Driving solid state relay input

coolro18

Joined Jul 10, 2020
17
Can a 4-20 mA (constant VDC) analog output on an industrial controller be used as an input on a solid state relay? I wasn’t sure how a device behaving as a variable current source might interact with a load such as the LED in an solid state relay that varies in its current draw. For example, what would happen if at a certain VDC, the controller output less mA than the LED drew at that voltage? Also, what are potential dangers of using this unconventional method?

Irving

Joined Jan 30, 2016
3,887
Not normally, the SSR is a voltage driven device (and may not be optical internally) and, typically has a switch on level of 3v and a drop out of 1.9v.

A 200ohm resistor across the SSR input might work, 20mA gives 4v, and 4mA gives 0.8v but you'd have to be sure of keeping it below say 8mA to be sure of switch off.

coolro18

Joined Jul 10, 2020
17
Not normally, the SSR is a voltage driven device (and may not be optical internally) and, typically has a switch on level of 3v and a drop out of 1.9v.

A 200ohm resistor across the SSR input might work, 20mA gives 4v, and 4mA gives 0.8v but you'd have to be sure of keeping it below say 8mA to be sure of switch off.
Ok, after checking, the specific controller I am using can actually output a specified amperage ranging from 0-21 mA. Would this be enough for switch-on and switch-off without adding a resistor across input terminals? And what would happen to the remaining voltage after the input drop?

Irving

Joined Jan 30, 2016
3,887
Your device is like a valve that can be open (21mA) or closed (0mA) or somewhere in between. But unless there's a tank of water (a voltage) connected to the valve and somewhere on the other side where the water can flow through, nothing actually happens.

The point is that the SSR input won't see a voltage unless that current acts against something. Most SSR are high impedance devices and it's the voltage that triggers them. The resistor provides a path for the current to flow to generate the voltage but there must be a voltage source in the system.

Voltages

coolro18

Joined Jul 10, 2020
17
There is a voltage source of 15VDC from the controller

crutschow

Joined Mar 14, 2008
34,453
Do you have a specific SSR in mind?

GetDeviceInfo

Joined Jun 7, 2009
2,196
Do you have a specific SSR in mind?
You need the bypass resistor to shunt the current away, yet generate a voltage. Without the bypass resistor, the device will apply the 15 volts to your high impedance input, at any signal strength, attempting to move the corresponding current.

Irving

Joined Jan 30, 2016
3,887
That's what I was saying. It depends on the input impedance of the SSR. Not all manufacturers specify this. Omron mention it in their "precautions for use" and give the calculation to do. But it's clearly in the "if you must" category because impedance varies with temperature too, and that's not a well specified parameter.

For many current output devices, off isn't 0mA, there is some leakage.

crutschow

Joined Mar 14, 2008
34,453
Most SSR are high impedance devices and it's the voltage that triggers them.
No necessarily.
Many have an opto coupler input consisting of an input LED in series with a resistor to limit the current.

Irving

Joined Jan 30, 2016
3,887
No necessarily.
Many have an opto coupler input consisting of an input LED in series with a resistor to limit the current.
I suppose it depends on your definition of 'high impedance'. Compared to a typical 4-20mA current circuit, Omron's input impedance of 1.6k upwards for their opto-inputs is arguably high, suffice to say they are concerned about leakage currents of 625uA giving a false turn-on. A typical SSR picked from the RS catalog has turn on at 4v, turn off at 1v and an input impedance of 1.6k. A 4mA signal would generate 6.4v, turning it on. To turn it off, you'd need to get down to <1v, or <625uA guaranteed. Its too tight a margin. A 250ohm resistor across the input at 4mA means 540uA through the SSR and 3.46mA through the resistor giving 0.864v, better but still tight. At 20mA its only just 4.32v which is just about OK for turn on. Its do-able but I'd not be happy, the margins are too close. I'd prefer to see an active drive circuit, where, say, <8mA = 0v and > 16mA = 15v (since that's the supply we have).

Something like: