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Analogue circuit question -advice needed please !

Discussion in 'Electronic Basics' started by [email protected], Jul 2, 2005.

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  1. Guest

    I should know how to do this, but please treat as a Newbie query :)

    I have a simple circuit question for any analogue experts out there !
    I'd be delighted if someone can suggest what I suspect is a simple
    solution. I have no problem with digital circuitry, or making the
    solution. My problem is a lack of understanding of analogue bits n
    pieces and lack of time to learn due to work pressures :(


    THE PROBLEM

    - I have to link a very modern fuel-tank level probe to an old
    fashioned car fuel gauge, via a simple circuit and it all runs of the
    cars 12v supply.


    THE DETAILS

    - The supply rail will be typically 12-14vs
    - The fuel probe is an expensive dedicated hyper modern capacitive
    device with a pic chip type device in it. Once calibrated it generates
    a high impedance output voltage of 0-5v linearly. Linearity isn't
    important to me. It cant source more than a milliamp.
    - The old fuel gauge is a fixed resistance 45ohm device with internal
    mechanical damping. It isn't very linear but this isn't important.
    The gauge reads empty to full with a voltage change across it of 0 -
    10volts, and current consumption of 0-230ma or so.
    - The 0 -> 5v input signal is pretty much a steady DC value so I cant
    have a capacitor on the input as in a conventional simple amplifier
    circuit.


    THE QUESTION

    - What simple FET based (or transistor!) circuit would link these two
    devices ? Ideally I'd like someone to advise on a schematic, with
    approx component values, that I can go off and build and tweak.


    Help !! If you want to email me offlist use amremote at hotmail dot
    com

    Thanks in advance.
    Andy / UK.
     
  2. kell

    kell Guest

    You should cross-post to sci.electronics.design
    I imagine you will get suggestions to use comparators or op-amps.
    Here's an idea using transistors:

    Vcc n-channel mosfet out
    _____________c e_________________________________________
    | |___| |
    | | /
    |_/\/\/\_____|_________c \
    2k2 \ / about 50K
    \ \
    | /
    npn |----------------|
    | |
    probe 100 / /
    signal-->____/\/\/\_____/e \
    | / 100K
    | \ adjustable pot
    | /
    | \
    | |
    | |
    |__________________|
    |
    |
    ground

    After you finish twiddling the resistor divider on the right it will
    impress a certain fraction (here, you will end up setting it at about
    half) of the output voltage on the base of the transistor. This gets
    compared to the probe signal voltage by means of the base-emitter
    junction of the bjt. If this (half) the output is .6 volt (that's the
    b/e drop) more than the probe signal, the bjt pulls down the bias on
    the mosfet gate. So the output voltage is regulated at (input +
    ..6volts) X (divider ratio). Linear, though not quite a direct
    proportion.
    You may have to amplify the input.
    If so: get another npn, connect its collector to Vcc, its base to the
    probe signal, and you will get amplified current out the emitter at a
    voltage one diode drop below the probe signal. This would be a good
    thing because the two diode drops we get in this circuit would
    thereupon cancel each other! Cute.
     
  3. kell

    kell Guest


    I forgot to mention: in my post with the ASCII schematic, use a
    proportional font to view it. If you are in Google, click options and
    "view original."
     
  4. Andrew Holme

    Andrew Holme Guest

    You want a DC amplifier with a voltage gain of 2, high input impedance, and
    high output current drive. You could use an op-amp non-inverting amplifier
    with an emitter-follower buffering the output e.g. (view in fixed font)

    ' +12V
    ' Vin |\ |
    ' ---------------|+\ |/
    ' | >---------| Q1
    ' .-|-/ |>
    ' | |/ |
    ' | |
    ' | |
    ' R1 | R2 |
    ' ___ | ___ | Vout
    ' .---|___|----o------|___|-----o--------------
    ' '
    ' ===
    ' GND
    '

    You need an op-amp that runs off a single supply, with inputs that work down
    to the negative supply rail.

    Q1 will dissipate up to 1 Watt.

