Editing Multivibrator (section)

===Multivibrator frequency===
==== Derivation ====
{{unreferenced section|date=January 2011}}
The duration of state 1 (low output) will be related to the time constant ''R''<sub>2</sub>''C''<sub>1</sub> as it depends on the charging of C1, and the duration of state 2 (high output) will be related to the time constant ''R''<sub>3</sub>''C''<sub>2</sub> as it depends on the charging of C2. Because they do not need to be the same, an asymmetric [[duty cycle]] is easily achieved.

The voltage on a capacitor with non-zero initial charge is:

:<math>V_\text{cap}(t) = \left[\left(V_\text{capinit} - V_\text{charging}\right) \cdot e^{-\frac{t}{RC}}\right] + V_\text{charging}</math>

Looking at C2, just before Q2 turns on, the left terminal of C2 is at the base-emitter voltage of Q1 (V<sub>BE_Q1</sub>) and the right terminal is at ''V''<sub>CC</sub> ("''V''<sub>CC</sub>" is used here instead of "+''V''" to ease notation). The voltage across C2 is ''V''<sub>CC</sub> minus ''V''<sub>BE_Q1</sub> . The moment after Q2 turns on, the right terminal of C2 is now at 0&nbsp;V which drives the left terminal of C2 to 0&nbsp;V minus (''V''<sub>CC</sub> - ''V''<sub>BE_Q1</sub>) or ''V''<sub>BE_Q1</sub> - ''V''<sub>CC</sub>. From this instant in time, the left terminal of C2 must be charged back up to V<sub>BE_Q1</sub>. How long this takes is half our multivibrator switching time (the other half comes from C1). In the charging capacitor equation above, substituting:

:''V''<sub>BE_Q1</sub> for <math>V_\text{cap}(t)</math>
:(''V''<sub>BE_Q1</sub> - ''V''<sub>CC</sub>) for <math>V_\text{capinit}</math>
:''V''<sub>CC</sub> for <math>V_\text{charging}</math>
results in:

:<math>
V_{\text{BE}\_\text{Q1}} = \left(\left[\left(V_{\text{BE}\_\text{Q1}} - V_\text{CC}\right) - V_\text{CC}\right] \cdot e^{-\frac{t}{RC}}\right) + V_\text{CC}
</math>

Solving for t results in:

:<math>
t = -RC \cdot \ln\left(\frac{V_{\text{BE}\_\text{Q1}} - V_\text{CC}}{V_{\text{BE}\_\text{Q1}} - 2 V_\text{CC}}\right)
</math>

For this circuit to work, V<sub>CC</sub>>>V<sub>BE_Q1</sub> (for example: V<sub>CC</sub>=5&nbsp;V, V<sub>BE_Q1</sub>=0.6&nbsp;V), therefore the equation can be simplified to:

:<math>
t = -RC \cdot\ln\left(\frac{-V_\text{CC}}{-2 V_\text{CC}}\right)
</math>

:which can be simplified to: 

:<math>
t = -RC \cdot \ln\left(\frac{1}{2}\right) = RC \cdot \ln{2}
</math>

The period of each ''half'' of the multivibrator is therefore given by <math>t = R C \cdot \ln{2} </math>.

The total period of oscillation is given by:

<math>T =  (R_2 C_1 + R_3 C_2) \cdot \ln{2}</math>

<math>f = \frac{1}{T}

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

\approx \frac{1.443}{R_2 C_1 + R_3 C_2}</math>

where...
* ''f'' is [[frequency]] in [[hertz]].
* ''R''<sub>2</sub> and ''R''<sub>3</sub> are resistor values in ohms.
* ''C''<sub>1</sub> and ''C''<sub>2</sub> are capacitor values in farads.
* ''T'' is the period (In this case, the sum of two period durations).

'''For the special case''' where
* ''t''<sub>1</sub> = ''t''<sub>2</sub> (50% duty cycle)
* ''R''<sub>2</sub> = ''R''<sub>3</sub>
* ''C''<sub>1</sub> = ''C''<sub>2</sub>

<math>f = \frac{1}{T}

= \frac{1}{2R C \cdot \ln{2}}

\approx \frac{0.72}{RC}</math><ref name=Fink75>Donald Fink (ed), ''Electronics Engineers' Handbook'', McGraw Hill, 1975 {{ISBN|0-07-020980-4}}, page 16-40</ref>

==== Output pulse shape ====

The output voltage has a shape that approximates a square waveform. It is considered below for the transistor Q1.

During [[#State 1|State 1]], Q2 base-emitter junction is reverse-biased and capacitor C1 is "unhooked" from ground. The output voltage of the switched-on transistor Q1 changes rapidly from high to low since this low-resistive output is loaded by a high impedance load (the series connected capacitor C1 and the high-resistive base resistor R2).

During [[#State 2|State 2]], Q2 base-emitter junction is forward-biased and capacitor C1 is "hooked" to ground. The output voltage of the switched-off transistor Q1 changes exponentially from low to high since this relatively high resistive output is loaded by a low impedance load (capacitor C1). This is the output voltage of R<sub>1</sub>C<sub>1</sub> integrating circuit.

To approach the needed square waveform, the collector resistors have to be low in resistance. The base resistors have to be low enough to make the transistors saturate in the end of the restoration (R<sub>B</sub> < β.R<sub>C</sub>).