How to force output low if incoming signal is lost (binary counter & 555)

Thread Starter

edgeb

Joined Jun 18, 2020
10
Hi folks

I‘ve designed a circuit, as shown, with a 5 volt AC input fed into a binary counter and from there into a 555 timer which is configured as missing pulse detector. The incoming feed is from a fuel flow transducer which changes frequency based on fuel flow. This circuit is being used to control a fuel transfer pump and it is imperative that the pump is shut off (by the 555 output going low) if there’s a slow-down of flow rate, or any kind of break in signal.

The system works as intended, in that an incoming frequency of over 90Hz will be enough to keep the 555 output high, whereas a lower frequency will take the 555 output low.

However, if the incoming connection is broken (by unplugging the incoming lead), the 555 output stays high.

How do I get the 555 output to go low if the incoming connection is broken?
View attachment 211674
View attachment 211674

Any input would be great.

Thanks
Edge
 

ericgibbs

Joined Jan 29, 2010
11,643
hi,
When using a 555 timer, if the Trig input is held Low [ due to say the counter not being clocked by the 50Hz source] the output will be high, thats your current problem.
To avoid this the Trig input is AC coupled as per my clip. [ to the +5V line]

I will look at the momentary switch problem.

E
 

ericgibbs

Joined Jan 29, 2010
11,643
hi edge,
Who designed the circuit,?
For the momentary switch to work and latch On the Relay the 555 has to be outputting a High even when the 90Hz input is disconnected.!

Is there any reason for using a momentary switch instead of a regular On/Off latching switch.?

E
 
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Thread Starter

edgeb

Joined Jun 18, 2020
10
hi edge,
Who designed the circuit,?
For the momentary switch to work and latch On the Relay the 555 has to be outputting a High even when the 90Hz input is disconnected.!

Is there any reason for using a momentary switch instead of a regular On/Off latching switch.?

E
Hi Eric

I designed the circuit. You’re right - if there’s no AC input to the counter, and from there into the 555, it won’t latch. I designed it that way. But by starting the pump, via the relay, fuel will flow and the AC input will be activated.

The momentary switch manually activates the 555, and via the transistor, it closes the relay, which not only powers the pump but also feeds current into counter and the 555. Doing so basically the puts whole thing live. I guess you could say it’s a latching circuit If you then release the momentary switch, it stays live until the first missing pulse sets the 555 output low, which then opens the relay, killing power to the counter and the 555, and that way the whole system dies until you start it again.

I didn’t want to use a pulse-based relay because I want the system to stop pumping if there is any fault at all - it’s for an aeroplane and theres no room for error. It’s better for fuel not to pump
Than for it to carry on pumping when it shouldn’t. So I only want it to pump while there’s current to the relay.

The circuit functions exactly as designed - which is a small miracle, given my inexperience.

I’m now trying to design some fault tolerance into it and the one potential pitfall I see is if there’s a break in the line from the fuel flow transducer into the circuit. (The transducer is right at the front of the fuselage and my instrument panel where this circuit will be housed is about 2m towards the tail).

Hence my post.

I hope that makes sense?

Thanks!
 

ericgibbs

Joined Jan 29, 2010
11,643
hi Edge,
That explanation makes it clear in what you are aiming to do.
I will look your circuit over and get back to you.
E
 

ericgibbs

Joined Jan 29, 2010
11,643
hi,
OK.
I have assumed HC4020 on the circuit.
More questions.
What time is the 555 Mono Output ON period set for.?
Which pin of the 4020 output are using is it Q7 as per your drawing.

I am drawing up a LTSpice simulation circuit to test the timing.
E

Update:
This a draft circuit of what we have so far, for ref only.
 

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Thread Starter

edgeb

Joined Jun 18, 2020
10
hi Edge,
That explanation makes it clear in what you are aiming to do.
I will look your circuit over and get back to you.
E
Thanks very much for the input Eric.
hi,
OK.
I have assumed HC4020 on the circuit.
More questions.
What time is the 555 Mono Output ON period set for.?
Which pin of the 4020 output are using is it Q7 as per your drawing.