    The two resistors (R1, R2) set the gain. R1=R2 for a gain of 2. I suggest
    R1=R2=10k

    I would include a 10uF power supply decoupling capacitor, and a 0.5A fuse in
    the power lead.

    I would also connect a large resistor (100k - 1M) from the input to ground,
    in case it's left floating.
     
  5. kell

    kell Guest

    D'oh! I labeled the mosfet like a bjt. Instead of c and e, it should
    be labeled d for drain (in place of the c) and s for source (in place
    of the e).
     
  6. kell

    kell Guest

    Edit: you need a bipolar as shown above instead of the mosfet. The
    2k2 resistor changes to a 1k 1/2 watt resistor or a few hundred ohms, 1
    watt.
     

  7. You mean a non-proportional font.

    If you use a proportional font, you have to know which one was used to
    create the drawing so you can get things to line back up.
     
  8. Just to be sure: Does 0V indicate empty or full?
    (snip)

    Again, does 0 volts indicate empty or full?

    No sense suggesting something that has a 50 50 chance of reading
    backwards.
     
  9. Guest

    Thanks to all for the suggestion so far.

    Couple of answers to the above :- I am not too concerned with absolute
    accuracy due to the horribly non linear fuel guage. But yes, when the
    tank is heading towards empty the probes output tends towards zero
    volts. (This is for a restored classic car with a new petrol tank and
    fuel level sensing, and I merely need some indication of the tanks
    contents, using the original dial ! )

    One thing I am wary of is loading the probes output, and the first
    suggestion above has a 100ohms to ground, where as that would have to
    be a few K to avoid dragging down the probes output. Maybe an input
    amplifier will be needed.

    Anyone else got ideas/comments ? I'll have a crack at trying it out
    tommorow.
    Thanks.
     
  10. Guest

    Thanks to all for the suggestion so far.

    Couple of answers to the above :- I am not too concerned with absolute
    accuracy due to the horribly non linear fuel guage. But yes, when the
    tank is heading towards empty the probes output tends towards zero
    volts. (This is for a restored classic car with a new petrol tank and
    fuel level sensing, and I merely need some indication of the tanks
    contents, using the original dial ! )

    One thing I am wary of is loading the probes output, and the first
    suggestion above has a 100ohms to ground, where as that would have to
    be a few K to avoid dragging down the probes output. Maybe an input
    amplifier will be needed.

    Anyone else got ideas/comments ? I'll have a crack at trying it out
    tommorow.
    Thanks.
     
  11. Chris

    Chris Guest

    Mr. Holme has provided a good, workable circuit for what you want. It
    might be able to use some input protection for the op amp, but other
    than that, it should work fine.

    Here's a redrawing with a bit of input protection and his in-line fuse
    (view in fixed font or M$ Notepad):
    ` VCC
    ` +
    ` |
    ` VCC o
    ` + 1/2 LM358 ( FU1
    ` | ) 1/2A
    ` D - VCC o
    ` ^ + |
    ` Vin ___ | ___ |\| |
    ` o----o-|___|-o-|___|--|+\ ___ |/
    ` 0-5V | 1.8K | 3.3K | >--|___|--|TIP31C
    ` .-. - .--|-/ 100 |>
    ` R| | D ^ | |/| | Vout 0-10V @ 0-250mA
    ` | | | | === o---------o-----.
    ` '-' === | GND | |
    ` | GND | .-. |
    ` === | 10K| | |
    ` GND | | | / \
    ` | '-' (M 1) 45 ohms
    ` | | \_/
    ` '-----------------o |
    ` | |
    ` .-. ===
    ` 10K| | GND
    ` | |
    ` '-'
    ` |
    ` ===
    ` GND
    (created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

    This should work well with a 13.8V supply. With low battery or with
    the key on ACC, a full tank may look just a little low, but that
    shouldn't be a problem. Choose whatever value for R that you feel you
    need for circuit input impedance, or leave it out altogether. The 1.8K
    and 3.3K input resistors and the diodes (use 1N4002 or better) provide
    protection for the non-inverting op amp input (use an LM358 or other
    single supply op amp).

