Load Current Rating vs Load Voltage Rating for a Relay

[darby_m]

Hey guys, I have a newbie question about electronics and I was wondering if I could get a hand.

I have a piece of computer equipment using 12VDC at 1.78Amps. I want to use a relay to switch this circuit on/off. I am interested in a solid state relay because of the size and (to my knowledge) lack of magnet used in its operation. However, a lot of the relays I was looking at say the load voltage rating is way up there between 200V and 350V with tiny load current ratings between 100mA and 150mA.

In the terms of what the relay can handle, with me only wanting to run 12V through one, would the load current rating go up? Not exactly sure how this works.

Any help in understanding this would be so super appreciated!

 

[GoldDigger]

No. The load current rating does not go up as the voltage goes down for a solid state relay. It might for a mechanical contact relay, but not be that large a factor. And the DC rating may be smaller still even if the SSR works for DC. Keep looking.

 

[darby_m]

Perfect. Thanks a ton! I'll look for a solid state relay with a load current rating of like 2amps then. Am I correct in assuming that the voltage being higher then what want is ok though? For example 20V 2Amps, would work for my 12V 1.8Amps?

 

[topgone]

Perfect. Thanks a ton! I'll look for a solid state relay with a load current rating of like 2amps then. Am I correct in assuming that the voltage being higher then what want is ok though? For example 20V 2Amps, would work for my 12V 1.8Amps?

When you talk about voltage rating of a "relay" (20V), what that means is that the relay can be used on a system that has 20V or so. The most important criteria you will have to watch is the current rating of the relay, "2A" in your case. Does "2A" in the data sheet is listed as "continuous" rating or just a peak value? Does your load not draw more current in excess of 2 amps? If not, then you're good to go.

 

[gar]

151121-0046 EST

darby_m:


There are many devices that can switch an electrical circuit. When some switching device has an actuator added to it, then it becomes a relay.

Two wires can constitute an electrical switch. But more usually a mechanical switch will will use some particular material or materials for the switch contacts that are suited to the application. For a mechanical relay the material, size, heat sinking, force, and opened space will determine the switching capability. Probably most mechanical relays will use an electro-magnetic as their controlling element. Physical spacing and insulation of the coil from the switched contacts will determine the isolation voltage rating.

Solid-state switches generally fall into two categories. Those that the control signal can turn the switch on and off, and those where the control signal determines when the switch turns on, and whether the switch is maintained on, but turn off is determined by when the load current drops below a holding level (Triac or SCR) after the control signal is removed. Normally these can not switch on and off a DC current without special external circuitry.

The load switching capability of the SSR is determined by the characteristics of the solid-state switch. Load over voltage or current can easily damage an SSR. Also the on state voltage drop is generally much higher than a mechanical switch. Typically a mechanical switch is millivolts, and an SSR is a volt or two.

See http://www.ixysic.com/productbriefs/graphics/PB-PowerRelays.pdf for some power SSRs useful for switching DC.

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[mgookin]

Every relay should have specs on min and max make & break. That's to actuate the contact.

Then there are load ratings for the load you are switching.

You have AC/ DC ratings for the switching circuit and you have AC/DC ratings for the load you're controlling, and they're both different.

Every relay is two different circuits. One circuit to switch the load and another circuit which is the load. Check your ratings on both.

 

[gar]

151121-1936 EST

darby_m:

There is another factor you need to consider that may not be selfevident.

Every circuit has an inductive component, large or small, but some. What happens to the voltage across a switch or the inductor when a current is flowing in an inductor and you open the switch? How do you solve this problem whatever it may be?

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[darby_m]

Thanks again everyone for sharing your knowledge, it has been extremely helpful.

I believe I found a relay that will do what I want and have acquired a couple to play with. My load was also overestimated in my first post and after checking with a meter it is truly between 12V DC 0.9Amps and 1.1Amps. The relay I have says it's load current in a DC configuration is 2Amps, and the load voltage is 60V. Here is a link to the relay's data sheet:

http://www.vishay.com/docs/81646/vo14642a.pdf

I have built a little bread board circuit to test the relay and have ran into a snag that I simply can't seem to figure out. I have given the relay 1.13V on the control pins and wired pins 4, 5, and 6 according to the dc only configuration in the data sheet. For now I am simply trying to switch on an LED to test that the relay is working. However, the LED lights regardless of applying the control voltage. Therefore, the LED is lit when there isn't the 1.13V on the SSR's Anode.