I am drawing up a LTSpice simulation circuit for test the timing.
E
Apologies, I’ve I’m using Q8 on the 4020 - I’m not in front of it now but I think that gave me a frequency of 2.11Hz.

Q7 on the drawing is just to an indicator LED (flashing at a human-friendly frequency) but you can ignore that for the purpose of this discussion.



Thanks
Edge
 
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ericgibbs

Joined Jan 29, 2010
11,643
The Mono output period of the 555 is adjusted with the POT so I’m not sure precisely what it is but it’s such that at >80Hz input from the 4020, the output stays high. Below 80Hz it goes low. That gives a period of approximately 0.0125 seconds.
Hi,
If your source frequency clock input to the 4020 is a nominal 90Hz , as per your drawing, then the signal on pin 12 of the 4020 will be many times lower, ie not 80Hz.??

E

This plot shows Q7 out with a 100Hz clock source.
 

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ElectricSpidey

Joined Dec 2, 2017
1,404
One issue I see is the inability to predict where the counter will stop if the signal is lost.

Personally I would drop the counter and directly drive the discharge transistor with the AC signal, perhaps with an analog buffer in between, so I could insure the proper output level if the signal is lost.

You could perhaps keep the counter in the circuit to give the “user friendly” LED flash.

I would also use a more conventional 555 missing pulse circuit.

Also I would get rid of the diode at the output of the 7805 because that is dropping the output down to a iffy voltage for a bjt 555.
 

ericgibbs

Joined Jan 29, 2010
11,643
hi edge,
Our values disagree, please check the attached images.
I make the 4020 as ~0.7Sec ie: 1.4Hz.?
Check that we are both using the same 4020 pin out.
E



Update:
Have you considered what the 555 output will do when the signal input frequency drops to a lower frequency.?
Looking at the circuit the relay and pump will switch On and Off at slow rate, as the 555 triggers every time Q6 goes high/low
 

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ElectricSpidey

Joined Dec 2, 2017
1,404
Yea, your missing pulse circuit is fine...

As far as the buffer, I was thinking maybe a follower just to keep the load on the AC signal low. Then you could place a pull resistor at the trigger to insure it goes high, when the signal is lost.

See what I'm thinking is...if the counter stops with the output low, the 555 output will never go low.

But there might be other better ways to deal with that, such as placing a diode in series with the output.

But, I'm going to leave you in the hands of eric, because things can become confusing, with too many spoons in the soup.
 

Thread Starter

edgeb

Joined Jun 18, 2020
10
hi edge,
Our values disagree, please check the attached images.
I make the 4020 as ~0.7Sec ie: 1.4Hz.?
Check that we are both using the same 4020 pin out.
E



Update:
Have you considered what the 555 output will do when the signal input frequency drops to a lower frequency.?
Looking at the circuit the relay and pump will switch On and Off at slow rate, as the 555 triggers every time Q6 goes high/low
Hi Eric

OK, here’s an updated circuit - please ignore the previous. You’ll see here I’m using Q8 on the counter and have fixed the size of the resistor on the 555, at 32k. I’ve also removed the diode from the output of the 7805.

Using this model, the relay latches and releases appropriately.

0DAFE506-D3EB-4ADB-8F7B-0E2900D9C3DA.jpegThe issue of what happens if the AC input is lost remains to be solved though...
 

ericgibbs

Joined Jan 29, 2010
11,643
hi edge,
Look at this addition of a transistor to the 555 RST pin.
PB push-button, energises the relay, removes the RST from the 555, if there is Trig from the 4020 the 555 will go High.
If the 90Hz fails or falls below a minimum frequency the 555 will be un-triggered and the relay will switch On the RST.
So the 555 cannot re trigger until the PB is pushed again.
OK.
R2 resistor is the Pump
E
AAA 425 19.26.gif
 

Alec_t

Joined Sep 17, 2013
11,813
I think you'll still need a capacitive feed to the Trig pin, as suggested in post #5. As is, if Tout got stuck low then 555 Out would stay high.
 
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