- #1
swraman
- 167
- 0
Hi,
Im am building a measurement device to measure the speed of a fan.
Essentially I have a diode on one side of the fan, and a phototransistor pointing to it on the other side of the fan. This results in an output signal resembling a square wave.
I am also trying to build an circuit that will output a voltage proportional to the frequency of the square wave. Is this possible? The input frequency of the square will always be around 50-300Hz, so I don't have to worry about extremely large frequencies or I can later tune the circuit to deal with saturation outside of these bounds.
I was thinking of using a differentiator op-amp circuit, which would simplify the signal into a set of positive and negative impulses at the edges of each square in the square wave, but I don't know any way to output a voltage proportional to the number of spikes/sec.
If it makes any diference, I am planning on making a line of LEDs, say 10, where the number of LED's glowing will be proportional to the frequency of the fan.
Thanks
Im am building a measurement device to measure the speed of a fan.
Essentially I have a diode on one side of the fan, and a phototransistor pointing to it on the other side of the fan. This results in an output signal resembling a square wave.
I am also trying to build an circuit that will output a voltage proportional to the frequency of the square wave. Is this possible? The input frequency of the square will always be around 50-300Hz, so I don't have to worry about extremely large frequencies or I can later tune the circuit to deal with saturation outside of these bounds.
I was thinking of using a differentiator op-amp circuit, which would simplify the signal into a set of positive and negative impulses at the edges of each square in the square wave, but I don't know any way to output a voltage proportional to the number of spikes/sec.
If it makes any diference, I am planning on making a line of LEDs, say 10, where the number of LED's glowing will be proportional to the frequency of the fan.
Thanks