Fuel warning light delay

Can anyone help with a delay device to be added to the circuit in the attached web link.

The delay is required so that the warning light is not illuminated until the sender signal at the set low level threshold has been solid for say 2 minutes. So that bumps bends or acceleration doesnt flash the light as the fuel sloshes around.

I am pretty inexperienced with electronics-I can understand and build a basic 555 circuit,to give an indication of my level!

Hope someone can help.

Patrick

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Reply to
Patrick
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I think you can use a 555 timer with the timing capacitor connected to both threshold and trigger inputs (don't connect discharge). Thew timing resistor is connected between the low fuel signal and the timing capacitor. This is basically an RC low pass filter connected to a high hysteresis Schmitt trigger. You could get better filtering by putting a second RC time constant upstream of this resistor, using about twice as large a capacitor and half as large a resistor.

You will have to protect the 555 from high voltage spikes from the electrical system (18 volt zener across the 555's supply terminals, in parallel with the bypass capacitor and a series resistor to +12).

The timing network might be something like this: (set the reader to a fixed with font, like Courier to see this as a picture)

Input---47k--+--100k--+----+--Trig +| | | 1000uF 470uF +-Thresh | | +---+----+ | GND

When the voltage on the threshold pin exceeds 2/3 of the supply, the output of the 555 output latches low.

When the voltage on trigger is less than 1/3 of the supply, the output latches high.

See:

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Reply to
John Popelish

--- View in courier:

]--+-->+12A | [1N5353] | GND

+12A is to be used for Vcc for the 4060 and the 4001, and as shown above.

The way this thing works is that as long as the collector of your NPN is high, which will be when the tank has over a certain amount of gas in it, the 4060 and the RS latch made from the NORs connected to Q13 of the 4060 will be RESET, the 2N4401 will be cut off, and the lamp won't be lit. When the level of gas in the tank drops to the point where your NPN turns on, the RESET will be taken off of the 4060 and it'll start counting. If the gas doesn't bounce around for a minute or so, the counters MSB will go high, SETting the latch, which will turn on the 2N4401 and light the lamp. Any disturbance causing your NPN's collector to go high will result in another system RESET, and at least another minute before the lamp can turn on again.

The two NORs shown with their inputs grounded are spares, and their inputs _must_ be grounded as show.

-- John Fields Professional Circuit Designer

Reply to
John Fields

Did I miss something or:

[.. same as before ....]

| |

[...] Also: I would put a resistor in the Vcc going to the CMOS parts and perhaps a 15V zener across them. This is going in a car and voltage spikes are common in cars.

I'm make R4 10K or use a power MOSFET as the output transistor or something. A CD4060 won't like putting out the current it is being asked for.

I started off with the diode facing the other way and running from Q13 but decided that if Q13 was really loaded with a 1K, I'd be better taking the collector of the transistor as the input to the clock stopper.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

--
I think you missed something....

Here\'s the whole thing with an added reference designator:


                                          ]--+-->+12A
                   |
                [1N5353]
                   |
                  GND


Notice that if we start Off with a full tank the collector of Q1
will be high, which will RESET the 4060 and kill its oscillator, so
the 1N914 you inserted isn\'t needed.
Reply to
John Fields

Thanks ,guys for your help. I think I am probably a bit out of my depth! I can just about understand Kens circuit, which looks slightly simpler. I am not familiar with what the NORs are in John's !!! Couple of queries

1 Is it a 4401(referred to in cicuit diag) or 4001 (in note at bottom) 2 I am not familiar with the pin out on the 4060, from a diagram of a 4060B I can see Ctc =9,Rtc=10,Rs=12,Q13=3,MR=16?,is that right? Do I need to do anything with all the unused pins? 3 In my application the tank sender unit is grounded in the tank and supplied off the back of the gauge,I'm not quite clear how that is connected to the Emitter of the PNP,where your diagram shows VBAT 4 I will probably be using a bright LED as the lamp,protected by a suitable resistor,hope that doesnt change the rest of your circuit. 5 Finally do you agree that if I protect the sender with something like a .25A fuse, I will save it, if I c*ck something up?!

