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So I have a relatively simple circuit. I have a little float sensor that is connected to a 24V circuit and when that float sensor is triggered (water is too low in the container) it connects the circuit, which triggers a solid state relay to connect a 120VAC circuit to turn on a pump which pumps up the container until the float sensor disconnects. I would like to add a status LED on the side of the 24V, probably on the low side of the SSR, so that I know when it is pumping and when it is off. Is there any clever way to add a 5V LED onto this circuit, or do I just need to bite the bullet and buy a 24V LED? I've included my circuit below:

enter image description here

UPDATE:

Here is the updated schematic based on the help I've received:enter image description here

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    \$\begingroup\$ LEDs are not voltage, but current devices. There is no such a thing as "5V" LED, unless it is not a pure LED. \$\endgroup\$ – Eugene Sh. May 18 '17 at 16:51
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    \$\begingroup\$ Series resistor. Look around this site about how to match a resistor to LED, there are hundreds of related questions. \$\endgroup\$ – Eugene Sh. May 18 '17 at 16:54
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    \$\begingroup\$ And your solid-state relay is backwards. The LED side of an SSR is the input, not the output. \$\endgroup\$ – WhatRoughBeast May 18 '17 at 17:01
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    \$\begingroup\$ The fact OP is using SSR's , drawn incorrectly, means he needs more help \$\endgroup\$ – Sunnyskyguy EE75 May 18 '17 at 17:03
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    \$\begingroup\$ pumps induce a negative spike on turn off. that's why it's needed and your SSR is likely a series pass high side switch using a Mosfet which may have a reverse output clamp zener diode internally. \$\endgroup\$ – Sunnyskyguy EE75 May 18 '17 at 17:27
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First, take note that your SSR is drawn backwards in your schematic.

Most LED's are current driven devices that will illuminate properly with 10 to 15 mA of current. When this much current passes through the LED, it generally has about 2 volts across it. This means that you need a resistor that will have the other 22 volts across it when ~10 mA of current is passing through it.

A simple Ohm's law calculation of 22 volts / 0.010 amps shows that a resistor of ~ 2200 ohms should do the job. Then calculate the wattage rating of the resistor as 0.010 amps * 0.010 amps * 2200 ohms and you have 0.22 watts. For reliability sake, you should use at least a 0.5 watt resistor.

The LED is connected in parallel with the LED of your SSR with the same polarity as the LED in the SSR (recalling that your schematic shows it backwards).

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  • \$\begingroup\$ Thanks Glenn. So I've added a new schematic representing what I took to be your answer. Does that match with what you intended? \$\endgroup\$ – clifgray May 18 '17 at 18:00
  • \$\begingroup\$ Yes that should do it. \$\endgroup\$ – Glenn W9IQ May 18 '17 at 18:08

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