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I have a device, Neptune Apex, that has programmatically controlled 24VDC outputs.

Using the 24V I would like to trigger a relay that is built into another component, that has the specs of: 3-12VDC, 3-30mA.

Since the current draw is very minimal and will be constant, what would be the cheapest and easiest way to reduce the 24VDC to 12V to trigger the relay?

I don't really want to have to purchase and introduce another component like a buck converter, so could I do this with a voltage-divider or some other simple components and not burn anything up?

Update:

The 3-12VDC device is a Power Switch Tail II (http://powerswitchtail.com) and it it does appear to use a opto-isolated relay.

PowerSwitch Tail Specifications

Does this help to nail down the best solution?

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    \$\begingroup\$ That's not a very clear input specification, but it sounds like this is a logic signal and not the actual power for the relay, so something like a voltage divider would probably work. \$\endgroup\$ Jul 29 '19 at 20:27
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    \$\begingroup\$ @ChrisStratton: sounds like an opto-isolated relay to me... \$\endgroup\$ Jul 29 '19 at 20:41
  • \$\begingroup\$ You would still need a voltage source for the other side of the opto-isolator. \$\endgroup\$
    – hotpaw2
    Jul 29 '19 at 21:45
  • \$\begingroup\$ The web site for Neptune says that the Apex already uses a 12 V supply (in their discussion about using a UPS to maintain operation in the presence of a power failure.) Do you have access to that 12 V supply rail? I'd imagine you would because they also discuss placing other 12 V devices tied up to this add-on they sell. And that suggests that the 12 V is accessible. \$\endgroup\$
    – jonk
    Jul 29 '19 at 21:56
  • \$\begingroup\$ @jonk I don't have access to a switched 12v supply. The 24V switched output is what I need to use to activate and deactivate my 12v device. \$\endgroup\$
    – Adam Ring
    Jul 30 '19 at 11:22
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Using the 24 V I would like to trigger a relay that is built into another component, that has the specs of: 3 - 12 VDC, 3 - 30 mA.

This sounds awfully like a solid state relay input which is, basically, an infrared LED and series resistor.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. SSR with external current limiting resistor.

An infrared LED will drop about 1.4 V. If the SSR draws 30 mA at 12 V then the internal resistor is about \$ \frac {V}{I} = \frac {12 - 1.4}{30m} = 353 \ \Omega \$ (where 'm' is shorthand for milli).

Addition of an external 1.8 kΩ resistor would limit the current to about 10 mA at 24 V. This is well above the 3 mA minimum on-current.

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  • \$\begingroup\$ I attached the devices schematic to my original post, it is an opto-isolated relay so would the above design be effective in powering the relay? \$\endgroup\$
    – Adam Ring
    Jul 30 '19 at 12:14
  • \$\begingroup\$ Yes, except that there's an indicator LED that we didn't know about previously. Measure the current drawn at 12 V and then calculate the resistance required to drop 12 V at that current. \$\endgroup\$
    – Transistor
    Jul 30 '19 at 12:34
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    \$\begingroup\$ current draw is probaly 30mA at 12V (like it says on the label) anything between 400 ohms and 3K should work fine \$\endgroup\$
    – Jasen
    Jul 30 '19 at 13:09
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You can use a Zener and a transistor, such as in Explaination of high current zener transistor regulator circuit

or use a resistor in series with your load, the relay. The resistor should have \$R = \frac{12 V}{30 mA} = 400 \Omega\$ and be able to dissipate more than \$ P = 30mA * 12V = 360 mW\$, I would sugest a \$ \frac{1}{2} W\$ one, for a safety margin.

You could also make a resistive divisor such as this one. When the relay is not drawing current it outputs at most 12V, when it draws current it reduces the voltage but it remains within your specifications. It is still nor well regulated, but it works for your application. Beware you will need that the resistor R1 have 400 Ohms and 1 W dissipation and the resistor R2 have 400 Ohms and have 1/2 W dissipation.

schematic

simulate this circuit – Schematic created using CircuitLab

Watch out that both resistive solutions are not well regulated and can lead to problems, prefer the zener method.

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  • \$\begingroup\$ More likely a simple resistive divider; it doesn't really seem like this is power the relay, just signaling it, so you can't use a series element alone unless you know the load impedance is something that can be relied on. \$\endgroup\$ Jul 29 '19 at 20:26
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    \$\begingroup\$ Well, 30 mA at 12V is quite a bit for a logic signal. And, if you do use a resistive divisor you will have to account for the current drawn by the relay. \$\endgroup\$
    – jDAQ
    Jul 29 '19 at 20:34
  • \$\begingroup\$ If it drew 30mA from a node with 24V across a resistor of 400 ohms, it would have 12V on the other side of the resistor. I agree that the "solution is not regulated and can lead to problems" \$\endgroup\$
    – jDAQ
    Jul 29 '19 at 20:43
  • \$\begingroup\$ It's not the 24v that is unpredictable, it's the actual load current that is unknowable from the specification. \$\endgroup\$ Jul 29 '19 at 20:57
  • \$\begingroup\$ So, a resistive divisor would have a Thevenin equivalent of a voltage source in series with a resistor. If the equivalent resistance is too high, you would have the problem that the voltage at the load varies with the current being drawn by the load. Yet, for the equivalent resistance to be low, you would have to make a divisor out of "shunts", wasting a lot of power. Any suggestions? \$\endgroup\$
    – jDAQ
    Jul 29 '19 at 21:08
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Since the current draw is very minimal and will be constant, what would be the cheapest and easiest way to reduce the 24VDC to 12V to trigger the relay?

A resistor would be best 400Ω ought to get you 12V with 30mA from 24V. The downside would be the heat burned up in the resistor (360mW so use a 0.5W resistor).

Other than that another awesome circuit might be an adjustable current source:

enter image description here Source: https://diyaudioprojects.com/Technical/Current-Regulator/

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  • \$\begingroup\$ Your sugggested "cool circuit" will produce just as much heat as the resistor solution. \$\endgroup\$ Jul 30 '19 at 16:14
  • \$\begingroup\$ I'll change that to awesome, I love constant current circuits, I think they are 'cool' \$\endgroup\$
    – Voltage Spike
    Jul 30 '19 at 16:17

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