555 timer

• posted

Hello,

Is it possible to gt 10% duty cycle with a 555. I went to this web site:

and I could not get les than 40-50%

Ken

• posted

Make R1 larger than R2 and the times will be at the same ratio.

Don

• posted

Yes you can get a 10% duty cycle out of a 555. Try these pages

which a couple of the many similar circuits you get if you Google "555 variable duty cycle"

I have used this sort of 555 circuit and it does work. The secret is in the diodes.

PH

Answer: Because it annoys Philthy Phil the Troll.

Question: Why do I top post?

• posted

Looking at the formulas posted on the website:

-Positive Time Interval (T1) = 0.693 * (R1+R2) * C

-Negative Time Interval (T2) = 0.693 * R2 * C

-Frequency = 1.44 / ( (R1+R2+R2) * C)

One extreme is R2 = 0 in which case the positive time interval is some finite amount and the negative time interval is zero for a duty cycle of

100%.

Another extreme is R1 = 0 (or R2 approaching infinity) in which case the positive and negative intervals are equivalent for a duty cycle of 50%

So the output-to-ground (positive logic) duty cycles vary between 50% and

100% which is equivalent to an output-to-positive (negative logic) duty cycle of between 0% and 50%. Inverting the output will therefore get you an output-to-ground (positive logic) duty cycle of 0% to 50%.

Dorian

• posted

Duh, R1 can never be zero - R1 connects to the positive supply rail. Making it zero forces the internal NPN discharge transistor (pin 7) to crowbar the power supply to ground. It's in the application notes.

The double diode duty cycle adjuster works really well . . .

I did see one application where they added a few gates to the output and used propagation delay to get a really short stable pulse output.

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• posted

yes.

possibly you can't do it like that. | | +--------+ | | | . . . .|. . . . | . VCC(8) . | . . +--RES(4) OUT(3)------+---- . 555 . | +--TH(6) DIS(7)-- | | . . | +--TR(2) CV(5)-- | | . . | | . GND(1) . | | . . . .|. . . . | | | | | | | +--[10K]-|-+-[100K]---+ | | | | `--||----+ `--|

• posted

Yes ' I tried he version with the diode and it works fine, thanks

ken

• posted

This is also a very unconventional hook-up for the 555 timer. Notice the timing capacitor is being charged and discharged through the 555 output and not connected to Vcc via the charge and discharge resistors, as is normally done. The diodes are allowing the timing capacitor to charge through one half of the pot and discharge through the other half. Also the discharge transistor is being used to drive the MOSFET instead of its normal use to discharge the timing capacitor. I would use a CMOS version of the 555 because the output gets closer to the supply rails and schottky diodes because of their low forward voltage and fast recovery feature.

Dorian

• posted

Take the standard circuit that has resistors between pins 6-7 and 7-8. Add one more component: a diode between pins 7 and 8 parallel to the resistor there.

1N4148 ought to do it. Attach the anode to pin 8 and the cathode to pin 7. This effectively takes the resistor on pin 8 out of the equation during the cap charging time, the time the 555 has a high output.
• posted

and short circuits the powersupply when pin 7 goes low.

a diode between pins 6 and 7 would be better. anode to 7,

```--
Bye.
Jasen```
• posted

Thanks for catching that. The diode does indeed go pins 6-7.

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