1
\$\begingroup\$

I'm trying to install a low-flow switch (circuit is open below 0.1 gal/min, closed above that) as a safety measure on an immersion heater, but I have some questions about the setup I've come up with (I've only taken an intro to electronics physics class). I have attached a picture of the diagram of the system and included links to the relevant product pages below. My questions are: Do I need to have some sort of resistor or load in the flow switch circuit between the fuse and switch so that there is something for the voltage to drop across when the flow is too low and the circuit is open? Similarly, would I need a resistor or load of some sort between SSR1 and Fuse 2 for the same reason? Lastly, I need the current in the flow switch loop to be below 0.7 Amps. I see from the SSR specs at the webpage below that the impedance rating is 36,000+/-20% Ohms, so with a 12 VDC power supply and 36 kOhms, I calculate a current of 0.3 mA. Am I calculating the current for the flow switch circuit correctly? Thank you!

Diagram

Update: Based on @Michael initial comment below, I'm thinking that the diagram below should work and it will get rid of one SSR and the 12 VDC power supply, please let me know if you think this will work. Thanks for the help everyone! Updated Diagram

\$\endgroup\$
  • \$\begingroup\$ There are a lot of AC power plugs. Do you need so many? Sometimes when Y is on and X is off bad things happen. \$\endgroup\$ – George Herold Jan 13 '15 at 2:51
  • \$\begingroup\$ Good point. Based on @Michael comment below I think I should be able to get rid of the 12 V power supply and one of the SSRs, which I outlined in an updated diagram above - hopefully this would be a better design. \$\endgroup\$ – ryanDavid Jan 13 '15 at 17:22
  • \$\begingroup\$ No, the voltage will drop across the SSR when the SSR is off, and across the heater when both SSRs are on. \$\endgroup\$ – user253751 Sep 5 '17 at 23:29
0
\$\begingroup\$

You do not need to place two SSRs in series to your heater. As a matter of fact two SSRs in series probably do not function very well any way.

Instead you should incorporate the flow detector output and on/off control switch logic into a common signal control with the single SSR that the PID controller operates. In all likelihood the PID controller itself may very well have an enable/disable input that can be used to support the extra shutdown logic.

If there are no extra control inputs to the PID controller then a simple small signal relay can be used to interrupt the control signal to the PID SSR.

\$\endgroup\$
  • \$\begingroup\$ I originally wanted to only have one SSR and have the flow switch in the AC loop, but the amperage limit for the flow switch is 0.7 Amps while the 1.5 kW heater will pull 12.5 Amps - so that wouldn't work. But since the temperature controller outputs a 12 V DC pulse to control the SSR, can I just put the flow switch into that 12 V DC pulse loop? I added an updated diagram to the original post. The controller has a manual enable/disable function, but we want the flow switch to not allow the heater to work if a leak happens when no one's there. \$\endgroup\$ – ryanDavid Jan 13 '15 at 17:18
  • \$\begingroup\$ @ryanDavid - Exactly what I suggested. You should look though to see if the PIC Controller has a logical enable/disable and take advantage of it if existing. \$\endgroup\$ – Michael Karas Jan 13 '15 at 17:31
  • \$\begingroup\$ It looks like the controller doesn't have a logical enable/disable input - only the RTD input, so it looks like I'll go with the latest diagram that I put up. Thanks so much for your help! \$\endgroup\$ – ryanDavid Jan 13 '15 at 18:11
0
\$\begingroup\$

You only need series resistors to limit the current flow if there is a short circuit or similar. However, sometimes power supplies require a certain minimum load in order to generate the correct output voltage. This is somewhat uncommon, though. In your case, the resistance of the SSR will be very high so very little current will flow when the flow switch is closed - an additional resistor would not be necessary. On the AC side of things, you don't want any additional series resistor as this will just act like another heater, except it will just be wasting heat instead of heating the water. 12 volts across 36 k ohms would be about 0.3 mA, yes. It's 1 mA per volt per k ohm, so 1/3 of 1 mA.

\$\endgroup\$
  • \$\begingroup\$ Good. I didn't think that I would need the resistors, but I felt like it would be good to check with someone more knowledgeable so that I don't mess something up. Thanks for your help! \$\endgroup\$ – ryanDavid Jan 13 '15 at 17:18

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.