In the previous tutorial we saw that the **555 Timer IC** can be connected either in its Monostable mode thereby producing a precision timer of a fixed time duration, or in its Bistable mode to produce a flip-flop type switching action.

But we can also connect the 555 timer IC in an Astable mode to produce a very stable **555 Oscillator** circuit for generating highly accurate free running waveforms whose output frequency can be adjusted by means of an externally connected RC tank circuit consisting of just two resistors and a capacitor.

The **555 Oscillator** is another type of relaxation oscillator for generating stabilized square wave output waveforms of either a fixed frequency of up to 500kHz or of varying duty cycles from 50 to 100%. In the previous 555 Timer tutorial we saw that the Monostable circuit produces a single output one-shot pulse when triggered on its pin 2 trigger input.

Whereas the 555 monostable circuit stopped after a preset time waiting for the next trigger pulse to start over again, in order to get the 555 Oscillator to operate as an astable multivibrator it is necessary to continuously re-trigger the 555 IC after each and every timing cycle.

This re-triggering is basically achieved by connecting the *trigger* input (pin 2) and the *threshold* input (pin 6) together, thereby allowing the device to act as an astable oscillator. Then the 555 Oscillator has no stable states as it continuously switches from one state to the other. Also the single timing resistor of the previous monostable multivibrator circuit has been split into two separate resistors, R1 and R2 with their junction connected to the *discharge* input (pin 7) as shown below.

In the **555 Oscillator** circuit above, pin 2 and pin 6 are connected together allowing the circuit to re-trigger itself on each and every cycle allowing it to operate as a free running oscillator. During each cycle capacitor, C charges up through both timing resistors, R1 and R2 but discharges itself only through resistor, R2 as the other side of R2 is connected to the *discharge* terminal, pin 7.

Then the capacitor charges up to 2/3Vcc (the upper comparator limit) which is determined by the 0.693(R1+R2)C combination and discharges itself down to 1/3Vcc (the lower comparator limit) determined by the 0.693(R2.C) combination. This results in an output waveform whose voltage level is approximately equal to Vcc – 1.5V and whose output “ON” and “OFF” time periods are determined by the capacitor and resistors combinations. The individual times required to complete one charge and discharge cycle of the output is therefore given as:

Where, R is in Ω’s and C in Farads.

When connected as an astable multivibrator, the output from the **555 Oscillator** will continue indefinitely charging and discharging between 2/3Vcc and 1/3Vcc until the power supply is removed. As with the monostable multivibrator these charge and discharge times and therefore the frequency are independent on the supply voltage.

The duration of one full timing cycle is therefore equal to the sum of the two individual times that the capacitor charges and discharges added together and is given as:

The output frequency of oscillations can be found by inverting the equation above for the total cycle time giving a final equation for the output frequency of an Astable 555 Oscillator as:

By altering the time constant of just one of the RC combinations, the **Duty Cycle** better known as the “Mark-to-Space” ratio of the output waveform can be accurately set and is given as the ratio of resistor R2 to resistor R1. The Duty Cycle for the 555 Oscillator, which is the ratio of the “ON” time divided by the “OFF” time is given by:

The duty cycle has no units as it is a ratio but can be expressed as a percentage ( % ). If both timing resistors, R1 and R2 are equal in value, then the output duty cycle will be 2:1 that is, 66% ON time and 33% OFF time with respect to the period.

An **Astable 555 Oscillator** is constructed using the following components, R1 = 1kΩ, R2 = 2kΩ and capacitor C = 10uF. Calculate the output frequency from the 555 oscillator and the duty cycle of the output waveform.

t_{1} – capacitor charge “ON” time is calculated as:

t_{2} – capacitor discharge “OFF” time is calculated as:

Total periodic time ( T ) is therefore calculated as:

The output frequency, ƒ is therefore given as:

Giving a duty cycle value of:

As the timing capacitor, C charges through resistors R1 and R2 but only discharges through resistor R2 the output duty cycle can be varied between 50 and 100% by changing the value of resistor R2. By decreasing the value of R2 the duty cycle increases towards 100% and by increasing R2 the duty cycle reduces towards 50%. If resistor, R2 is very large relative to resistor R1 the output frequency of the 555 astable circuit will determined by R2 x C only.