    The op amp, TIP31C power transistor and resistors provide a gain block
    with a precise, linear voltage gain of 2, and the ability to source up
    to 1/4A into your 45 ohm load.

    Try Mr. Holme's circuit, you should like the results. An op amp
    circuit is much easier, more accurate and less expensive than cobbling
    together a circuit with discrete transistors.

    Good luck
    Chris
     
  12. colin

    colin Guest

    Simple circuit idea ..

    +12v +---+-------+
    | | |
    | R=2k @=meter
    opt=R | |
    | | |C
    | +-----| T2=any med power (to220) npn just simply bolted to
    something metal
    | T1| |E
    22k| |E |
    o--R-+--| R=6.8 ohm
    | |C |
    10k=R | | t1= any smal signal (high gain) pnp
    | | |
    -----+---+-------+

    The 0-5v sensor voltage is divided by the 22k/10k to ~ 0-1.6v
    This voltage then apears accros the 6.8 ohm resistor where it sets the
    curent through T1 and hence the meter to ~ 0-230ma
    T1 counteracts the vbe drop acros T2 and provides curent gain.

    This should be moderatly acurate. bear in mind this is untested and at full
    scale T2 is just about saturating, wich may mean 'full' scale is not quite
    acheived depending on type of transistor.
    ov input might not be quite zero current, this may corespond better to the
    the guage anyway, the optional R is if you want to increase the curent on
    empty. (try with 1meg)
    you can always tweak it ie lower the 10k or the 6.8 ohm etc.

    Colin =^.^=
     
  13. colin

    colin Guest

    oops line wrap

    +12v +---+-------+
    | | |
    | R=2k @=meter
    opt=R | |
    | | |C
    | +-----| T2
    | T1| |E
    22k| |E |
    o--R-+--| R=6.8 ohm
    | |C |
    10k=R | | t1= any smal signal (high gain) pnp
    | | |
    -----+---+-------+


    T2=any med power (to220) npn just simply bolted to something metal (but
    insulated!)


    Colin =^.^=
     
  14. kell

    kell Guest


    Chris got it right. The op-amp circuit is much better. Piece o' cake,
    prolly work first time out.

    Kell
     
  15. arem_29

    arem_29 Guest

    hi!...i hope im not too late, try checking this link regarding your
    problem,
    http://www.tpub.com/electronics.htm
    well, if ever you already solved it, this still can help you, in soe
    other case!.. not just for electronics, also other topics like
    engineering and others!..
     
  16. Guest

    Follow up :-

    Many thanks for all the ideas submitted. I now have a fully working
    solution based on the op-amp idea and it works a treat.

    I only wish my knowledge of analogue electronics was better so I could
    contribute something back to this particular forum.

    Cheers !
     
  17. Chris

    Chris Guest

    One other small thing -- you _will_ need a heat sink for your pass
    transistor, if you don't have one. With a 14V supply, about 2 watts of
    power will be dissipated by that transistor when the gas tank is about
    half full. That's well into the "groan zone" for a TO-220 package by
    itself.

    Just about any heat sink made for a TO-220 package will do, but the
    bigger, the better -- especially in an automotive environment, where
    you can get elevated temperatures. A heat sink rated for 4 watts or
    more is probably a good idea. A thin coat of silicone heat sink
    compound (so you can just barely see through it when applied) is
    optional.

    Thanks for an interesting question, and taking the time to give
    feedback. That's contribution enough.

    Good luck
    Chris
     
  18. Rich Grise

    Rich Grise Guest

    Ah, contribute whatever you've got - we run the gamut. :)
    Hopefully, it'll be under the dash, and not out in the crud.
    No, it's NOT! That's almost as bad as not using a heat sink at
    all. Personally, I prefer that pasty white stuff, but probably
    only because I "inherited" about a 16-oz jug of it from the
    Air Force. ;-)

    Cheers!
    Rich
     
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