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*10E-6 ~= 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/ --