The problem with this basic astable 555 oscillator configuration is that the duty cycle, the “mark to-space” ratio will never go below 50% as the presence of resistor R2 prevents this. In other words we cannot make the outputs “ON” time shorter than the “OFF” time, as (R1 + R2)C will always be greater than the value of R1 x C. One way to overcome this problem is to connect a signal bypassing diode in parallel with resistor R2 as shown below.

By connecting this diode, D1 between the *trigger* input and the *discharge* input, the timing capacitor will now charge up directly through resistor R1 only, as resistor R2 is effectively shorted out by the diode. The capacitor discharges as normal through resistor, R2.

An additional diode, D2 can be connected in series with the discharge resistor, R2 if required to ensure that the timing capacitor will only charge up through D1 and not through the parallel path of R2. This is because during the charging process diode D2 is connected in reverse bias blocking the flow of current through itself.

Now the previous charging time of t_{1} = 0.693(R1 + R2)C is modified to take account of this new charging circuit and is given as: 0.693(R1 x C). The duty cycle is therefore given as D = R1/(R1 + R2). Then to generate a duty cycle of less than 50%, resistor R1 needs to be less than resistor R2.

Although the previous circuit improves the duty cycle of the output waveform by charging the timing capacitor, C1 through the R1 + D1 combination and then discharging it through the D2 + R2 combination, the problem with this circuit arrangement is that the 555 oscillator circuit uses additional components, i.e. two diodes.

We can improve on this idea and produce a fixed square wave output waveform with an exact 50% duty cycle very easily and without the need for any extra diodes by simply moving the position of the charging resistor, R2 to the output ( pin 3 ) as shown.

The 555 oscillator now produces a 50% duty cycle as the timing capacitor, C1 is now charging and discharging through the same resistor, R2 rather than discharging through the timers discharge pin 7 as before. When the output from the 555 oscillator is HIGH, the capacitor charges up through R2 and when the output is LOW, it discharges through R2. Resistor R1 is used to ensure that the capacitor charges up fully to the same value as the supply voltage.

However, as the capacitor charges and discharges through the same resistor, the above equation for the output frequency of oscillations has to be modified a little to reflect this circuit change. Then the new equation for the 50% Astable 555 Oscillator is given as:

Note that resistor R1 needs to be sufficiently high enough to ensure it does not interfere with the charging of the capacitor to produce the required 50% duty cycle. Also changing the value of the timing capacitor, C1 changes the oscillation frequency of the astable circuit.

We said previously that the maximum output to either sink or source the load current via pin 3 is about 200mA and this value is more than enough to drive or switch other logic IC’s, a few LED’s or a small lamp etc and that we would need to use a bipolar transistor or MOSFET to amplify the 555’s output to drive larger current loads such as motor or relays.

But the **555 Oscillator** can also be used in a wide range of waveform generator circuits and applications that require very little output current such as in electronic test equipment for producing a whole range of different output test frequencies.

The 555 can also be used to produce very accurate sine, square and pulse waveforms or as LED or lamp flashers and dimmers to simple noise making circuits such as metronomes, tone and sound effects generators and even musical toys for Christmas.

We could very easily build a simple 555 oscillator circuit to flash a few LED’s “ON” and “OFF” similar to the one shown, or to produce a high frequency noise from a loudspeaker. But one very nice and simple to build science project using an astable based 555 oscillator is that of an Electronic Metronome.

Metronomes are devices used to mark time in pieces of music by producing a regular and recurring musical beat or click. A simple electronic metronome can be made using a 555 oscillator as the main timing device and by adjusting the output frequency of the oscillator the tempo or “Beats per Minute” can be set.

So for example, a tempo of 60 beats per minute means that one beat will occur every second and in electronics terms that equates to 1Hz. So by using some very common musical definitions we can easily build a table of the different frequencies required for our metronome circuit as shown below.