I am sure this is all very elementary stuff and I appreciate your help.

Patrick

Reply to
Patrick

I missed the latest replies,while I was putting those questions together! Patrick

Reply to
Patrick

In article , John Fields wrote: [....] I think I was right:

This point goes high and stays that way

The 1N914 I put in kills the oscillator when Q13 goes high and stops the

4060 in that state so the flip-flop isn't needed. [...]

I'd expect to be picking bits to the 15V zener out of the surounding plastic. In a "load dump" case there are many amps available.

The output transistor can be a high enough voltage one to take the voltage but the CMOS need protection, hence the resistor in just its power line.

See above. Remember, the diode goes to the transistor's collector and thus pulls down on the oscillator only when Q13 is high.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

In article , Patrick wrote: [...]

IIRC: The CD4001 is the NOR gate he is refering to.

As my circuit shows, you don't need it but what John is making is an RS-latch. Try it as sort of equations like this:

X = not(A or Y) Y = not(B or X)

Start by assuming A, B, and Y are all low. Now observe what happens if A goes high. Y will be high. At that point taking A low, will leave Y high. You can only get Y low by taking B high.

He has a 2N4401 transistor, over at the bottom right. The is a modest NPN transistor. A 2N2222 would also work in his circuit.

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All the unused outputs should be left disconnected.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

I guess I would just do the whole thing with a small PIC.

PIC - the 555 of the 21st Century.

Reply to
Richard Henry

In article , Richard Henry wrote: [...]

You'd need an LM7805.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

Here is a response from the gauge manufacturer about the sender specification. How does this effect anything? Patrick, The resistance range of the telemetrix fuel sender is 10 ohms at empty, to 180 ohms at fuel. Hence, as the fuel level decreases, the voltage at the top of the sender decreases, opposite to that required by your circuit. Apart from sorting that out, don't see too many probelms as your impedances are high relative to that of sender and gauge. Regards, Ian

Ian John Design/Application Engineer Caerbont Automotive Instruments Ltd

Reply to
Patrick

Here is a response from the gauge manufacturer about the sender specification. How does this effect anything? Patrick, The resistance range of the telemetrix fuel sender is 10 ohms at empty, to 180 ohms at fuel. Hence, as the fuel level decreases, the voltage at the top of the sender decreases, opposite to that required by your circuit. Apart from sorting that out, don't see too many probelms as your impedances are high relative to that of sender and gauge. Regards, Ian

Ian John Design/Application Engineer Caerbont Automotive Instruments Ltd

Reply to
Patrick

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Yup.
Reply to
John Fields

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_And_ some discretes to tame the spikes.
Reply to
John Fields

--
I just redrew your ASCII drawing showing the 12V connection to the
sender, the sender as a variable resistor, and the connection to the
emitter of the PNP as you showed it.
Reply to
John Fields

--
So your gauge circuit looks like this:?

       +12V
         |
      [GAUGE]
         |
         +----E2
         |
      [SENDER]
         |
        GND

Where E2 is the voltage going into the emitter of your PNP?
Reply to
John Fields

John Yes,gauge is also grounded ,if that makes any difference

See

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I'm amazed at the interest that this has generated!! Even though most of it is way over my head. Any one fancy,drawing up the final circuit ,so a newbie can build it? Why is it I cant seem to view the circuits in Courier on home PC, but on work one ,I can? Many Thanks Patrick

Reply to
Patrick

In article , John Fields wrote: [...]

Since the frequency is low, we can just put a 100K resistor and a couple of 1N914s in there to protect it. I really didn't look at that part of the circuit.

I assume that the purpose was to make a comparitor with a large backlash. The suggested circuit looks more complex than using a comparitor IC. A quick count has it having about 20 solder joints vs about 18 for using an LM211.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

In article , Patrick wrote: [.....]

Theory #1: Bill Gates has it in for you personally and is messing with your mind.

Theory #2: You aren't actually getting it set to Courier at home either bause you don't have that font installed or you clicked something in some settings box that overrides the font.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

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