Monday 21 March 2011

Multivibrator period (frequency)Derivation

The total period of oscillation is given by:
T = t1 + t2 = ln(2)R2 C1 + ln(2)R3 C2
f = \frac{1}{T}

= \frac{1}{\ln(2) \cdot (R_2 C_1 + R_3 C_2)}

\approx \frac{1}{0.693 \cdot (R_2 C_1 + R_3 C_2)}
where...
  • f is frequency in hertz.
  • R2 and R3 are resistor values in ohms.
  • C1 and C2 are capacitor values in farads.
  • T is the period (In this case, the sum of two period durations).
For the special case where
  • t1 = t2 (50% duty cycle)
  • R2 = R3
  • C1 = C2
f = \frac{1}{T}

= \frac{1}{\ln(2) \cdot 2RC}

\approx \frac{0.721}{RC} [3]
The duration of each state also depends on the initial state of charge of the capacitor in question, and this in turn will depend on the amount of discharge during the previous state, which will also depend on the resistors used during discharge (R1 and R4) and also on the duration of the previous state, etc. The result is that when first powered up, the period will be quite long as the capacitors are initially fully discharged, but the period will quickly shorten and stabilise.
The period will also depend on any current drawn from the output and on the supply voltage.

No comments:

Post a Comment