Fresh water level sensor.

tjbrace

tjbrace

Well-Known Member
Joined
8 Mar 2007
Messages
622
Location
West Mersea, UK
Visit site
I have built a variation of the attached circuit but the indicator leds keep dying. The variations are:
Monitors 2 tanks so 6 of the 7 Darlington pairs are used.
There is no buzzer
I used 12 volt white indicator leds and eliminated the in-line resistors
I have a red and a green led for port and starboard tanks.

Could it be the varying voltage of the battery bank killing the leds, eg when charging more than 14 volts is seen.
View attachment 71812
 
tjbrace said:
Could it be the varying voltage of the battery bank killing the leds, eg when charging more than 14 volts is seen.
Click to expand...

I struggle to see what else it could be. But if you used regular LEDs and something like the 1k series resistors instead of hybrid units, you should have good headroom: allowing say a 2V drop across the LED, at 12V supply that’s only 10mA through the LED, which is about half what you’d usually put through one, leaving loads of room for safety against an irregular supply voltage.

You probably put a momentary switch in the supply line to the circuit, so it’s not draining your battery except when you want to read the 'meter'.
 
tjbrace said:
Thanks for the input. I will try regular leds with resistors next.
Click to expand...

Actually after I wrote my post I rather liked Norman_E's buck convertor idea which is clever and cheap and would be less work to implement than replacing the LEDs (unless you've already blown them all anyway!) I don't know whether the buck converter's output voltage level is absolute or relative: whether once you've set its output voltage and the supply voltage rises (the challenge you're apparently facing) the output voltage will remain as you've set it or also rise. Norman_E or someone else better informed than me might. If it sets an absolute level, then for a quid (but a month's wait for delivery from Shenzhen!) you'd have an effective and easy solution.

EDIT: I see from another buck converter circuit that the output voltage does stay stable despite rises in input voltage, so perhaps Norman's idea is not just a good way to test whether input voltage variation is your problem, but also the best solution.
https://www.amazon.co.uk/Converter-Adjustable-Stabilizer-Transformer-Motorcycle/dp/B00IJ353V6
 
I would also use a DC-DC converter, set to 12v. There are plenty on Ebay with UK stock. I'd avoid the really cheap ones. Lots of XL6009 buck/boost converters on there for less than three quid.
 
I rebuilt the circuit using 3 volt leds and 1k resistors as mentioned previously, no luck though as 3 of the 6 indicators failed. The red and green leds were ok, but they are not driven through the ic. The surviving indicator leds had 3volts across them whereas the failed ones showed 12 volts (at both ends of each resistor). Finally, I removed all power from the gate pins and the surviving leds stayed on!
I think my chip was fried. The suggested circuit must be flawed.
 
tjbrace said:
I rebuilt the circuit using 3 volt leds and 1k resistors as mentioned previously, no luck though as 3 of the 6 indicators failed. The red and green leds were ok, but they are not driven through the ic. The surviving indicator leds had 3volts across them whereas the failed ones showed 12 volts (at both ends of each resistor). Finally, I removed all power from the gate pins and the surviving leds stayed on!
I think my chip was fried. The suggested circuit must be flawed.
Click to expand...

The failed LEDs will have gone open-circuit, so you would expect there now to be 12 volts across them.

Your resistors had colour codes brown, black, red, right, for 1kΩ? (Ignore the 4th band for this purpose.)

You know which way round to wire the LEDs, right? If you wired an LED backwards you'd get the same open-circuit behaviour as if you had fried it: the LED wouldn't light and you would see the whole 12V across it. The cathode (negative side) always has the short wire. That's the side with the cross bar and the light coming out of it on the diode symbol. As the particular chip you have is a negative driver, as the diagram shows the cathode goes to that side and the long wire, the anode, to +12V in your circuit.

All very strange. The circuit is fine. Always possible the Darlington pair chip is fried, but as its supply voltage isn't limited to anything like 12V it's hard to see how that would have been done. I'm sorry. Perhaps some brighter spark than me will come along and suggest.
 
BelleSerene said:
The failed LEDs will have gone open-circuit, so you would expect there now to be 12 volts across them.
Ok

Your resistors had colour codes brown, black, red, right, for 1kΩ? (Ignore the 4th band for this purpose.)
Yes, straight from Rapid Electronics

You know which way round to wire the LEDs, right? If you wired an LED backwards you'd get the same open-circuit behaviour as if you had fried it: the LED wouldn't light and you would see the whole 12V across it. The cathode (negative side) always has the short wire. That's the side with the cross bar and the light coming out of it on the diode symbol. As the particular chip you have is a negative driver, as the diagram shows the cathode goes to that side and the long wire, the anode, to +12V in your circuit.
I’m sure these were correct.