Here is a quick little diagram to demonstrate what my pin configuration looks like:


http://oi64.tinypic.com/qz0hag.jpg

I feel like I am missing something simple, but can't seem to see it. Please let me know what you think, and thank you all again so much for the help.

 

[mgookin]

Thanks again everyone for sharing your knowledge, it has been extremely helpful.

I believe I found a relay that will do what I want and have acquired a couple to play with. My load was also overestimated in my first post and after checking with a meter it is truly between 12V DC 0.9Amps and 1.1Amps. The relay I have says it's load current in a DC configuration is 2Amps, and the load voltage is 60V. Here is a link to the relay's data sheet:

http://www.vishay.com/docs/81646/vo14642a.pdf

I have built a little bread board circuit to test the relay and have ran into a snag that I simply can't seem to figure out. I have given the relay 1.13V on the control pins and wired pins 4, 5, and 6 according to the dc only configuration in the data sheet. For now I am simply trying to switch on an LED to test that the relay is working. However, the LED lights regardless of applying the control voltage. Therefore, the LED is lit when there isn't the 1.13V on the SSR's Anode.

Here is a quick little diagram to demonstrate what my pin configuration looks like:

View attachment 13909
http://oi64.tinypic.com/qz0hag.jpg

I feel like I am missing something simple, but can't seem to see it. Please let me know what you think, and thank you all again so much for the help.

The spec sheet wants a jumper from pin 4 to pin 6. Did you do that?

Only other thing I can suggest is take it apart and look for continuity from pin 5 to 6 with nothing connected. It's a FormA relay so it should be open.

 

[gar]

151203-2051 EST

darby_m:

First, do not apply a voltage with low source impedance to the input LED. Rather apply a current, my suggestion 10 to 15 mA, with a maximum open circuit voltage of 5 V ( the maximum reverse voltage rating of the LED ). So use a series resistance from a 5 V source of about (5-1)/0.015 = 270 ohms.

The SSR data sheet is not really very clear.

I believe internally you will find that there is a diode between 6 and 5 and another one between 4 and 5. These are opposing each other from 6 to 4. Basically a reverse biased shunt diode across each MOSFET.

From the symbolism being used I would assume that conduction of the diode between 6 and 5 would occur when 6 is more negative than 5 by the diode drop of the diode. And no conduction thru that diode when 6 is more positive than 5.

This also would mean conduction thru the MOSFET would occur when 6 was more positive than 5 and the MOSFET was turned on. And no conduction when the MOSFET was turned off.

Your drawing seems to imply that 6 is more positive than 5. Thus, when the MOSFET is off the koad LED should be off. The leakage current of the MOSFET is supposed to be less than 1 microamp at 60 V. With any typical load LED 1 microamp should produce virtually no light. However, some bright LEDs I use produce a lot of light at 500 microamps. LED brightness is approximately proportional to current.

Test one of your SSRs with an ohmmeter. Use a Simpson 260. When not on Rx1 the source voltage is either 6 or 9 V (if I remember correctly) depending upon the age of the meter. Put the polarity switch in DC + and the + terminal will be positive relative to COMMON. Connect + to 6 and COMMON to 5. Should read infinite. With a 260 on Rx100 a 1 A silicon diode reads about 6 on the scale or 600 ohms. Using Rx100 switch to DC - and you should see conductivity.

My thought is that you are forward biasing the shunt diode in your SSR. Either your polarity is reversed or my understanding of the datasheet is wrong.

.

 

[gar]

151104-1622 EST

The 14642 has a considerably lower current rating for AC than DC.

In the DC connection this is because the two FETs are in parallel.

Whereas in AC operation only one FET is used at a time plus it is in series with a diode. I don't see any good information on maximum AC current. One table shows a gross difference in some apparent resistance for AC vs DC. All devices are power limited. With more resistance and the same current there is more power dissipation and thus higher temperature rise, but we are temperature limited. Thus, a higher resistance in AC means a lower maximum current.

Note: an FET looks much like a resistor when turned on, a diode does not. An FET is a three terminal device. The gate (control electrode) voltage relative to the source determines conductivity. When the gate is driven to saturation the drain to source resistance becomes lowest. To see some characteristics of a single FET see https://www.google.com/url?url=http...gguMAA&usg=AFQjCNFUgbyW53OYU4EbYJLoLKUjrY3_gA
You have to download the .pfd .

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