Divide by 2 circuit implementation

Discussion in 'General Electronics Chat' started by servion, Mar 30, 2010.

  1. servion

    Thread Starter New Member

    Mar 30, 2010
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    Hello, I am hoping for get your help with implementing a very simple (I hope) divide by 2 circuit. I am not an electronics guy by any means, so please bear with my ignorance. I did a bunch of searching first, and the threads I found seemed to be regarding implementing more complex devices and started to go over my head. I'm hoping this will be simple and quick for me to build/buy myself.

    The problem is making the tachometer on my motorcycle read correctly. I have installed an aftermarket engine management controller to replace the original one, and the tach output signal from this computer is exactly double of what the tachometer expects. I need to cut this output in half to get the tachometer to read correctly. Ideally, I would like to build/get a little device that I can wire up inline with the output wire between the tach and the computer. The trouble I'm having is where to find such a device.

    Through my googling, it seems like I'll need a "d-type flip-flop" circuit, is that correct? If possible, I'd like to build this with either stuff I have lying around on-hand, or something I could go to my local electronics store and pick up.

    I currently have a few chips lying around that I was wondering if they would work for what I need. "SN74HC373N", the datasheet I was able to find with searching: http://www.chipsinfo.com/TI/SN74HC373N.htm Does this have what I need? Or would there be a more effficient/simpler way to go about this? I don't know what sort of signals or wires I'd need besides the source pin from the tach output. I have readily available ground, +12V and +5V pins.
     
  2. bertus

    Administrator

    Apr 5, 2008
    15,638
    2,343
    Hello,

    Do you know what levels are present at the computer output?
    Do you know what levels the tachometer wants to have as input?

    The 7400 series is made for 5 Volts.
    There is also the 4000 series that can be used on 12 Volts.

    Greetings,
    Bertus
     
  3. servion

    Thread Starter New Member

    Mar 30, 2010
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    By level, are you inquiring as to the voltage output of the tachometer output signal? I'm sorry, I do not. I do have both +5V and +12V poewr sources available. I can make a post on a forum dedicated to the engine management computer that I am using, but I would like to be able to ask specifically for what information you need.
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    A D flip flop that has the Q^ (Q not) output tied to the D input will change state at 1/2 the clock rate.
     
  5. servion

    Thread Starter New Member

    Mar 30, 2010
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    Yes, that is like what I was finding as well. Like this?

    http://www.falstad.com/circuit/e-divideby2.html

    However, the trouble I am having is finding out how/where to get one of these. I have wire that is putting out a signal with a varying frequency and I need that cut in half and output to another wire, if that makes sense.
     
  6. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    If it's compatible with logic levels, it should work. You might state the magnitude of the signal and the upper limit on the frequency.

    For instance, 4000 series CMOS will handle up to 15 volts with no problem, while most TTL will only go to 5 volts. Frequencies in the tens of megahertz can be problematic.
     
  7. servion

    Thread Starter New Member

    Mar 30, 2010
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    Well, the link that I found was just a teaching app/description of what a d-type flip-flop is. not a specific part number or anything like that. I am hoping to get guidance how to figure out what to buy and where, etc. to actually build this thing.

    The tachometer expects 2 pulses per revolution, and the engine will spin up to 11000rpm, so that looks like a max of 367Hz.

    I am still not sure of the voltage that this signal will be output with, I will try to figure that out ASAP. I would "assume" it would nto be over 12V because pretty much everything in this system 5V or 12V.
     
  8. Norfindel

    Active Member

    Mar 6, 2008
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    servion likes this.
  9. servion

    Thread Starter New Member

    Mar 30, 2010
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    Thank you!!! Alright, so I'm looking at the tech data sheet for part # MM74C74, and this is what I'm thinking I'll need to do (based on the divide by 2 info on the link I posted earlier). I understand this is very simple for most of you, but this is all new to me, so I would very much appreciate confirmation or any further suggestions!

    Data sheet: http://pdfserv.datasheetpro.net/FairchildSemiconductor/MM74C74.pdf

    I should get a little generic PCB to connect the chip to. then:

    Connect my source signal (the one I need cut in half) to pin 3
    Connect pin 6 to pin 2
    Connect pin 5 to my tachometer
    Connect GND to a sensor ground (provided though the engine computer).

    My first question,
    Connect 14 (Vcc) to... a +12V source? Is this necessary/correct?

    Next question, the document says "Voltage at Any Pin (Note 2) −0.3V to VCC +0.3V" - what does this mean? I read that to mean that a +12V signal is too much, but elsewhere in the document it says handles up to 15V.

    Third question, if I build this device as described here, will the output signal (from pin 5), which is now half of my source frequency, have the same voltage as the source signal?

    Thank you all!
     
  10. Norfindel

    Active Member

    Mar 6, 2008
    235
    9
    Yes, your input signal goes to the clock (pin 3), pin 6 /Q goes to pin 2 data. You take the output from Q at pin 5. Also, connect pin 4 and 1 to Vcc.

    Pin 14 is Vcc and goes to the positive terminal of your power source, while GND at pin 7 goes to the negative terminal.

    This IC has 2 flip-flops. It's good practice if you aren't using the second flip-flop, to connect pin 13, 10, 11 and 12 (clear, preset, clock and data for the 2nd flip-flop), to Vcc, to avoid it from taking noise, and uselessly change states.

    You can take a look at the datasheets from fairchild's site itself, no need to search the internet to find it.

    The output voltage will be very near Vcc and GND (maybe 1v difference or so).

    It's expected that the signal on the clock pin is at least 8 volts when it goes to high, and less than 2v when it goes low.
     
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  11. servion

    Thread Starter New Member

    Mar 30, 2010
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    Alright, sounds good. So just to clarify, I connect pins 1, 4 and 10 through 14 inclusive to the voltage source (+12V).

    Why am I connecting pins 1 and 4 (the first flip-flop's "clear" and "preset" pins)? What do those do for me?
     
  12. Norfindel

    Active Member

    Mar 6, 2008
    235
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    Yes, that's how you must connect it.

    The Preset and Clear pins can be used to set the state of the flip-flop, and they take precedence over everything else. Fairchild's datasheet shows a truth table of how this signals work. The only case when they don't change the output is when both are in 1. If Preset is made low at any time, Q will be set to 1, ignoring clock, and data inputs.
     
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