5V is all you can get. Do you have something specific in mind that you'd like to do?

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