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 TL082 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 on
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and give suggestions to improve this cicruit and to determine component values.
On a sunny day (Tue, 27 Feb 2007 13:56:33 +0100) it happened "tekniq" wrote in :
Yes, use a cap on the input of the opamp to AC couple. But more then that, use a cap in the INPUT, and short out R9, and remove D9. The zener will then cause a -0.7 to + zener voltage.
A sensor ---||------------- R ---------||--- R11 |k | zener === rest same |a |
------------------------------------------
For low signals A carries the full sensor signal, AC coupled to R11 For very high signals (speeds if inductive sensor) the voltage in A is limited by the zener.
There may be other and better ways to do this. Is the sensor grounded on one side? if not go differential to reduce noise, with offset compensation.
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, this should remove the negative half and clamp the positive voltage to the 5.1 zener level, right?
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?
I'd lose D9; just put C6 directly to R8. The zener will work; my personal preference would be diode clamps to the rails:
+5 | | K [diode] | input ----[C6]----[R8]----+--+----[R11]---etc. | | K [diode] | [GND]
Something like 1N4148 or 1N914 would be faster switching, but, lessee - with 50V peak at the end of a 1K, that's like 50 mA, which might be a bit much ... Oh, sure. The 1N4148 is good for 150 mA:
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I'd put the diodes "after" R8, because I'd want the signal to see a resistance before the diode shorts it out.
Also, put a 1K or 2.2K pullup directly from IC5A-1 (the output) to Vcc. The LM324 is a comparator, with an open-collector output.
Of course, your supply is protected from load dumps and all that, right?
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, this should remove the negative half and clamp the positive voltage to the 5.1 zener level, right?
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?
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Thanks
I'll continue my posts here in basics where i still belong (sigh :)) I find your comments very useful, i thought by restricting the signal to
+0.6 V would be a bit small and more susceptible to noise, but as the comparator swiches at very low signal levels this should make no difference. As for the current tru the diodes, i could make the 1K resistor a lot bigger (i have just estamated some component values) but i thought this would interfere with the hysterisis feedback loop?
The 1N4001 i choose because i still have a lot of them, I would like to know the swiching speed of these but none of the datasheets states them( guess its that bad) So faster diodes it is.
I'm still a bit in the dark about the input filters and how and when to use what. As i see it al i have left now is a coupling cap and a input resistor. What are the functions of the input resistor here? It limits current trough the diodes but otherwise it seems redundent. Is there still something missing here? some improvements maybe?
My supply is bypassed with a 10uF elco and a 100n cap, does it need more protection against load dumps from what?
Very curious about these types of circuit, would like to know more.
Another thing when i come to think about it; by putting the diodes rail to rail the signal would still be -0.6v +0.6v which is below the negative (GND) supply rail of the comparator this is unacceptable according the datasheets, the same happens with the zeners. I thought the diodes would take care of this, but know?
+5 | | K [diode] | input ----[C6]----[R8]----+-------[R11]---etc. | | K [diode] | [GND]
[followups-to is now only s.e.basics] On Wed, 28 Feb 2007 01:15:45 +0100, tekniq wrote: ...
It's always worked for me, but I've never really subjected an input to such drastic swings.
The logical choice, then, would be Schottky diodes, but someone else would have to come up with a part number. A google search on Schottky diodes might be fruitful.
On a sunny day (Tue, 27 Feb 2007 22:15:07 +0100) it happened "tekniq" wrote in :
C1 A
The peak current is limited by the inductance of the pickup coil, and its resistance, in series with the impedance of C1.
Just try above circuit :-)
Are those VOLTS? In that case you have only 200mV pp, and the whole issue changes.
Then I would perhaps do it like this:
-- R3 ---------- DC block | | CD block lowpass C3 | | sensor ---||------------- R1 ---------||--------------- + | C1 |a |k | | opamp ------- diode diode === C2 bias R2 |k |a | |---- - GND - /// /// /// | | 1/2 V (from you trim pot)
Added R2, to make sure the + input of the opamp is closer then 100mV to 1/2V too, else it wont trigger. If you make R2 say 10 kOhm, then R3 should be very very high, because if the output flips to +12 (supply) thene you have a divider R3 / (R2 + R3) that should yield less then 100 mV, so R3 should be at least 60 x R2, or better 1 MOhm.
If you want a comparator, USE A BLOODY COMPARATOR!!
You can find comparators that are designed to work on 5 volts, and to give a logic level output.
--
Peter Bennett, VE7CEI Vancouver BC, Canada
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