I have built water level indicator circuit as shown in this figure.


If the inputs of IC 74148 is connected directly to negative supply through pull up resistor it works, but if it is connected to negative supply through wires immersed in water it doesn't. Any Suggestion?


First bit of advice - ditch that circuit. It almost certainly won't work. As you've discovered, water won't do to make the connections. 7400 TTL circuits need a nominal 1.6 mA at 0.8 volts or less to produce a valid low input, or the equivalent of about 500 ohms. Water simply is not that good a conductor (unless maybe it's seawater). The pullup resistors add another .4 to .5 mA to the requirement, dropping the effective resistance requirement to about 400 ohms.

This appears to be a classic example of somebody having a neat idea and putting it on the internet without actually sitting down and analyzing it, let alone trying it.

You might try replacing the 74148 with a 74HC148, and raising the pullup resistors to 1 M. And maybe adding a little salt to the water if you have to.

However, there's a basic problem with this approach. Applying DC to contacts which are near water is a great way to accelerate corrosion, and the corrosion products often act as insulators. This has obvious bad effects. For long-term reliability you've got to use gold-plated contacts, and the voltage across them has to be AC.

  • \$\begingroup\$ today i learned \$\endgroup\$ – Matt Clark Mar 18 '16 at 4:06
  • \$\begingroup\$ Can i use microcontroller instead of 74148 and 7447? \$\endgroup\$ – Umair Mar 18 '16 at 4:18
  • \$\begingroup\$ @Umair - Oh, sure. It's possible (but not guaranteed) that your microcontroller inputs are sensitive enough, and have weak enough pullups, that you can directly sense the presence of water. Keep in mind the corrosion issue, though - that doesn't go away. \$\endgroup\$ – WhatRoughBeast Mar 18 '16 at 4:27
  • \$\begingroup\$ Be aware that that the current through the water should be limited (especially in fault conditions) when using DC and salt, as the setup leads to electrolysis. Common salt (NaCL) will split into Sodium and Chlorine. Chlorine being a possible health hazard in water and air. Don't ingest that water. Details of risks: chemistry.stackexchange.com/questions/14055/ \$\endgroup\$ – Grebu Mar 18 '16 at 14:48
  • \$\begingroup\$ @Grebu - The gates themselves limit the current to a couple of mA or so. For CMOS, the pullup resistors would limit it to microamps. More important, without partitioning, a simple electrolysis will not produce chlorine. At the anode, NaOH reacts with chlorine, so overall, only hydrogen is produced. See chemistry.stackexchange.com/questions/41182/… \$\endgroup\$ – WhatRoughBeast Mar 18 '16 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.