555 Astable Oscillators and Mark-Space Ratio

From One-Shot to Continuous Oscillation

The previous lesson showed how the 555 timer in monostable mode fires a single timed pulse when triggered. In astable mode, there is no stable state at all — the output switches continuously between HIGH and LOW, producing a repeating waveform without any external trigger. This makes the 555 astable a simple and reliable oscillator, used in LED flashers, alarm sounders, clock pulse generators and many other applications.

The Astable Circuit

The 555 astable uses two external resistors ($R_1$ and $R_2$) and one capacitor ($C$). The capacitor charges through both $R_1$ and $R_2$ in series, but discharges through only $R_2$. This asymmetry means the time the output spends HIGH is longer than the time it spends LOW, producing a waveform where the mark time and space time are different.

The timing equations for the two half-cycles are:

Time HIGH (mark time): $T_{ON} = 0.7 \times (R_1 + R_2) \times C$
Time LOW (space time): $T_{OFF} = 0.7 \times R_2 \times C$

Period and Frequency

The total period of one complete cycle is the sum of the two half-cycles: $T = T_{ON} + T_{OFF}$. Substituting the expressions above:

Key Fact

555 astable frequency:

$$f = \frac{1.44}{(R_1 + 2R_2) \times C}$$

where $f$ is in hertz, $R_1$ and $R_2$ are in ohms, and $C$ is in farads. Note that $R_2$ is multiplied by 2, not $R_1$.

Period $T = 1/f$. You can also find the period directly:

$$T = 0.7 \times (R_1 + 2R_2) \times C$$

Mark-Space Ratio

Mark-space ratio (sometimes called mark-to-space ratio) describes the shape of the output waveform:

$$\text{Mark-space ratio} = \frac{T_{ON}}{T_{OFF}} = \frac{R_1 + R_2}{R_2}$$

Because the capacitor charges through both $R_1$ and $R_2$ but discharges through only $R_2$, $T_{ON}$ is always greater than $T_{OFF}$ when $R_1 > 0$. This means the mark-space ratio is always greater than 1 in a standard 555 astable. A mark-space ratio of 1:1 (equal HIGH and LOW times) is called a 50% duty cycle. The standard 555 astable cannot achieve exactly 50% duty cycle unless $R_1 = 0$ Ω (which would damage the IC), so a modified circuit using a diode is needed for a 1:1 ratio.

A 555 astable has $R_1 = 10$ kΩ, $R_2 = 47$ kΩ and $C = 100$ nF. Calculate the frequency and mark-space ratio.

Step 1 — Convert units to SI base units:

$R_1 = 10 \times 10^{3} = 10{,}000$ Ω; $R_2 = 47 \times 10^{3} = 47{,}000$ Ω; $C = 100 \times 10^{-9}$ F.

Step 2 — Calculate the denominator:

$R_1 + 2R_2 = 10{,}000 + 2 \times 47{,}000 = 10{,}000 + 94{,}000 = 104{,}000$ Ω

Step 3 — Substitute into the frequency formula:

$$f = \frac{1.44}{104{,}000 \times 100 \times 10^{-9}} = \frac{1.44}{104{,}000 \times 10^{-7}} = \frac{1.44}{0.0104} \approx 138\text{ Hz}$$

Step 4 — Mark-space ratio:

$$\text{Mark-space ratio} = \frac{R_1 + R_2}{R_2} = \frac{10{,}000 + 47{,}000}{47{,}000} = \frac{57{,}000}{47{,}000} \approx 1.21$$

The output is HIGH for about 1.21 times as long as it is LOW.

Reading an Oscilloscope Trace

An oscilloscope displays voltage on the vertical axis and time on the horizontal axis. Two key controls determine the scale: volts per division (V/div) and time per division (time/div or timebase).

To read a 555 astable output waveform from an oscilloscope trace:

Amplitude: Count the number of vertical divisions from the bottom of the waveform to the top of the HIGH level. Multiply by V/div. This is the amplitude of the waveform. Do not confuse amplitude with peak-to-peak voltage — for a waveform that swings between 0 V and +9 V, the amplitude is 9 V and the peak-to-peak voltage is also 9 V (but for a waveform centred on 0 V, peak-to-peak is twice the amplitude).
Period: Count the number of horizontal divisions for one complete cycle (one HIGH + one LOW). Multiply by time/div. This gives the period $T$ in seconds (or ms or μs, depending on the timebase setting).
Frequency: $f = 1/T$. If $T$ was measured in ms, convert to seconds before dividing: $1$ ms $= 10^{-3}$ s.