Making small AC signal from DC, Cycle Speedo

Thread Starter

Motardo

Joined Sep 21, 2011
22
Hello,
I want to generate a periodic signal to drive my bicycle computer speedometer. The computer speed sensor is a two wire device that I guess is an inductor. When a magnet on the bike wheel passes nearby, it sends a pulse (through the two wires) to the computer. I'm guessing that the pulse is in the micro to millivolt range, and it has positive and negative components, and no DC component.
I have made oscillator circuits with 555's and comparators, and they output DC pulses in the several volts range. I would like to plug the output of the oscillator into the cycle computer, but first I think I need to convert it to a very small AC pulse. Can you help suggest an easy way to do this?
P.S. The point is to make a pocket portable frequency counter.
Thanks
 

SgtWookie

Joined Jul 17, 2007
22,210
Here is one way to get bipolar pulses from a 555; see the attached.

C2 allows the rising/falling edges of the output to pass, and R3/VR1 keep the average voltage output centered on 0v. VR1 allows adjustment from 0v p-p to ~9.1% of the supply voltage. It's basically a differentiator circuit.
 

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debe

Joined Sep 21, 2010
1,169
First i would check what your bike computer actualy uses for a pick up. If its 2 wires like the ones i have they are reed switches. If 3 wires then will be a hall switch pickup.
 

Thread Starter

Motardo

Joined Sep 21, 2011
22
Here is one way to get bipolar pulses from a 555; see the attached.

C2 allows the rising/falling edges of the output to pass, and R3/VR1 keep the average voltage output centered on 0v. VR1 allows adjustment from 0v p-p to ~9.1% of the supply voltage. It's basically a differentiator circuit.
Thank you Sgt. for this schematic. I had an idea that I would need a capacitor to block the DC and pass the AC, but I wasn't real sure how to do it. The graphs you provided next to the schematic really helped me understand how this circuit is working and where the negative pulses are coming from. The voltage on VR1 is just following the current through it; positive when the cap charges and current flows into ground, negative when the cap discharges and current flows up from ground :)
 
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