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A solid state relay specifies a trigger current of "7.5mA/12V". How much current does it draw at 5V?

This is the relay. I'm controlling it with an Arduino which has a maximum current draw of 40 mA per pin. Am I correct in calculating it multiplying 7.5 by 12 then dividing by 5 (the new voltage)? That gives me 18 mA, well within the limits of the Arduino pin - is it really that simple, or am I missing something important?

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  • \$\begingroup\$ A relay is not a constant power device, it will not draw more current at lower voltage. In fact it may not trigger at all at 5V if it's specified as a 12V relay. \$\endgroup\$ – John D Jul 18 '15 at 22:49
  • \$\begingroup\$ @JohnD: the linked datasheet shows that it is an SSR, rated for 3 - 32 volt input. \$\endgroup\$ – Peter Bennett Jul 18 '15 at 22:55
  • \$\begingroup\$ @PeterBennett Thanks, Peter, I was in a hurry hence the comment vs. an answer. You're correct, so the relay should work at 5V, but it's still not a constant power device so current shouldn't be MORE than the current at 12V. What it will actually be really depends on the internal circuitry. \$\endgroup\$ – John D Jul 19 '15 at 2:47
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I have a comparable "hockey puck" SSR handy. It also has the 3V to 32V input, although it's a different model (ESR5102401000Z). I made some quick measurements.

enter image description here

I was curious how these solid state relays (SSR) manage to accept a control voltage with a fairly wide range: from 3V to 32V. Of course, the datasheets for these "hockey puck" SSR don't provide the details about the input side. I can think of 3 schemes. Different models of SSRs may use different schemes.

Current limiting resistor

The LED and resistor are such that the current is low enough at 32V not to destroy the LED, but still high enough at +3V to activate the relay. In this arrangement, the input current will increase linearly with input voltage.

Series current limiter (active)

The input current shouldn't vary much with input voltage. The constant current source may be more or less stiff. The plot in Spehro's answer suggests a series constant current regulator in his SSR.

Shunt current limiter (active)

The input current will increase linearly with input voltage. Excess current shunted around the LED, but still drawn from the input. Here's an example of what a shunt current limiter may look like (source).

enter image description here

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    \$\begingroup\$ If there is one common factor in all these possibilities it is that the current should be <= 7.5mA at 5V. Until someone comes up with one that uses a SMPS constant power regulator... \$\endgroup\$ – Spehro Pefhany Jul 19 '15 at 2:46
  • \$\begingroup\$ Load is missing in the triac side? \$\endgroup\$ – GR Tech Jul 19 '15 at 3:44
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It will draw about the same current at 5V. The IR LED used in the SSR requires a certain amount of current to switch reliably and the typical input circuit for a 3-32V input SSR is a simple constant-current regulator made with discrete BJTs.

Here is another manufacturer's graph showing input current vs. voltage:

enter image description here

Of course this is just an educated guess on my part as the datasheet is characteristically inscrutable and/or mute on the point (as well as lacking safety agency approval markings), but every similar SSR I've seen works similarly to this.

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    \$\begingroup\$ Apparently, not all SSRs regulate the input current like that. Have a look at my answer in this thread. \$\endgroup\$ – Nick Alexeev Jul 19 '15 at 2:43
  • \$\begingroup\$ Well, nothing beats measurements! +1 to you. That one could also meet the datasheet spec for the OP's SSR> \$\endgroup\$ – Spehro Pefhany Jul 19 '15 at 2:45
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Your calculation is wrong - it should be 7.5 * (5/12), as we would expect a lower current with a lower voltage if there is just a resistor in series with the LEDs..

However, the datasheet shows an "input circuit" between the input terminals and the LEDs, so there may be something more than just a resistor in series with the LED - perhaps a constant current circuit, so the LEDs won't get excessive current with a 32 volt input.

I'd guess the current at 5 volts might be a little less than 7.5 mA, but probably not more than 7.5 mA. The only way to be sure is to measure it.

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  • \$\begingroup\$ perhaps the input circuit is some sort of current-sourcing buck regulator. infrared LEDs need less than 1V to operate. \$\endgroup\$ – Jasen Jul 18 '15 at 23:34

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