4
\$\begingroup\$

I wanted to build a circuit that gives 50%-duty-cycle square wave. I built the circuit of figure 4.4.6 from this link which is shown below:

I used the NE555 chip, and used a potentiometer instead of the resistor. The resulting waveform is shown below:

You can see that it's not a 50%-duty-cycle as expected.

What is the reason of that? Charging and discharging are done using the same resistor and capacitor, so on-time and off-time should be equal.

\$\endgroup\$
6
  • \$\begingroup\$ We can see the picture, but the link to the page where you got the picture seems to be missing. \$\endgroup\$
    – Andrew Morton
    Apr 11, 2016 at 17:30
  • \$\begingroup\$ Oh sorry. I added the link. \$\endgroup\$
    – ammar
    Apr 11, 2016 at 17:48
  • 3
    \$\begingroup\$ If you use the CMOS version instead of the ancient bipolar version, you will get very close to 50% duty cycle, provided you don't overload the output. \$\endgroup\$
    – Spehro Pefhany
    Apr 11, 2016 at 18:17
  • 5
    \$\begingroup\$ A common old trick is to just run the 555 at twice the frequency you really want and connect its output to a flip-flop connected as a divide-by-two. The output will be completely symmetric for all ordinary purposes. \$\endgroup\$
    – Jamie Hanrahan
    Apr 11, 2016 at 20:25
  • \$\begingroup\$ Ditto that! If it's important for the duty cycle to be as close as possible to 50%, then a divide-by-two stage will remove any asymmetry from the waveform. \$\endgroup\$
    – Solomon Slow
    Apr 11, 2016 at 21:21

5 Answers 5

Reset to default
9
\$\begingroup\$

If you look at the 555 internal diagram you can see that the output you use is not symmetrical: the top side is a darlington, so it has (at least) two Vbe drops. The low side is a common emitter, so it can be saturated.

enter image description here

In short: the output low will be closer to ground than the output high will be to Vcc.

One way to compensate would be a low-valued resistor between the output and Vcc.

\$\endgroup\$
8
  • \$\begingroup\$ Aha, thanks. But can you please explain how connecting that resistor will compensate? \$\endgroup\$
    – ammar
    Apr 11, 2016 at 17:49
  • 1
    \$\begingroup\$ It will lower the voltage drop when the output is high. But the cost is a higher current when the output is low. \$\endgroup\$
    – Wouter van Ooijen
    Apr 11, 2016 at 17:50
  • 3
    \$\begingroup\$ I'm not sure how this answer has anything to do with duty cycle. \$\endgroup\$
    – tcrosley
    Apr 12, 2016 at 2:01
  • 2
    \$\begingroup\$ @tcrosley Instead of the default 555 configuration, which uses the discharge pin and a pull-up resistor, the OP uses the output pin to drive the charge and discharge of the timing capacitor. He hoped this would lead to a (more) symmetrical waveform, but the drive of the output pin is not symmetrical: high is less high than low is low, hence the high part will take longer than the low part (as can be seen on the scope output). \$\endgroup\$
    – Wouter van Ooijen
    Apr 12, 2016 at 5:27
  • 1
    \$\begingroup\$ 1k becomes too hot?? What is your supply voltage? (And what you do mean by 'too hot': to touch? Does it smoke?) \$\endgroup\$
    – Wouter van Ooijen
    Apr 24, 2016 at 15:28
4
\$\begingroup\$

The Original Source explains this.

Most 555 astable oscillator designs use two resistors and depend on Vcc. This typically means a Duty cycle > 50%.

This design uses Vout, so the timing circuit acts as a load on the output, which can effect frequency and mark to space ratio. An idiosyncrasy of the design.

Try Adjust 555-Based Generator's Duty Cycle Without Affecting Frequency from Electronic Design. They walk you through how to set up resistors and capacitor.

50% duty cycle 555 Edit...

For completeness, I've included the formulas from Electronic Design and modified procedure. $$p = \frac {R_2} {R_1}\ \ \ \ \ \ q = \frac {R_3} {R_1}$$ $$t_1 = R_1\ C\ (p+1)\times ln(2)\ \ [1]$$ $$t_2= R_1\ C\ \left ( {p + \frac {q}{q+1}} \right ) \times ln\left ( {\frac {q-2}{2q-1}} \right )\ \ [2]$$ For 50% Duty Cycle: $$t_1 = t_2$$ $$R_1\ C\ (p+1)\times ln(2) = R_1\ C\ \left ( {p + \frac {q}{q+1}} \right ) \times ln\left ( {\frac {q-2}{2q-1}} \right ) \ \ [3]$$ $$f = \frac {1}{2 t_1}= \frac {0.7213}{R_1\ C\ (p+1)}\ \ [4]$$

The procedure is as follows:

  1. Set the desired frequency and select a value for C.
  2. Calculate R1 from Equation 4.
  3. Pick a R3.
  4. Calculate q.
  5. Calculate the value of p from Equation 3.
  6. Calculate R2 = pR1.
\$\endgroup\$
2
  • \$\begingroup\$ Thanks. But the link you provided is of the picture you put. \$\endgroup\$
    – ammar
    Apr 11, 2016 at 20:00
  • \$\begingroup\$ Fixed. I initially checked it, but must of fooled it up along the way. \$\endgroup\$
    – StainlessSteelRat
    Apr 11, 2016 at 20:08
2
\$\begingroup\$

The only sure-fire way I know of to get a perfect 50% duty cycle is to send the 555 output to a flip-flop of your choice. Make sure it is compatible with the 555's working voltage.

Any pulse stream the 555 puts out is divided by 2, with a perfect 50% duty cycle. I would prefer that over several trim pots (which drift with temperature by as much as 200 ppm). The flip-flop also sharpens the rising and falling edge of the square wave.

For 12 volt circuits you can use the CD4013, for 5 volt circuits you can use a 74HC74, which will work down to 3 volts. Do not use 74AC74, as its rise/fall time is so fast it may cause ringing in the outputs. By itself the 555 has its limits without adding a lot of parts.

\$\endgroup\$
1
  • \$\begingroup\$ According to the website the OP used for that design, it will work but many factors can make the duty-cycle other than 50%-50%. It would never be used in a production run where a 50% duty cycle had to be accurate and stable over time. \$\endgroup\$
    – user105652
    Apr 15, 2016 at 0:57
0
\$\begingroup\$

According to the datasheet I found at: http://www.ti.com/lit/ds/symlink/lm555.pdf, page 11,

$$D=\frac{R_B}{R_A+2R_B}$$

where R_A is between pin 7 and +V_CC and R_B is between pins 6 and 7. I don't see either resistor in your diagram.

\$\endgroup\$
1
  • 4
    \$\begingroup\$ That's an entirely different way of using the 555 as an astable. The OP's way is equally valid, and simpler. \$\endgroup\$
    – user16324
    Apr 11, 2016 at 17:43
-1
\$\begingroup\$

To get a squarewave from a 555; - I run the 555 at some HIGHER frequency ...and feed it into a 74393 dual counter chip ...versatility and accuracy goes THROUGH the ROOF!!! http://www.ti.com/lit/ds/symlink/sn54ls393-sp.pdf

\$\endgroup\$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.