555 Astable Oscillator
++Vcc
 
 /
 \ Ra
______  /
0[1 8]+  Rb
+[ ]+/\/\+
o[3 ]+
 +[______] 
  
++
 
Vcc 
 Ct

+ 0V
Output:
T2
...+ ++ +...
   
++ ++
T1
T1=0.7*(Ra+Rb)*Ct (output pin3 is 1)
T2=0.7*Rb*Ct
Total period = T = T1 + T2
for 50hz clock T=1/50 = 0.02 seconds
the constant 0.7 is acually ln2 (natural log of 2) which is 0.6931...
Hence T = 0.7*(Ra+2*Rb)*Ct
For the case of T1=T2 we can omit these resistors, and put another
between pin2 and pin3
in this case the T becomes 1.4*Rt*Ct
all units are SI, that is:
R in ohms
C in farads (not microfarads)
T in seconds
F in hertz
Maybe in my notes I used uF instead of F
Now, for T=0.02 using T=2*ln(2)*R*C ~= T=1.4*R*C
RC=0.01443
for 3300 ohm C becomes 0.01443/3300 = 4.37*10E6 ~= 4.4uF
and other values for standard C values are:
uF ohm
 
C=1 R=14426 *
C=1.5 R=9617
C=2.2 R=6557
C=3.3 R=4371
C=4.7 R=3069
C=6.8 R=2121
C=10 R=1442 *
note: marked ones can be used with %1 ranges resistors 14.3kohm and 1.43kohm
Anyway, maybe building the first version should fix the unknown problem.
in this case, we use T1=T2=0.01 seconds:
0.01=0.7*Rb*Ct
take Ct as 4.7uF for example:
Rb=0.01*1000000/(0.7*4.7)
Rb=3Kohm
and using T1=0.7*(Ra+Rb)*Ct
Ra=10000/(0.7*4.7)3000
Ra=40ohm
actually, if we equate the equations we see that Ra goes to zero as
T1 becomes closer to T2.
But we cannot take Ra as 0 in this circuit. Because 555 works by
comparing the voltages.
Ehm, this is not a perfect solution. But it may suit the needs.
Standard resistor values are here:
http://www.rfcafe.com/references/electrical/resistor_values.htm
My regards,
Ilker Ficicilar
ilker(@)ekitap.gen.tr
ilker(@)ficicilar.name.tr
http://cbm.ficicilar.name.tr/
