Ive bin using the 555 timer IC for years and ive never questioned the
formulas. I blindly plug what i need into the formula and get the
results.
Most on this forum are already familiar with the formulas in question
but for the novices, these are the ones im referring to:
Astable Mode: charge time T1 = 0.693(Ra + Rb)C

Discharge time T2 = 0.693RbxC
Monostable Mode T = 1.1RC
Can someone fill me in on where the numbers .693 and 1.1 come from ?
Thanx.

I believe these factors are the times in relation to the RC fomulae against
the difference between the top and bottom switching thresholds. Have to look
at an RC curve and plot 1/3 to 2/3 (I think) voltage change.

Hi,
I also have a question about 555 timers. If you take Ra out of the timer
circuit (ie set it to 0), does this mean the mark-space ratio will be
exactly 5? Will the circuit still operate?
regards,
-Duncan

When you're discharging a capacitor, the factor is exp(-t/T) where
t is time and T is the RC product. Since exp(-ln(2)) is exactly 1/2,
at t = .693RC is the point where the voltage has fallen to 1/2 of
it's initial value. On the 555, the internal comparitors are set at
2/3 and 1/3 of Vcc, so the "initial" value here is 2/3 Vcc and the
halfway point is 1/3 Vcc.
The 1.1 comes from the internal "turn off" comparitor's setpoint at
2/3 Vcc. The trigger "discharges" the comparitors (turning on the
output) so we need to get back to the "turn off" point. The charge time
curve is (1 - exp(-t/T)) and so we need 2/3 = (1 - exp(-t/T)) or
t = -ln(1/3)RC.

Thanks Rich, thats exactly what i was looking for. I just tried that
out on my "scientific" calculator and it worked out just like you
suggested.
t = -1n(0.333)RC
= 1.0987 RC
Rounding off gives 1.1 RC for the mono....perfect.
Im still abit sketchy on the logic behind the astable but it works:
t = exp(-1n(2))RC
= 0.693RC
Thanx again much appreciated.

Hmmm... case where having fancy fonts and formatting would have come in
handy in a usenet posting. We'll do this without HTML, tho...
That expression is (voltage at time t) = V0 * e to the (- t/T). To
switch states, the voltage has to fall (rise) from 2/3 Vcc to 1/3 Vcc
(or vicy versa), a factor of 2. So, for the falling voltage case,
(e ^ (-t/T)) must equal 1/2.
Manipulate that to (-t/T) = (log base n) of 1/2, or t = - ln(2) * T.
The T is the RC time constant.

Thanx again Rich, that has cleared up a few misunderstandings and
blunderings on my part.
Im beginning to see why ive never seen the question raised about the
formulae used in the 555 timer. Appears one would require some
knowledge of calculus and differencial equations to get a full grasp
on the concepts involved. The last time ive even looked at a calculus
book (or algebra for that matter) was over 13 years ago, so im abit
rusty ( no surprise ;-) )
Anyway after an exhaustive search and postings to various news groups
and forums, heres what ive got ( appart from a pounding headache ):]
For the 555 Monostable:
2/3 = 1-e ^ (-t/RC)
-1/3 = -e ^ (-t/RC)
1/3 = e ^ (-t/RC)
1n (1/3) = -t/RC
-1.0986123 = -t/RC
t = 1.0986123 RC
Rounded off gives t = 1.1RC used in the 555 Monostable.
For the 555 Astable:
( Charging time ) t1 = -1n (1/2)(Ra+Rb)C
= 0.693(Ra+Rb)C
( Dicharging time ) t2 = -1n(1/2)(Rb)C
= 0.693(Rb)C
With a bit of thought it can be seen that this idea applies, not just
to the 555 timer, but to timer and oscillator design in general.
Thanx again...i an go away a little wiser.
Net.

FYI: Your equations show 1n (one-n) instead of ln (L-n).
It seems like you can't see the forest from the trees. The concept of the 555 is
dirt simple if you keep in
mind the internal voltage divider. All 3 internal resistors are the same value
(or damn near the same values).
Because of this the it forms a voltage divider of two types: an upper limit (for
the upper voltage cutoff) and
a lower limit (for the lower voltage cutoff).
Since the lower voltage divider is
[R/(R+R+R)]V = 1/3V
and the upper voltage divider is
[(R+R)/(R+R+R)]V = 2/3V
these form the comparator 'window' of operating voltages.
The external capacitor and resistor form the charging and relaxation circuit.
The time constant as mentioned
(t) is the the time of charging and decay of the external resistor and capacitor
circuit. The time constant is
only important in respect to how long it takes for the capacitor to charge up to
the upper limit to trigger the
output to go low. Once the output is low, it drains off the capacitor to the
lower voltage (1/3)V and is then
set high (to V) to recharge the capacitor. The time constant is *only* in
respect to the external resistor and
capacitor values, since that is what determines the time it will take to reach
the upper limit 2/3V and the
time that it will take to reach the lower limit 1/3V.
If the internal resistors were not equal, then the numbers 1.1 and 0.7 are
totally meaningless!!!
Please keep this in mind. Hope this helps.
:-)

Ack! yes that should have bin ln , which also looks logical since it
means " natural logarithm". Thnx for catching that, hadnt noticed.
is dirt simple if you keep in
(or damn near the same values).
(for the upper voltage cutoff) and
The time constant as mentioned
capacitor circuit. The time constant is
to the upper limit to trigger the
lower voltage (1/3)V and is then
respect to the external resistor and
the upper limit 2/3V and the
totally meaningless!!!
I totally agree.

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