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I have a capacitive sensor connected to a 555 timer wired as an astable multivibrator and the free running frequency is 60kHz. When the capacitance of the sensor changes, the frequency of the 555 also changes. The maximum frequency deviation from the center frequency is 20kHz, i.e lowest frequency (60kHz-20kHz) is 40kHz and I want to convert this frequency deviation to voltage. I need a simple circuit using a 555, OPAMP, transistor etc.

I would prefer not to use the LM331 frequency to voltage converter.

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  • \$\begingroup\$ Does the duty cycle change? \$\endgroup\$ – Ignacio Vazquez-Abrams Mar 2 '14 at 6:32
  • \$\begingroup\$ @IgnacioVazquez-Abrams almost constant duty cycle of 50% \$\endgroup\$ – yogece Mar 2 '14 at 6:33
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One simple method is to add a single-shot / monostable vibrator that is triggered by the oscillator. If the mono time is slightly shorter than the period time of the highest frequency (\$T = \dfrac{1}{60\text{kHz}}\approx 16.7\text{ms}\$), you effectively have a PWM signal that is proportional to the input frequency.

Then all you need is a low pass filter that filters out the high frequency and outputs a DC-voltage, averaged from the PWM.

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  • \$\begingroup\$ The OP asked for a voltage proportional to the deviation in frequency from the nominal 60 kHz, so I don't think this solution quite does that. On the other hand, perhaps the OP just worded the question poorly and wants a simple F/V converter. \$\endgroup\$ – Joe Hass Mar 2 '14 at 14:33
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Try this one using the PLL IC565.

enter image description here

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  • \$\begingroup\$ thanks vishnu but i don't like to use 565 and expecting simple solutions like @jippie given \$\endgroup\$ – yogece Mar 2 '14 at 9:41

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