• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Solid state relay questions

Not open for further replies.


New Member
Hello everyone,

I have a 12V DC load that requires ~500mA. I also have a 3.4V DC source and am about to purchase a Crydom DO061B solid state relay for the input voltage/current and switch, respectively. I'm looking to switch the 3.4V closed (or turn it "on") which activates the relay which in turn turns a light on (the 12V load). In the DO061B datasheet there is a "Max. On-State Voltage Drop @ Rated Current" and my question is does this affect the Input voltage or the Output voltage? For example, does the 3.4V drop to 1.9V at the "-DC CONTROL" leg or does the 12V drop to 10.5V at the "-DC LOAD" leg? I'm going to set this up to have a switch between the input source (3.4V) and "+DC CONTROL", "-DC CONTROL" to ground, 12V to "+DC LOAD" and "-DC LOAD" to the positive terminal of the light. Does anyone have any comments or suggestions? Can anyone answer my question about the "Max. On-State Voltage Drop @ Rated Current"? Thanks a lot!

- PrL

DO061B Datasheet: http://www.crydom.com/pdf/crydom_do.pdf (see this for +/- DC LOAD/CONTROL diagram)


I wouldn't expect it too, it is afterall supposed to act like a relay. If it did, it should probably be called a transisitor. :lol:

I guess your gonna have to experiment and see for yourself


Active Member
Hi PrL,

These solid state relay units are principally for use where the
input has to be isolated from the output, that is where two
systems are independent with no common wiring.

The voltage drop you refer to is in the output side and is a
maximum at full current. Very often the drop incurred is less
than the maximum given by the manufacturers.

This device will not be expecting to handle a current in excess
of 1.0 Amperes. Its normal operating range is intended to be
between 0.2 Amperes and 1.0 Amperes. However, it is rated to cope
with up to 5.0 Amperes for one second, no longer. So if your load
is inductive, then check it out carefully, those ratings reflect
typical changes in filament lamps, which take more current when
cold, and less current when running. Inductive loads normally
have a much higher surge factor, inductive loads also give a 'kick'
when switched off depending on how much current is moving at the
moment of switching off. This 'kick' can be quite a spike so if
an inductive load is intended for this device then appropriate
quenching should be considered.

From your description of the use you intend for this device, you
will be using a switch to operate it, presumably from the 12 volt
supply that will operate the lamp. (through a suitable resistor)

Since this device could only handle 1 Ampere, the max lamp would be
12 Watts.

Have you considered an ordinary relay, or maybe running the lamp
through the switch ?

Accompanying is a data sheet on the DO061B

Best of luck with it, John :) :)



Active Member
For 1A i'd put it straight on the switch.
(unless its one of those tiny little switches that only do signal current)


New Member
Hello all,

To clarify a bit the 3.4V will be coming from a PlayStation controller and the switch will be a button (X, O, square, triangle, etc.). The 12V will be coming from a separate power supply. When a button is pressed it registers as a button press on the console and also turns a light on, in this case the light is a cold cathode fluorescent light (CCFL). The 12V will be converted to 620V through an inverter which will then power the CCFL.

john1 - I've considered an ordinary relay, but the switching speed and the sound from the relay deterred me from using it. As you can see from the explanation above I can't use a switch for the light itself.

Russlk - I've also considered using transistors as a switch but I'm pretty sure I need the isolation.

Are there any more suggestions? Thank you all for your help.

- PrL


Active Member
you are doing it properly.
For that you will need the isolation.
And also the small drive from the game may not drive
a standard relay, but should work this ok.

These devices make use of the curious effect that light affects the
semiconductor material used.

A small light is placed close to a light sensitive device like an SCR
and when the light is operated, the device conducts.
The gap between the two is often rated at thousands of volts (breakdown)
this provides the isolation between the output and the control.

Similar devices use light sensitive devices like transistors in the
same way, to pass signals between circuits or parts of circuits that
have to remain isolated from each other.

Best of luck with it, John :) :)


Active Member
The output voltage drop 1,5V with max (1A) output load, and not depend from input voltage. Most of case the input contain a current-generator for optocoupler LED, so the input current not depend from input voltage.
Not open for further replies.

Latest threads

EE World Online Articles