Musical Definition |
Rate | Beats per Minute |
Cycle Time (T) |
Frequency |

Larghetto | Very Slow | 60 | 1sec | 1.0Hz |

Andante | Slow | 90 | 666ms | 1.5Hz |

Moderato | Medium | 120 | 500ms | 2.0Hz |

Allegro | Fast | 150 | 400ms | 2.5Hz |

Presto | Very Fast | 180 | 333ms | 3.0Hz |

The output frequency range of the metronome was simply calculated as the reciprocal of 1 minute or 60 seconds divided by the number of beats per minute required, for example (1/(60 secs / 90 bpm) = 1.5Hz) and 120bpm is equivalent to 2Hz, and so on. So by using our now familiar equation above for calculating the output frequency of an astable 555 oscillator circuit the individual values of R1, R2 and C can be found.

The time period of the output waveform for an astable 555 Oscillator is given as:

For our electronic metronome circuit, the value of the timing resistor R1 can be found by rearranging the equation above to give.

Assuming a value for resistor R2 = 1kΩ and capacitor C = 10uF the value of the timing resistor R1 for our frequency range is given as 142k3Ω at 60 beats per minute to 46k1Ω at 180 beats per minute, so a variable resistor (potentiometer) of 150kΩ would be more than enough for the metronome circuit to produce the full range of beats required and some more. Then the final circuit for our electronic metronome example would be given as:

This simple metronome circuit demonstrates just one simple way of using a 555 oscillator to produce an audible sound or note. It uses a 150kΩ potentiometer to control the full range of output pulses or beats, and as it has a 150kΩ value it can be easily calibrated to give an equivalent percentage value corresponding to the position of the potentiometer. For example, 60 beats per minute equals 142.3kΩ or 95% rotation.

Likewise, 120 beats per minute equals 70.1kΩ or 47% rotation, etc. Additional resistors or trimmer’s can be connected in series with the potentiometer to pre-set the outputs upper and lower limits to predefined values, but these additional components will need to be taken into account when calculating the output frequency or time period.

While the above circuit is a very simple and amusing example of sound generation, it is possible to use the **555 Oscillator** as a noise generator/synthesizer or to make musical sounds, tones and alarms by constructing a variable-frequency, variable-mark/space ratio waveform generator.

In this tutorial we have used just a single 555 oscillator circuit to produce a sound but by cascading together two or more 555 oscillator chips, various circuits can be constructed to produce a whole range of musical and sound effects. One such novelty circuit is the police car “Dee-Dah” siren given in the example below.

The circuit simulates a warble-tone alarm signal that simulates the sound of a police siren. IC1 is connected as a 2Hz non-symmetrical astable multivibrator which is used to frequency modulate IC2 via the 10kΩ resistor. The output of IC2 alternates symmetrically between 300Hz and 660Hz taking 0.5 seconds to complete each alternating cycle.

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Can the two 555’s be squenced together to produce a …|.|…|.|…

basically rigger, trigger pause, repeat?

Wanting to trigger a pair of transistors squentially, with adjustable timing between the two and adjustable pause.

Thanks

Yes two 555 timers can be sequenced together. The NE556 or LM556 dual timer can be used for this.

Clear and to the point. Thanks

Excellent explanation of the 555 timer used in astable mode.

I’ve seen the 50% duty cycle circuit being used to measure the EC of solutions.

Is there any way to keep T1 constant witdth and only vary T2 with nonpolar Caps

connected to probe in solution….Then also connect pin 3 to [ constant current source -lm344+R ] to charge a cap an hopefully get a linear DC voltage of the frequency and determine the EC of solution , of course the probe K needs to be known….

Thank you for your reply,

As a matter of fact, I did consider the use of a resettable monostable circuit, however by doing so I’m missing a pulse each time the 555 is timing out. Le met explain.

I have a light signal, expected to flash between 1 and 1.5 secs, and by using appropriate buffer, I feed the pulse signal to a 4017 (decade counter).

The 555 will rise the 4017 RST pin if no pulse was detected within 2 secs. Doing so, this will reset the 4017 counter, which is what is wanted. Since the 555 doesn’t oscillate in a monostable configuration, the 4017 RST pin remains HIGH.

