More than a half-century old, the ubiquitous and mind-bogglingly useful 555 analog timer has become a perennial object for both kudus and criticism.  Most of the former and some of the latter is justified, but sometimes a supposed shortcoming will grow to the status of “common knowledge” just because a simple fix has been overlooked, even for a part in such long, wide, and popular use as the 555.  A case in point is the oft stated (yet mistaken) notion that it’s complicated to coax the original (bipolar) version of the 555 to oscillate with a symmetrical square wave 50:50% duty cycle.

The origin of this myth seems to be early 555 datasheet application notes that illustrated the topology shown in Figure 1 for astable oscillation.  Its expression for output duty cycle =

(R1 + R2):R2 tells us 50:50 will be impossible unless R1 = 0.  Since a short circuit from Discharge to Vs would seem more likely to make the part output smoke than square waves, the conclusion follows: 50:50 square waves are, indeed, impossible.

However, fortunately, as illustrated in Figure 2, the classic two-resistor topology isn’t the only way available to defrock this feline.  Specifically, if a third resistor (R3) is added in series with the Discharge pin, then a new signal level = Vd = Vs(R3/(R1+R3)) is introduced as the asymptotic limit for the exponential discharge of C1, thus stretching the T2 phase of the oscillation cycle.  In addition, the RC time constant for T2 is increased from R2C1 to (R2 + R1R3/(R1+R3))C1.

Figure 2 New “three resistor” topology.

The net effect, if R2 = R1 and R3 = 0.225*R1, then 

T1 = Ln(2)*(2R1)C1 T2 = Ln((2/3Vs – Vd)/(Vs/3 – Vd))*(R1 + R1R3/(R1+R3))C1 = Ln(2)*(2R1)C1 T1 = T2 Fosc = 0.36/(R1C1) Duty = 50:50

The method also works for many R2/R1 ratios different from 1.0   Some examples appear in the Table below:

In every case Fosc = 0.72/((R1+R2)C1) And duty cycle = 50:50

As noted, the method works all the way down to R2 = 0 (i.e., leaving only R1 and R3 with R2 replaced with a direct connection).  However, R3 = 0.423*R1 makes Vd = 0.297*Vs which is uncomfortably close to the 555’s nominal Trigger voltage = 0.333*Vs and may seriously exaggerate the effects of offset voltages and on-chip resistor tolerances on Fosc and duty cycle.

Meanwhile, what’s the moral to our story?  Even after 50 years, myths are made for busting!

The two-resistor configuration suggests an interesting variation on the idea (Figure 3a).  The circuit shown allows duty cycle to be varied over 0 to 100% with a single trimmer, while maintaining a constant positive pulse width (Figure 3b):

Figure 3a The new circuit.

Figure 3b How the duty cycle varies with R2 setting.

Stephen Woodward‘s relationship with EDN‘s DI column goes back quite a ways. In all, a total of 64 submissions have been accepted since his first contribution was published in 1974.

Many engineers now chose one of the CMOS replacements instead of the classic bipolar 555. These are faster, use less power, and don’t have the ugly supply-current glitch. They also have a rail-to-rail output stage, which allows for a much more simple 50% duty-cycle oscillator. Tie the comparator pins 6 and 7 together, with the usual capacitor to ground, and connect a resistor to the output. The comparator pins cycle between 1/3 and 2/3 supply voltage.

That’s certainly a valid solution and saves 2 resistors, provided of course you don’t need the more robust output current rating (4x for sink and 20x for source) of the bipolar part, its ability to work with higher supply voltages, or its stable Fosc and duty cycle that are independent of output loading. Moral: Sometimes even dinosaurs know a useful trick or two.

There’s lots of value in your exercise, and we appreciate seeing it. (I’ve always treasured your contributions.) One big advantage of the classic 555, is it’s (a) really cheap, (b) widely available, and (c) very rugged. So it’s nice to have a solution that doesn’t depend on the output. And (d), it was designed by our other hero, Hans Camenzind.

I have an ALD555 running at 1 Mhz that is wired for 50% duty cycle as per most datasheets. Pins 2 to 6 tied together, with R connected from 2/6 to 3 and C connected 2/6 to ground. 7 open.

I want to try your suggestion.

“Tie the comparator pins 6 and 7 together, with the usual capacitor to ground” – 6/7 to C to ground.

I’m not following the rest . When you say output you mean pin 3? R gets connected from 3 to what? What about pin 2 ?

Perhaps inelegant, but I’d always dispense with R1 & R2, and simply add a resistor from pin 3 to pin 6 leaving ping 7 unconnected. Of course any load on pin 3 would vary the duty cycle.

Thank you Stephen Woodward! This is another bright Design Idea which adds to the many that you published and I enjoyed reading over the years.

And about the NE555 current spike. This comes from page 17 of the TI CMOS LMC555 datasheet….

“Power Supply Recommendations …Adequate power supply bypassing is necessary to protect associated circuitry. Minimum recommended is 0.1 μF in parallel with 1-μF electrolytic.”

Interestingly, that cap’ combination would probably be adequate for the bipolar part, too.

I didn’t include this trick in the DI per-se, but an interesting (and maybe even useful) situation arises when R2=0 and R3 is adjustable (e.g., a rheostat-connected trimpot) from 0 to R1/2. Then T1=Ln(2)R1C1 independent of R3, while T2 varies from zero to infinite as R3 is twiddled from 0 to R1/2.

Thus you’d have constant T1 pulse duration and duty cycle adjustable from 100% to 0%.

One utility-limiting caveat would be that substantial and increasing amounts of T2 jitter should be expected as R3 approaches R1/2 and duty cycle < 50%.

For those interested in the 0-100% variable duty cycle circuit, please see the new addendum.

The LM555 National Semiconductor data sheet from July 2006 shows a “50% Duty Cycle Oscillator” with two resistors. R2 from above is shorted, thus R2/R1=0.

Interesting reference. Thanks! Surprisingly, it turns out the title you supplied was enough for Google to find it. Here’s the link:

https://datasheet.octopart.com/LM555CM-NOPB-Texas-Instruments-datasheet-7283993.pdf

See Figure 14. Note that the ratio of their Ra/Rb = 0.431 is about as close to the R3/R1 = 0.423 given in my text for R2 = 0 as you can get with standard 5% resistors.

Oops. Typo. Make that Rb/Ra = 0.431.

Using the R3 resistor to get 50 % duty cycle is known for many years. A paper from October 2007 studies the circuit; see https://www.electronicdesign.com/archive/article/21765225/adjust-555based-generators-duty-cycle-without-affecting-frequency Unfortunately the web site does not provide link to the drawings and the tables. A printed version of the magazine would do the job. Out of that, the 555 circuit is amazing indeed. Originally designed as a “timing machine” with three modes of operation, it has accumulated dozens of other applications during the years. The 741 opamp is probably the only other example of such a great success.

Hi, do you have a photo of that printed article with diagrams? Thanks.

Try a Google with the search terms “555” and “Jordan Dimitrov”, I just did and among the hits was a version of that article with intact figures.

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