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

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

Andy / UK.

2. ### kellGuest

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 \
\ \
| /
npn |----------------|
| |
probe 100 / /
signal-->____/\/\/\_____/e \
| / 100K
| /
| \
| |
| |
|__________________|
|
|
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. ### kellGuest

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 HolmeGuest

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

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

5. ### kellGuest

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. ### kellGuest

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. ### Michael A. TerrellGuest

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. ### John PopelishGuest

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. ### ChrisGuest

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

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. ### colinGuest

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. ### colinGuest

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. ### kellGuest

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

Kell

15. ### arem_29Guest

hi!...i hope im not too late, try checking this link regarding your
problem,
http://www.tpub.com/electronics.htm
other case!.. not just for electronics, also other topics like
engineering and others!..

16. ### Guest

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. ### ChrisGuest

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 GriseGuest

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