Frequency signal multiplier - repost.

Hello all,

Repost with this extra info signal is 5.5 / 6.5 volts to vehicle earth and is 1Hz per rev i.e. 650 rpm tickover = 0.65kHz.

I'm after advice for a simple frequency multiplier, I need to increase the rev counter signal on my car by 1.5 i.e. if revs are 2000 I need a signal of

3000. So it needs to be 12 vdc supply 0 - 7000 rpm input 0 - 11000 output. Capable of easy build on a bit of Vero board.

Its to give the correct rev signal to a LPG system I have fitted but for some reason it only reads 2/3 of the rev's. It may be something to do with the fact that its a V6 working on a lost spark system. I.e. the plugs fire every rev not just on the compression stroke.

TIA.

George.

Reply to
George Gosbee
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It might help get a good answer if you post the schematic, or a link to a copy of the schematic for your existing circuit. You might only need to adjust values in your existing circuit to get the correct response.

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Dave M
MasonDG44 at comcast dot net  (Just substitute the appropriate characters in 
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Reply to
DaveM

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What was wrong with Ken Smith\'s post last time around?


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Reply to
John Fields

A simple approach since your system doesn't have to be 100% accurate (1.45x to 1.55x should be fine) is to use a frequency-to-voltage converter and a voltage-to-frequency converter. There are many ways to implement each.

The digital form is to - using a reference clock - count the number of clocks in a pulse (n). Use that value to generate a pulse every 2*n/3 clocks. A PIC microcontroller or CPLD would perform this function rather well. Just make sure your input and output values are compatible.

The output should end up as a smooth signal just like the input - no glitches or significant jitter to confuse the system down line.

Reply to
John_H

Maybe nothing but being the simple A/C controls man that I am, I could not make much of the diagram (as drawn on this page, maybe if it was drawn out by hand on paper I could understand). If I look up the spec sheet for the

4046 will it all become obvious? George.
Reply to
George Gosbee

You can buy v to f chips that work both ways. That means, convert your frequency to a voltage, add some dc bias, then convert the result back to a frequency. Two chips and a resistor... ;)

Look at this:

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  Bob Monsen
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Reply to
Bob Monsen

Note, the divide by three needs to be one of the forms that gives a square-wave output, which requires that it internally uses clock transitions from both edges of the input. We've had several such circuits posted as ASCII schematics here on s.e.d. over the years; I posted the one I designed for Sea Data Corp over 30 years ago, it had the property that its first full output cycle finished exactly three full clocks after being reset. Ah, the good old days.

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Reply to
Winfield Hill

Thanks for the link,

Have printed this out to give me somthing to read at work. I can understsnd the input and output chips, but the dc bias? does this mean that the output chip has an extra input that will have a fixed voltage and that multiplys the voltage so that the output chip givis a relitive output. ie 200Hz in

300Hz out and 5kHz in 7.5kHz out.

Just a little more info and I should get the principal.

George

Reply to
George Gosbee

That one gets my vote. Although it is not an added dc bias, just arrange the f-v and v-f Gains to be in the correct ratio.

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Tony Williams.
Reply to
Tony Williams

I think I understand i.e. kHz in = 1v out (input chip) 1v in = 1.kHz out.

Is this correct?

And is there a chip that contains both a f-v and v-f that would do the job?

Thanks George.

Reply to
George Gosbee

Did I suggest the wrong phase detector? I thought PD-II was the flip-flop one that tri-states. If you use that one, no funny counter is needed.

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Reply to
Ken Smith

Looking at fig 24 on page 11, I see that a frequency in to a f to vp chip will give a voltage out this is then fed into voltage devider the output of this will be lower than the input to the voltage devider. So the v to f chip must be set up diferently to the f to v chip to give the multiplcation. I would think that bothe chips could be set up the same if a frequency devider was required.

I now have to see if I can get my head around how to set up the chips so that I can get 3kHz out if I feed it with 3kHz.

George

PS the TC9402 looks the best bet due to its input voltage.

Reply to
George Gosbee

Oh, right, I guess I skimmed the original... he is multiplying, not adding a fixed frequency. Two TC9400s should do him. He can probably sample them for free from microchip...

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  Bob Monsen
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Reply to
Bob Monsen

Thanks for the replies.

All I need to do is get a 5.5v 0 to 7kHz signal multiplied by 1.5. Using the cars 12v supply.

George.

Reply to
George Gosbee

Yes. Using your given frequencies, the F/V would be scaled to give an output of V volts at 7000Hz in. The V/F would be scaled for an output of 11000Hz at V volts input.

BTW: Are those frequencies correct, or just examples?

Bob Monsen's post pointed to a pair of Microchip devices, which I suspect is the minimum chip count..... I was actually heading for a 4-chip solution, possibly 5 or 6 when twiddly bits had been added.

I still don't understand why the revcounter changed by 2/3rds just because of a conversion to LPG.

Is it perhaps that you are using a revcounter from a 4 cylinder engine on your V6? Have you looked at the revcounter to see if it can be easily modified? That would be the easiest solution.

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Tony Williams.
Reply to
Tony Williams

Its not the original rev counter that's the problem that works OK on petrol and lpg, its the signal the lpg ecu picks up it is only 2/3 of the true. If I set the lpg ecu to 4 cylinders all is OK with the rev reading but only 4 lpg injectors fire (cars a V6). Set on 6 cylinders 6 injectors fire but the rev pickup is only 2/3. This is all to do with the wasted spark system the car comes with it has 3 ht coils firing each plug twice on the 4 stroke cycle. The same is true with the petrol injectors.

The lpg ecu rev input is not adjustable apart from changing the # of cylinders as above.

George.

Reply to
George Gosbee

In article , George Gosbee wrote: [...]

Does the pulse rate have to be uniform or will the system work with (for example) bursts of 3 pulses for every other input pulse?

My suggested 4046 circuit still sounds the best to me but making 3 pulses for every 2nd input can be an all digital circuit if you prefer that.

You can also make one pulse on every other input and two on the other every. This would be a little more even than bursts of 3.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

Are you saying in your last paragraph that on a digital circuit for every 2 pulses in I can get 3 out? If so that would seem to be the easy way to go. Would it work over the frequencys I require say 0.4 kHz to 7.5 kHz (400 to

7500 rpm) and at the 5.5v signal input and output? Would this be done using gates?

George.

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Reply to
George Gosbee

Not necessarily "obvious", but more clear than just the ASCII diagram - that's more of a block diagram than a schematic (below), but there's really not much more to it - I'd think the challenge would be, as has been mentioned, a symmetrical divide by 3. Couldn't that be done with a 3-stage Johnson ring, or am I hallucinating again? ;-P

Thanks, Rich

Reply to
Rich Grise

In that case will someone kindly take a look at

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Fig 7-2: on page 13 and Fig 5-2: on page 11, these look like they would do more or less for my project. If it will do 3 questions that are obvious to me do I connect Vout of the input stage directly to the input voltage (0 to

10v) of the output stage? second on the output stage pin 4 Vss seems to go to the neg of the car but pin 9 is called Digital Ground will these both be the neg of the car on my project? Last The out put stage has a gain control R3 at 100k will this be ok to trim the output to give my overall frequency gain of 1.5? I intent to use the middle line of the F/S Frequency table shown on Fig 5.2:

Thanks for any help, George.

Reply to
George Gosbee

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