# 555 RC time constant formula question.

Discussion in 'Math' started by atrumblood, Jun 19, 2012.

1. ### atrumblood Thread Starter Member

May 13, 2012
59
2
Just quick question about the formula to calculate the frequency of the 555 timer.

It reads as f = 1.44/(R1+R2)*C

Where does the 1.44 come from? I can't seem to find what that value represents.

Thanks,
Atrum

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2. ### MrChips Moderator

Oct 2, 2009
13,381
3,749
It is just what's left over when you take all the other constants in the analysis into consideration.

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May 13, 2012
59
2
Thanks man.

4. ### WBahn Moderator

Mar 31, 2012
19,149
5,177

Are you sure you got that correct?

First off, you've made a classic, common, easy-to-make mistake of not realizing that your 'C' is NOT in the denominator as you have written it. What you have written is the same as

f = 1.44C/(R1+R2).

But what I'm sure you meant to write was:

f = 1.44/((R1+R2)*C)

Get used to spotting these kinds of mistakes because you will make them, for instance when entering formulas into a spreadsheet.

But even with what you meant to write, it is not quite correct (if I understand the circuit you are discussing correctly).

I assume you are talking about the classic astable multivibrator circuit in which the three external timing components, R1, R2, and C, are connected in series (and in that order) from the positive rail to the negative rail, the discharge pin connect between the resistors and the trigger and threshold pins connect between R2 and C (and the control pin is simply bypassed)?

Keep in mind the following:

The output is set whenever the trigger pin falls below Vcc/3 while the output is reset when the threshold rises above 2Vcc/3. The discharge is active whenever the device is in reset.

So the device will discharge until the voltage on the capacitor reaches Vcc/3. At this point it will set the output HI and stop the discharge. Here is where we will start keeping track of time.

The device will charge toward Vcc with a time constant of (R1+R2)C. Since it is starting at Vcc/3, the voltage change (if allowed to go all the way) would be 2Vcc/3, but we want to know how long it takes to get halfway there.

The general form of a first order response is:

$
v(t) \ = \ V_{fin} \ - \ \left( V_{fin}-V_{ini} \right) e^{-\frac{t}{\tau}}
$

To go half way, we need

$
e^{-\frac{t}{\tau}}\ = \ \frac{1}{2}
\
t \ = \ \ln(2) \tau
$

So the amount of time it takes to charge upward is:

$
T_U \ = \ \ln(2) \left( R_1+R_2 \right)C
$

Once it reaches 2Vcc/3, the threshold is reached and the circuit begins discharging. Once again, it discharges according to a first order response and, once again, only discharges for half of the voltage excursion before triggering. So the same expression applies, but now the time constant is uses R2 and not R1.

$
T_D \ = \ \ln(2) \left( R_2 \right)C
$

The total time it take to complete one cycle is therefore:

$
T \ = \ T_U \ + \ T_D \ = \ \ln(2) \left( R_1+R_2 \right)C \ + \ln(2) \left( R_2 \right)C
\
T \ = \ \ln(2) \left( R_1+2R_2 \right)C
$

The frequency is just one over the period:

$
f \ = \ \frac{1}{\ln(2)} \ \frac{1}{\left( R_1+2R_2 \right)C}
$

Anyone care to guess what 1/ln(2) happens to be?

5. ### absf Senior Member

Dec 29, 2010
1,618
427
1.44 of course.

6. ### Wendy Moderator

Mar 24, 2008
20,877
2,655
You guys are over complicating it. It is the inverse of the time for the charge to go from 1/3 Vcc to 2/3 Vcc. Sure, it can take a small amount of calculus, but the formula has been derived many times. I tend to go with the graphic approach myself.

%
...
.............TC = R X C

555 HYSTERETIC OSCILLATOR formula for "f" incorrect

Last edited: Jun 19, 2012
7. ### WBahn Moderator

Mar 31, 2012
19,149
5,177
I'm sorry, but I absolutely do not get the point of the criticism here.

Someone that is trying to understand where a constant in a formula is coming from and apparently what they should be told is, "That's been derived many times. Just accept it."

How is an unscaled graph supposed to show them where a specific value came from? While the vertical scaling can be inferred, the horizontal scaling cannot. If they know enough to figure out the horizontal scaling, then they don't need the graph.

There is absolutely nothing wrong or overly complicated about the explanation I gave. I had to assume a circuit configuration because one wasn't given, so I clearly stated the standard ASM circuit I was assuming. I established that the voltage is cycling between 1/3 and 2/3 Vcc. I established how long it takes for a first order response to make such a transition. I established the two different time constants involved. I added them together to get the period and I then took the recipricol to get the frequency.

