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The high time from each pulse is given by:

and the low time from each pulse is given by:

where R1 and R2 are the values of the resistors in ohms and C is the value of the capacitor in farads.
The power capability of R1 must be greater than
.
Particularly with bipolar 555s, low values of
must be avoided so that the output stays saturated near zero volts during discharge, as assumed by the above equation.
Otherwise the output low time will be greater than calculated above.
The first cycle will take appreciably longer than the calculated time, as the capacitor must charge from 0V to 2/3 of VCC from power-up, but only from 1/3 of VCC to 2/3 of VCC on subsequent cycles.
To have an output high time shorter than the low time (i.e., a duty cycle less than 50%) a small diode (that is fast enough for the application) can be placed in parallel with R2, with the cathode on the capacitor side.
This bypasses R2 during the high part of the cycle so that the high interval depends only on R1 and C, with an adjustment based the voltage drop across the diode.
The voltage drop across the diode slows charging on the capacitor so that the high time is a longer than the expected and often-cited ln(2)*R1C = 0.693 R1C.
The low time will be the same as above, 0.693 R1C. With the bypass diode, the high time is

where Vdiode is when the diode's "on" current is 1/2 of Vcc/R1 which can be determined from its datasheet or by testing.
As an extreme example, when Vcc= 5 and Vdiode= 0.7, high time = 1.00 R1C which is 45% longer than the "expected" 0.693 R1C.
At the other extreme, when Vcc= 15 and Vdiode= 0.3, the high time = 0.725 R1C which is closer to the expected 0.693 R1C.
The equation reduces to the expected 0.693 R1C if Vdiode= 0.
The operation of RESET in this mode is not well-defined.
Some manufacturers' parts will hold the output state to what it was when RESET is taken low, others will send the output either high or low.
The astable configuration, with two resistors, cannot produce a 50% duty cycle.
To produce a 50% duty cycle, eliminate R1, disconnect pin 7 and connect the supply end of R2 to pin 3, the output pin.
This circuit is similar to using an inverter gate as an oscillator, but with fewer components than the astable configuration, and a much higher power output than a TTL or CMOS gate.
The duty cycle for either the 555 or inverter-gate timer will not be precisely 50% due to the fact the timing network is supplied from the devices output pin, which has different internal resistances depending on whether it is in the high or low state (high side drivers tend to be more resistive).
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