# Calculating t1 and t2 on a 50% duty 555 astable

Discussion in 'General Electronics Chat' started by tracecom, Mar 30, 2013.

1. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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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.

Last edited: Mar 30, 2013
2. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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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.

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3. ### WBahn Moderator

Mar 31, 2012
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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.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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?

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5. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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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.

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6. ### WBahn Moderator

Mar 31, 2012
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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

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

7. ### WBahn Moderator

Mar 31, 2012
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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.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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

9. ### tracecom Thread Starter AAC Fanatic!

Apr 16, 2010
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So, I guess that means that my calculation for t2 was correct since I arrived at the same answer?

Feb 17, 2009
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