8. ### atrumblood Thread Starter Member

May 13, 2012
59
2
I appreciate all the information guys. It helps to know how this formula is put together. I have always been the type that needs to know what's inside something, and what's inside that, and so on.

I do need to brush up on my math it seems. I have not done math of that level in some years. None the less, thank you for the detailed information, and of course the simplified explanations as well.
My apologies for not specifying the type of 555 configuration I was referencing. You are correct that I was referring to the astable mode of operation.

So if I understand correctly. the 1.44 is derived from 1/0.693 which is roughly the time it takes for a capacitor to charge?

9. ### WBahn Moderator

Mar 31, 2012
19,149
5,177
At the risk of making things too complicated, the notion of "the time it takes for a capacitor to charge" is too vague to be of much use when trying to understand, quantitatively, what is going on.

If a capacitor is charging (or discharging) through a resistor from some starting voltage toward some final voltage, then the voltage as a function of time is a decaying exponential that is characterised by a time constant that is equal to τ=R*C (track the units and you will find that ohms * farads yields seconds.

The expression that is governing this is

$
(1-e^{\frac{t}{\tau}})
$

When t = τ (=RC), we say that elapsed time is one time constant.

In one time constant, the voltage will make it

$
(1 \ - \ e^{-1})=63.2%
$

of the way. In two time constants it will make it

$
(1 \ - \ e^{-2})=86.5%
$

Theoretically, it will never completely charge. But, from nearly any practical viewpoint, it can be considered to have reached its final value after five time constants

$
(1 \ - \ e^{-5})=99.3%
$

What we are looking for is how many time constants it takes for it to make it halfway from its starting point to it's final value.

$
(1 \ - \ e^{\ln(0.5)})=50.0%
(1 \ - \ e^{-\ln(2)})=50.0%
$

So it takes ln(2) time constants to charge halfway.

Since this is the time (related to the period), the frequency will go as the recipricol of this, or 1/ln(2), which is 1.4426950....

At this point, you need to consider the circuit configuration more carefully. It is charging and discharging through the same RC time constant, or is the charging time constant different from the discharging one.

If they are the same, then the total time taken is

T = ln(2)RC + ln(2)RC = 2ln(2)RC

and the frequency is

f = 1/(2ln(2)RC) = 0.72/(RC)

If they are not the same, but instead as I showed in my derivation, the you get what I got.

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10. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,819
1,736
As I understand the circuit we are looking for the time to charge from 1/3 to 2/3 Vcc. Where did "halfway from its starting point to it's final value" come from?

(Yes the answer is ln(2) but I don't agree with how you got there.)

11. ### WBahn Moderator

Mar 31, 2012
19,149
5,177
When it is going up, it is charging from Vcc/3 to Vcc, an excursion of 2Vcc/3. We stop when it has gotten to 2Vcc/3, which is a change of Vcc/3. So the fraction of its eventual change is (Vcc/3)/(2Vcc/3) = 1/2.

You can certainly crank the math without making that observation and get the same result:

$
v(t) = V_i + (V_f -V_i)e^{-t/\tau}
\
V_i = \frac{1}{3}V_{CC}
\
V_f = V_{CC}
$

Then solve for t = Tu such that

$
v(t=T_U) = \frac{2}{3}V_{CC}
\
\frac{2}{3}V_{CC} = \frac{1}{3}V_{CC} + (V_{CC} \ - \ \frac{1}{3}V_{CC})e^{-T_U/\tau}
\
2 = 1 + 2e^{-T_U/\tau}
\
0.5 = e^{-T_U/\tau}
\
\ln(0.5) = -T_U/\tau
\
T_U = -\frac{1}{\ln(0.5)\tau}
\
T_U = \frac{1}{\ln(2)\tau}
$

A similar line of reasoning applies for the downward ramp.

12. ### atrumblood Thread Starter Member

May 13, 2012
59
2
Wow guys thank you so much for all of this information. It is extremely helpful for me.

Could someone list what all the Variable symbols represent. Such as V$_{}i$ and V$_{}f$ ?

13. ### WBahn Moderator

Mar 31, 2012
19,149
5,177
$
V_i \ : \ \text{initial voltage}
V_f \ : \ \text{final voltage}
\tau \ : \ \text{time constant}
T_U \ : \ \text{time to ramp from lower limit up to upper limit}
T_D \ : \ \text{time to ramp from upper limit down to lower limit}
V_{CC} \ : \ \text{positive supply voltage (negative supply is assumed to be 0V)}
$

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14. ### atrumblood Thread Starter Member

May 13, 2012
59
2

I noticed the symbol $e$. Am I correct in assuming this is Euler's Constant?

15. ### WBahn Moderator

Mar 31, 2012
19,149
5,177
Yes, it is.

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