Calculating t1 and t2 on a 50% duty 555 astable

Thread Starter

tracecom

Joined Apr 16, 2010
3,944
I found a circuit in the National datasheet for the 555 that is supposed to be a 50% duty cycle for a 555 astable. I built the circuit and it seems to work as expected. So I thought I would try to calculate the t1 (HIGH) and t2 (LOW) time for the output according to the formulae given. The HIGH calculation was pretty easy, and I think I got it right, but the LOW calculation is wrong. My work is attached; where did I go off track?

Thanks.

Never mind. I found my error.
 
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Thread Starter

tracecom

Joined Apr 16, 2010
3,944
Here are my (hopefully) correct calculations. As you can see, the duty cycle is not 50%; it's close, but not exact. Is that due to a calculation error that I have not found?

Thanks.
 

Attachments

WBahn

Joined Mar 31, 2012
29,978
I can't open the png file you attached to your last post and I don't see any attachment to your first post. So I don't even have any idea what the circuit is you are looking at.
 

Jony130

Joined Feb 17, 2009
5,487
The circuit diagram and the equation looks like this



But there is a one think that I don't understand.
The book give as the equation for discharge a capacitor.

\(\Large V_c = V_s*(e^{\frac{-t}{RC}})\)

And if we solve this for time we get this

\(\Large t = RC *ln(\frac{V_s}{V_c})\)

In this case Vs = 2/3Vcc and Vc = 1/3Vcc and R = RA||RB

So we end up with

t = RC•ln ( 2/3 / 1/3) = ln2•(R•C) = 0.693•RC

So why this equation don't work in this case?
 

Attachments

Thread Starter

tracecom

Joined Apr 16, 2010
3,944
I can't open the png file you attached to your last post and I don't see any attachment to your first post. So I don't even have any idea what the circuit is you are looking at.
The circuit in question is the one posted by Jony130, and my calcuation sheet is attached to this post as a PDF. My question is, "I found a circuit in the National datasheet for the 555 that is supposed to be a 50% duty cycle for a 555 astable. I built the circuit and it seems to work as expected. So I thought I would try to calculate the t1 (HIGH) and t2 (LOW) time for the output according to the formulae given. As you can see, the duty cycle is not 50%; it's close, but not exact. Is that due to a calculation error that I have not found?"

Thanks.
 

Attachments

WBahn

Joined Mar 31, 2012
29,978
The circuit diagram and the equation looks like this



But there is a one think that I don't understand.
The book give as the equation for discharge a capacitor.

\(\Large V_c = V_s*(e^{\frac{-t}{RC}})\)

<snip>

So why this equation don't work in this case?
Because your equation is for a capacitor that is discharging toward 0V. This one isn't. It is discharging to the voltage set up by the Ra,Rb voltage divider.

The general form of a first order response is

\(v(t) = Vf + (Vo-Vf)e^{\frac{-t}{\tau}}\)
 

WBahn

Joined Mar 31, 2012
29,978
The circuit in question is the one posted by Jony130, and my calcuation sheet is attached to this post as a PDF. My question is, "I found a circuit in the National datasheet for the 555 that is supposed to be a 50% duty cycle for a 555 astable. I built the circuit and it seems to work as expected. So I thought I would try to calculate the t1 (HIGH) and t2 (LOW) time for the output according to the formulae given. As you can see, the duty cycle is not 50%; it's close, but not exact. Is that due to a calculation error that I have not found?"

Thanks.
My first thought is that it would be somewhat remarkable if two values from the 5% resistor sequence would give exactly a 50% duty cycle.

My second thought is that I would not be surprised if you couldn't get closer to 50% that what these appear to give.

I used to keep a spreadsheet that had all the ratios of the values in the E24 series sorted so that I could easily look up the closest pair to whatever ratio I needed. That was several computers ago.
 

Jony130

Joined Feb 17, 2009
5,487
Because your equation is for a capacitor that is discharging toward 0V. This one isn't. It is discharging to the voltage set up by the Ra,Rb voltage divider.

The general form of a first order response is

\(v(t) = Vf + (Vo-Vf)e^{\frac{-t}{\tau}}\)
Ok I think I understand now.

So the solution is

\(t = RC*ln {\frac{Vo - Vf}{Vc - Vf}\)

When

Vo - initial voltage across capacitor. In our example Vo = 2/3Vcc.

Vf - finale voltage across capacitor, Vf = Vcc * RB/(RB +RA) = Vcc *22/73

Vc - Voltage level we want to discharge our capacitor.

Vc = 1/3Vcc in this case.


So we have

t = RC * ln * ( 2/3 - 22/73)/(1/3 - 22/73) = RC * ln (80/7) = RC * 2.436.
So for RB = 22kΩ; RA = 51kΩ and C = 100μF we have
t = RB||RA * 100μF * 2.436 = 15.369kΩ*100μF * 2.436 = 3.7442s
 
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