Add one resistor to give bipolar LM555 oscillator a 50:50 duty cycle

[Harald Kapp]

I just found this easy method described in EDN to create a 50:50 waveform from a 555. No need for diodes as usually shown.

 

[Alec_t]

EDN doesn't like my ad-blocker, so can't see that .
Anything like this?

 

[bertus]

Hello,

Here is a "printerfriendly" page:
https://www.printfriendly.com/p/g/PBJWVy
I hope it works.

bertus

 

[Harald Kapp]

EDN doesn't like my ad-blocker

Maybe put EDN on a whitelist? Lots of ads, agreed, but also lots of useful information.

 

[crutschow]

Yes, I have previously used that technique (simulation below):
It uses a third resistor in series with the discharge transistor to equalize the charge and discharge times of the timing capacitor.

 

[Alec_t]

I hope it works.

It does. Thanks.

 

[CircutScoper]

Yes, I have previously used that technique (simulation below):
It uses a third resistor in series with the discharge transistor to equalize the charge and discharge times of the timing capacitor.

View attachment 54658

Your simulation would be more realistic if it took into account that the effective discharge (output low) timeconstant isn't just R2C2 but is in fact (R2+R1R3/(R1+R3))C2. So, as shown in my EDN article, to get a 50:50 duty cycle, R3 should = 0.225R1 = 11.2k. Not 13.7k.

 

[crutschow]

Your simulation would be more realistic

My simulation is realistic in that it shows what the circuit does with the 13.7kΩ resistor I used.

That value gave a low time of 668.96552ms and a high time of 665.51724ms.

Using 11.2kΩ gave a low time of 576.8459ms and a high time of 667.98169ms.

So for whatever reason, the value I used gives much closer to 50% duty-cycle in the simulation than the value you calculated.

 

[CircutScoper]

Okay. Please try this experiment. What output waveform does your simulation predict if R2 = 0, all other values left unchanged?

 

[crutschow]

What output waveform does your simulation predict if R2 = 0, all other values left unchanged?

Do you mean R2 or R3?

 

[CircutScoper]

"Do you mean R2 or R3?"
R2. The resistor that connects to C2, THRS, and TRIG. Make it zero ohms and see if what your simulation then predicts for the "high" and "low" times look realistic.

 

[crutschow]

"Do you mean R2 or R3?"
R2. The resistor that connects to C2, THRS, and TRIG. Make it zero ohms and see if what your simulation then predicts for the "high" and "low" times look realistic.

Below is the simulation with R2 removed:
The low time is 88.91455ms, and the high time is 332.79446ms from the plot.
Both seem close to the expected for the component values used.

 

[CircutScoper]

Thanks for satisfying my curiosity.

Just for fun, you might want to try R3 = 22k, no other change.

 

[crutschow]

Just for fun, you might want to try R3 = 22k, no other change.

Whose fun?
That would make the discharge level close to the THRS trip point and make the high period very long.
It's not something one would do in a real circuit, so see no reason to simulate that.

 

[CircutScoper]

Actually, it should (if your simulation works accurately) make (pretty close) to a 50% duty cycle. Wouldn't that be fun?
See figure 14: https://datasheet.octopart.com/LM555CM-NOPB-Texas-Instruments-datasheet-7283993.pdf

 

[crutschow]

it should (if your simulation works accurately) make (pretty close) to a 50% duty cycle. Wouldn't that be fun?

Here's the simulation.
The duty-cycle is not close to 50%.
I trust the simulation, so I think Figure 14 is in error.

Having fun yet?

Better yet, for even more fun you should learn to do you own simulations with LTspice, so I don't need to be the middle-man.

 

[CircutScoper]

"I trust the simulation..." Well, that makes one of us.

In other news, did you know that every 555 datasheet on the planet says the Trigger voltage is 1/3V+, therefore 1.667V when V+ = 5V, not ~1.85V as shown by your "accurate" simulation?

 

[Martaine2005]

Trigger voltage is 1/3V+, therefore 1.667V when V+ = 5V, not ~1.85V as shown by your "accurate" simulation?

The data sheet says “typical”, not set in stone. Nothing is perfect. You should know that.

Martin

 

[CircutScoper]

Hi, Martin

What you say is of course true. I do wonder, however, what, in this context, the rationale for deliberately using an arbitrary and atypical Vtrig might be, especially without bothering to say so?

Hence, maybe it's more plausible to simply conclude it's that simulation that's neither "perfect" nor "set in stone?" Hmmmm?

 

[Martaine2005]

Simulations and calculations will differ.
Calculations don’t adhere to real world implementation of imperfect components. That’s a REAL reason for tolerances.
That can be, ie 1% to 20% tolerance.


Martin