All very strange. The circuit is fine. Always possible the Darlington pair chip is fried, but as its supply voltage isn't limited to anything like 12V it's hard to see how that would have been done. I'm sorry. Perhaps some brighter spark than me will come along and suggest.
The chip takes an unrestricted positive on pin 9.
The gate supplies are also unrestricted.
So several 12 volt sources attaking the chip
Click to expand...
Many thanks for your input
 
For £40 you can buy a domestic water-meter off Ebay with pulsed output at 1litre per pulse. £7 buys you an lcd pulse counter. Install in the output side of tank and wire the pulse counter to somewhere convenient. Fill your tank to the brim, set the pulse counter to zero and empty the tank. The reading on the pulse counter is the capacity of your water tank. Refill the tank and count out the litres. You know exactly how much water you have in your tank to the nearest litre. All done for about £50 with no electronics to sort
 
geem said:
For £40 you can buy a domestic water-meter off Ebay with pulsed output at 1litre per pulse. £7 buys you an lcd pulse counter. Install in the output side of tank and wire the pulse counter to somewhere convenient. Fill your tank to the brim, set the pulse counter to zero and empty the tank. The reading on the pulse counter is the capacity of your water tank. Refill the tank and count out the litres. You know exactly how much water you have in your tank to the nearest litre. All done for about £50 with no electronics to sort
Click to expand...
Ok, but I need 1 for each tank. My solution would be <£5 for both tanks if the circuit worked
 
tjbrace said:
I have found a more refined version of the circuit. It has 47k resistors protecting each gate connection. Still no protection on pin 9!
Click to expand...

I still think we can find you a solution.

I don’t think it’s to do with having no external base resistor. There’s an internal one of 2.7k on the ULN2003, and that would limit the base current to ~4mA from your ~12V supply; the max base current is specified as 25mA so although adding an external resistor would reduce unnecessary current drain down the transistor b-e junctions, it’s not going to harm the device.

The device is specified to run off up to 50V, so your supply of nominally 12V supply isn’t going to be frying it no matter how erratic your engine’s alternator output is.

Even if you did fry the device, it has no way of applying more than your supply voltage (say 12-14V) across your LED & series resistor, so it’s not going to bust your LED.

What’s your evidence that you blew one of your new LEDs? Did it light and later not work? Are you sure you’ve wired the non-working LEDs the right way round? (Anode or long terminal towards the +12V side; cathode or short terminal or the clipped part of the LED’s circumference towards the chip side.) You haven’t ever connected one of the LEDs straight across a battery to test it without using a series resistor, have you? (That would fry it.)

Test them: if you disconnect one of the resistors from the chip’s output and take it to earth instead, its LED should light and you should find 2-3V across it (depends on its colour) and the rest of the supply voltage across the resistor. If it doesn’t, either you’ve wired it the wrong way round or it’s broken. (If it still measures those voltages, it’s working but it’s an infrared LED so you can’t see its light - unlikely but I can’t know where you got them from!) When it does, reconnect the resistor to the chip’s output, and then when you take the chip’s corresponding tank sensor input to 12V, that LED should also light.

If you’re still in trouble, test the chip. Disconnect one of those resistors from its LED and connect it to +12V instead. The resistor should now be between one of the chip’s outputs and +12V. Turn the power on and you shouldn’t measure any appreciable voltage across the resistor. Correspondingly, the voltage between that output and earth should be about 12V. Now take the corresponding tank sensor input to +12V. The output voltage should fall to 0V (that’s from earth of course) and you should read 12V across the resistor. Now if you reinsert an LED in series with that resistor it’s going to light.
 
Last edited:
I replied to this yesterday but my post hasn’t been saved?
When first tested after the rebuild the white indicator leds all worked as did the red and green indicators. I think this shows the correct polarity of the leds.
The leds were never connected to the 12 volt source without a series resistor, so not damaged by this.
After a few minutes of testing the white indicators began to fail. 3 of the six stopped. The voltage across the failed leds was 12, whereas the working leds showed 3 volts.
I then disconnected all six gate connections from the chip and the 3 working leds still illuminated?
On the strength of your positive feel for this project, I have retrieved the ic from the dustbin and will try your suggestions
over the next few days.
Many thanks for your interest. T.
 
tjbrace said:
On the strength of your positive feel for this project, I have retrieved the ic from the dustbin and will try your suggestions
over the next few days.
Many thanks for your interest. T.
Click to expand...

Good luck. As I describe, separate the testing of the LEDs with their series resistors, from testing of the IC and its driver (input) connections.
 
I tested each pair of the ic with a working led plus resistor. I found that 3of the pairs would not drive the led and 4 of the pairs were ok.
This bears out my earlier test where 3 leds failed, 3 leds were ok, and the last pair was unused.
I will return the ic to the dustbin
 
You must log in or register to reply here.

Share this page

Top