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#1
11-15-2012, 09:33 AM
 SNVHKC New Member Join Date: Jun 2012 Location: India Posts: 4
555 timer 40khz 50%duty cycle

Could someone help me to generate a 40khz square pulses(Astable multivibrator) using 555timer IC,with 50% duty cycle.
#2
11-15-2012, 10:50 AM
 JDT Senior Member Join Date: Feb 2009 Location: UK Posts: 577

This is not the correct forum for this!

A 555 cannot generate 50% duty cycle. Also I'm thinking that a 555 is not the best device for operation at this frequency.

If a 50% duty cycle is important I would have an oscillator that runs at twice the required frequency and follow it with a divide by 2 flip-flop. Easily done in 4000 series CMOS.

Edit: In fact have a look at the datasheet for a HEF4047B. It's all in there!
 The Following User Says Thank You to JDT For This Useful Post: SNVHKC (01-04-2013)
#3
11-17-2012, 07:01 PM
 BillO Senior Member Join Date: Nov 2008 Location: Canada Posts: 974

Here is link to the data sheet. Page 10 shows one method of doing 50% duty cycle.

You can also google "555 50% duty cycle" and you'll get a few hundred thousand hits.
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 The Following User Says Thank You to BillO For This Useful Post: SNVHKC (11-19-2012)
#4
11-19-2012, 03:51 PM
 tracecom Senior Member Join Date: Apr 2010 Location: West Tennessee Posts: 3,234

In the attached schematic, make R1 = 910 Ω, C1 = .001 μF, and R2 = 17545 Ω. (You may use a 15 or 17 kΩ fixed resistor and a series trimpot for R2.) Mathematically, this should produce a 40 kHz output with a 51.26% duty cycle. Of course, component tolerances will cause variation.
Attached Images
 NE555 Astable Schematic.png (16.8 KB, 117 views)
 The Following User Says Thank You to tracecom For This Useful Post: SNVHKC (01-04-2013)
#5
11-19-2012, 07:00 PM
 spark8217 Member Join Date: Aug 2011 Location: uk Posts: 57

Formula: T = 0.7 * (R1 + (2 * R2)) * C1 (without diode)
f = 1.4 / ((R1 + (2 * R2)) * C1) (without diode)

T = 0.7 * (R1 + R2) * C1 (with diode)
f = 1.4 / ((R1 + R2) * C1) (with diode)

Symbols: T = Time period (Units: Seconds)
f = Frequency (Units: Hertz)
Mark = High time of output waveform (Units: Seconds)
Space = Low time of output waveform (Units: Seconds)
Duty Cycle = % of period T in which output is high (mark)

Data: Mode: Astable
Calculate component values
Frequency = 40 Kilohertz
Duty Cycle = 50 %

Results: Period = 25 Microseconds
Mark = 12,5 Microseconds
Space = 12,5 Microseconds
R1 = 1 Kilohms
R1 nearest preferred value = 1 Kilohms (E24 / 5%)
R2 = 1,786 Kilohms
R2 nearest preferred value = 1,8 Kilohms (E24 / 5%)

Diode D1 should not be fitted
 The Following User Says Thank You to spark8217 For This Useful Post: SNVHKC (01-04-2013)
#6
11-20-2012, 01:09 AM
 THE_RB Senior Member Join Date: Feb 2008 Posts: 5,377

Don't use pin7, drive the cap from pin3 which is bipolar (through one resistor), and you just need 1 cap and one resistor and it automatically makes a 50% duty oscillator.

I have not used pin7 in a while, unless I specifically need a 555 with a duty that is NOT 50%.
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 The Following 2 Users Say Thank You to THE_RB For This Useful Post: claudewang (11-10-2013), SNVHKC (01-04-2013)
#7
11-10-2013, 01:41 PM
 claudewang New Member Join Date: Oct 2013 Posts: 1

Quote:
 Originally Posted by THE_RB Don't use pin7, drive the cap from pin3 which is bipolar (through one resistor), and you just need 1 cap and one resistor and it automatically makes a 50% duty oscillator. I have not used pin7 in a while, unless I specifically need a 555 with a duty that is NOT 50%.
Hi, there. I have been looking for a way of using ne555 as a 50% duty oscillator with only one resistor tuning for frequency. Your method sounds awesome, but it's still not very clear to me how to implement it. Could you please do me a favor to show me an example circuit that can do exactly what you said? Many thanks.
#8
11-10-2013, 02:39 PM
 MikeML Senior Member Join Date: Oct 2009 Location: AZ27 Posts: 1,437

Here is an example of using the 555 to generate a square wave. If doing this at 5V Vcc or lower, use the CMOS version of the 555. Take the output as shown so as not to screw up the timing. Note that V(rc) turns around at 1/3Vcc and 2/3Vcc.
Attached Images
 DF116.jpg (230.5 KB, 64 views)
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Last edited by MikeML; 11-10-2013 at 02:44 PM.
 The Following User Says Thank You to MikeML For This Useful Post: claudewang (11-10-2013)
#9
05-02-2014, 11:23 AM
 sakar New Member Join Date: May 2014 Posts: 2

spark8217 can you please post the circuit diagram relative to which you are providing the values.
#10
05-02-2014, 11:25 AM
 sakar New Member Join Date: May 2014 Posts: 2

Quote:
 Originally Posted by spark8217 Formula: T = 0.7 * (R1 + (2 * R2)) * C1 (without diode) f = 1.4 / ((R1 + (2 * R2)) * C1) (without diode) T = 0.7 * (R1 + R2) * C1 (with diode) f = 1.4 / ((R1 + R2) * C1) (with diode) Symbols: T = Time period (Units: Seconds) f = Frequency (Units: Hertz) Mark = High time of output waveform (Units: Seconds) Space = Low time of output waveform (Units: Seconds) Duty Cycle = % of period T in which output is high (mark) Data: Mode: Astable Calculate component values Frequency = 40 Kilohertz Duty Cycle = 50 % Results: Period = 25 Microseconds Mark = 12,5 Microseconds Space = 12,5 Microseconds C1 = 10 Nanofarads R1 = 1 Kilohms R1 nearest preferred value = 1 Kilohms (E24 / 5%) R2 = 1,786 Kilohms R2 nearest preferred value = 1,8 Kilohms (E24 / 5%) Diode D1 should not be fitted

Can you please post the circuit diagram relative to which you are provoding the values.

 Tags 40khz, 50%duty, 555, cycle, timer

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