While the 4017 RST is high, the counter doesn’t increase with incoming pulses.

Afterwards, the first pulse received, after the 555 has “detected a missing pulse”, will reset the 555 cycle to get it ready for the next occurence. However, since the 4017 RST pin was HIGH when that pulse came in, the counter didn’t advance, but does advance with the next pulses, as long as the timeout delay is respected. Overall, the counter will be off by one each time the 555 will timeout, which is something I want to avoid.

This is why I originally opted for an astable configuration, making the 4017 reset to zero process a bit easier to handle.

Regards.

First of all, thank you for your great tutorials. They’re really well done and clear.

I’m actually doing a simple project which involves the use of a 555. For what I need to do, the circuit in the “Improved 555 Oscillator Duty Cycle” section, seems to be the one I need.

In my setup, R1 is 10K, R2 is roughly 200K and C1 is 22uF. For D1 and D2 I’m using 1N914 as I don’t have 1N4148 handy at this moment. This generate a short pulse every 3.5 secs or so, I’m using a trimpot so that I can adjust it later.

More likely because I didn’t have a complete understanding of the RESET pin, I designed my circuit in such a way that the RESET pin is brought LOW when needed via another portion of the project. I was under the impression that doing so would reset the cycle, but it seems to pause it instead. Let say the cycle is 3 secs long. If the RESET pin is brought LOW for 0.5 secs at t=1.5sec , then when brought back HIGH, it will take roughly 1.5secs for the cycle to complete, instead of the full 3secs as expected.

I have a strong feeling that this is due to the fact that C1 didn’t have enough time to discharge, therefore the circuit triggered “sooner” than what I anticipated.

Would it be advisable to add a NPN BJT (connecting C1 between the collector and emitter) and switch on that transistor at the same time as I bring the RESET pin LOW, in order to drain out C1 ?

Or would it be preferable to use that BJT to drain C1 and leave the RESET high all the time ?

Is there another way to drain C1, or is it my approach that is wrong ?

Your input would be much appreciated.

Best regards

Daniel,

It pauses instead of restarting because the discharge pin is not used in this circuit, so the capacitor keeps its charge after you bring reset pin to low and then high. If you keep reset low long enough, then yes, the cap gets discharged but it will take a while.

If you want a proper quick reset, use a circuit that discharges the cap through discharge pin.

Pin 4, the Reset pin, is used to control the state of the output on pin 3. The reset pin is activated when a low, (logic level “0”, 0V, or ground) condition is applied to the pin. The reset pin will force the the output to go “low” (logic “0”) no matter what the state of the other pins or timing process the 555 is in. Being an Astable connected 555, once pin 4 is high again, the timer will continue as before. Normally in Astable mode, to prevent unwanted false resetting of the 555, pin 4 is connected to Vcc.

You can make a resettable monostable by connecting pin 4 and pin 2 (trigger) together which will stop the 555 in the middle of its timing cycle and restart again when the signal goes high.

In the schematic for the 50% duty cycle oscillator, why isn’t pin 7 used? Will not using it possibly cause issues?

Also, could you explain how I would go about finding the R and C values for the 50% oscillator for a frequency of 2 Hz?

The timing capacitor, C1 charges and discharges through timing resistor, R2 rather than discharging through pin 7. Please use the formula above to find your R an C values for the required frequency of oscillation.

I want to use a 3.7V battery. Do you know any IC similar to lm555?

Hi,

it is possible to obtain a circuit capable of providing a 1Hz pulse with 1 % of duty cycle?

Referring to:

50% Duty Cycle Frequency Equation

Section: 555 Oscillator Applications

The diagram use the following values

R2 = 27 Kilo Ohm

C1 = 10 micro F

To help me with the calculations I created the following formula in excel. However I get a value of 2.67 Hertz whereas in your diagram you get 2 Hertz. Where am I making my mistake?

A1= 27 kilo ohm

A2= 10 mico F

=1/(0.693*(2*+A1*1000)*+A2/1000000)

Hello Fancois, you are not making a mistake, I rounded of the value of 2Hz. If you want an exact mathematical value then R would be 36kohm.