I was just wondering what the equation is to find the 555's output frequency is, when a control voltage is applied to pin 5. That would be very useful for me to know!
[Edit By OP]
So based on what Spehro Pefhany said, the equation for the output frequency (Substituting the equations for the high and low times) would be:
Where:
\$V_{Control}\$ is the control voltage
\$C \$ is the timing cap
\$V_{cc}\$ is the supply voltage
\$R_{1}\$ and \$R_{2}\$ are the timing resistors
\$f\$ is the output frequency
Answer
simulate this circuit – Schematic created using CircuitLab
The time when the output is high is \$ T_H = \tau_1ln(1- \$ \$V_C\over 2Vdd - Vc \$)
(it charges from \$V_C/2\$ to \$V_C\$)
The time when the output is low is \$T_L = \tau_2 ln(2)\$
(it discharges from \$V_C\$ to \$V_C/2\$)
frequency is f = \$1\over T_H + T_L\$
Where
\$ \tau_1 = (R1 + R2)\cdot C\$
\$ \tau_2 = (R2) \cdot C\$
The above ignores propagation delays and saturation voltages, so it's more accurate for low frequencies, fairly high resistance values, and a CMOS 555.
Here is an example plot with R1 = 1K, R2 = 10K, C = 10\$\mu\$F, Vcc = 10V and \$V_C\$ varied from 0.5V to 9.5V.
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