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VR-sensor opamp input conditioner circuit


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I'm working on a circuit to condition the AC signal from a variable
reluctance crank angle sensor.
The signal waveform amplitude varies from +- 100 mV to +- 50V depending on
The output must be a 0V-5V square wave and switch on the negative going zero

I'm using a standard TL084 opamp as a comparator with positive feedback
resistors R10 R11 to provide say 100mV hysterisis, I can  set the trip-point
on the - input of the comp. with a trimpot.

This is the part i'm not too sure about.
As the opamp rail-rail voltage is 0-5V the imput signal must stay in
I eliminate the negative half of the wave with a standard 1N4004 rectfier
diode and clamp it to max 5V with a zener diode and resistor.
I added another resistor R8 and capacitor C6 to filter the noise on the
input? but i doesn't look right to me. maybe i should put in an AC coupling
cap instead.
Also i don't know if the input resistor R8 conficts with the hysterisis

Please examine the circuit and give suggestions to improve this cicruit and to determine component values.




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Hi Martin,
A TL084 is not a rail-to-rail opamp. Its output voltage goes down to about 1V above ground to about 1V less than its positive supply, when it has a low current load.
The minimum supply voltage for a TL084 is 7V.

The sensor has an AC voltage that swings above and below ground. Therefore your circuit must be capacitor coupled. With capacitor-coupling then the 5V limiting must occur to both polarities of the input.

If the value of R8 is low enough then it won't affect the amount of hysteresis.

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Thanks for the reply's

I have corrected the schematic. I haven't got much exp. with opamps circuits yet, so ur help is very apreciated.

I replaced the opamp with a LM324, it has a supply voltage of min 3V, input and output can swing rail to rail, which should be suitable.

I added an AC-coupling cap and corectly reversed the D9 diode, also i removed the C6 capacitor.

Below is the new schematic, i included the estmated component values, is this about right?

One side of the pickup is connected to ground the other goes into the circuit input.

The frequenty of the waveform is expected to vary between 220 and 5000 Hz.
Below is a picture of the waveform which i captured with my laptop.

I don't realy understand what is involved with designing the signal input side of the circuit, maybe someone can explain this a bit?

-I only have to clamp the positive rail with the zener because D9 gets rid of the negative?

-The coupling cap blocks DC voltage and acts as a lowpass filter?
-Do resistor r8 and C6 also function as a filter?
-Then what is the use for R8?

-By removing the current limiting resistor from the zener doesn't it get too much current?

-What components do i need to include and omit?

[img width=680 height=321]

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