tekniq Posted February 27, 2007 Report Share Posted February 27, 2007 Hi,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 RPM.The output must be a 0V-5V square wave and switch on the negative going zero crossing.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 between.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 feedback.Please examine the circuit and give suggestions to improve this cicruit and to determine component values.Regards,Martin Quote Link to comment Share on other sites More sharing options...
audioguru Posted February 27, 2007 Report Share Posted February 27, 2007 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. Quote Link to comment Share on other sites More sharing options...
indulis Posted February 27, 2007 Report Share Posted February 27, 2007 D9 is backwards to clamp the negative half of the waveform. Get rid of R9 and move D10 to the other side of R8 Quote Link to comment Share on other sites More sharing options...
tekniq Posted February 27, 2007 Author Report Share Posted February 27, 2007 Thanks for the reply'sI 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] Quote Link to comment Share on other sites More sharing options...
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