what the equation is to find the 555's output frequency is, when a control voltage is applied to pin 5
By my calculations, the accepted answer and the formula echoed in the question are wrong. I believe the correct formula for frequency when a control voltage is applied is:
\$ f = { 1 \over C \cdot (R_1 + R_2) \cdot ln({1 + { v_{cont}\over {2 \cdot ( v_{cc} - v_{cont} ) } } }) + C \cdot R_2 \cdot ln(2) } \$
To run this formula in WolframAlpha, use this link.
With constituent components:
\$ t_h = C \cdot (R_1 + R_2) \cdot ln({1 + { v_{cont}\over {2 \cdot ( v_{cc} - v_{cont} ) } } }) \$
\$ t_l = C \cdot R_2 \cdot ln(2) \$
Why am I challenging the accepted answer?
I needed to run this calculation today, tried the formula suggested ... and got really weird results (like negative frequencies, and a trend that seems inversely proportional to that expected).
The reasoning in the approved answer is sound, and the chart seems correct, but the formula seems to have suffered a transcription/transposition error specifically in relation to the calculation of \$ t_h \$.
For example, if I use the formula provided to calculate R1 = 1K, R2 = 10K, C = 10μF, Vcc = 10V and VC = 9.5V I get an answer of -5.2816 Hz (when it should be ~3Hz as the chart suggests).
I'm posting my run of the calculation from scratch here as a new answer. If Spehro, OP and all agree with my calcs, I'm happy to see the original question and accepted answer updated (I'm not a rep whore).
NB: I'm using the TI NE555 datasheet for reference as it has more internal details than others I've seen.
In astable configuration, charge discharge follows these rules (from the datasheet):
- THRES > CONT sets output low and discharge low
- TRIG < CONT/2 sets output high and discharge open
Conventionally when pin 5 is unused (cap to ground), CONT = VCC * 2/3 due to the three stage voltage divider.
Given the complete RC response is
\$ v_t = v_\infty + (v_0 - v_\infty)e^{-t/\tau} \$
Then when pin 5 CONT has a voltage \$ v_{cont} \$ applied, our full charge boundaries are defined by:
\$ v_\infty = v_{cc} \$
\$ v_t = v_{cont} \$
\$ v_0 = {v_{cont}\over 2} \$
So plugging that back into the complete response formula:
\$ v_{cont} = v_{cc} + ({v_{cont}\over 2} - v_{cc})e^{-t/\tau} \$
Simplifying and re-arranging to derive a formula for \$ t = t_h \$:
\$ v_{cont} - v_{cc} = ({v_{cont}\over 2} - v_{cc})e^{-t/\tau} \$
\$ {v_{cont} - v_{cc} \over {v_{cont}\over 2} - v_{cc}} = e^{-t/\tau} = {1 \over e^{t/\tau}} \$
NB: I think this is the missing step. If we don't invert here, we derive the formula as currently listed in the Q & A.
\$ {{v_{cont}\over 2} - v_{cc} \over v_{cont} - v_{cc} } = e^{t/\tau} \$
\$ {1 + { v_{cont}\over {2 ( v_{cc} - v_{cont} ) } } } = e^{t/\tau} \$
\$ ln({1 + { v_{cont}\over {2 ( v_{cc} - v_{cont} ) } } }) = t/\tau \$
\$ t = \tau ln({1 + { v_{cont}\over {2 ( v_{cc} - v_{cont} ) } } }) \$
So I'm concluding the formula for \$ t_h \$ is actually:
\$ t_h = C \cdot (R_1 + R_2) \cdot ln({1 + { v_{cont}\over {2 \cdot ( v_{cc} - v_{cont} ) } } }) \$
So if I go back and revise the calculation of R1 = 1K, R2 = 10K, C = 10μF, Vcc = 10V and VC = 9.5V I now get an answer of 3.0491 Hz. That's much more reasonable, and matches Spehro's chart.