# Smallest Magnetic Relay and how to choose the Ampere rating [closed]

I am building an automation system, where I don't have much space, so I require really small relays, 3.3 or 5 VDC controllable relays.

I need to connect it to say a lamp, or a fan, equipment's that draw less than ampere and work at 230 VAC. The 5 V 10 A relays are really big, and I need some advice on choosing the smallest relay and want to know the minimum ampere rating they should have.

## closed as off-topic by JIm Dearden, ThreePhaseEel, Dave Tweed♦Dec 29 '16 at 20:28

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• I think you'll find shopping questions aren't allowed. – JIm Dearden Dec 27 '16 at 15:37
• i want to know, how to choose amperage, how it differs from products, please be kind enough to help me, thanks – Niveth Dec 27 '16 at 15:41
• @JimDearden, read the question properly, daft accusations aren't allowed either. Niveth, take a look at Solid-State Relays (SSRs). An internet search will explain them and how they avoid the switching transients of mechanical relays. You should be able to find 230Vac/400Vdc or higher devices for 2A and you should put a 1A fuse in series. – TonyM Dec 27 '16 at 15:58
• @TonyM Thanks for that. No harm done. Have a great New Year. – JIm Dearden Dec 27 '16 at 17:24
• A magnetic reed switch with a coil wound around it. – Optionparty Dec 27 '16 at 18:42

SSRs are usually rated for more current than you need (I gather you need $\le 1\:\textrm{A}$) and their packaging is often designed with a heat sink in mind. So they are usually not small devices -- perhaps rather larger than the size of an equivalent-purpose mechanical relay. They also tend to drop up to $2\:\textrm{V}$ across their leads, so in use you might see dissipation of $\le 2\:\textrm{W}$. Which probably means that you actually don't need a heat sink, of course. But they tend to build them with that in mind, anyway. So they usually aren't small.

It's possible to make an SSR using a pair of SCRs and tiny 8-pin ICs like the MOC3023 or MOC3063. But even if you source some smallish parts quite capable of the modest dissipation you need, doing all that just complicates matters and you'd also need to work out the packaging and safety details, too. So I think that's off the table here.

That said, you can find SSRs in your current rating. They will look something like these:

But I'm not sure why you can't find a small relay. Here's one:

This can be found for sale at Digikey: Potter & Brumfield Through-Hole Relay SPST-NO. The contacts are rated for $250\:V_{AC}$ at $2\:\textrm{A}$, which seems to meet what you say you need. They are also small, at $10\: \textrm{mm}\times 6\: \textrm{mm}\times 5.65\: \textrm{mm}$. And not terribly expensive, either.

(I haven't checked if that relay is UL approved. You need to check that out, if it matters to you.)

The idea of a reed relay is also possible. Here's what one of those would look like:

This can also be found for sale at Digikey: Cynergy 3 S2-05EU. The contacts are rated for $300\:V_{AC}$ at $1\:\textrm{A}$, which also seems to meet what you say you need. They are also small but larger than the above telecom unit, now at $7.62\: \textrm{mm}$ diameter and $22.86\: \textrm{mm}$ length -- about 3 times the volume as the earlier device. They are more expensive, too. But in this case, designated as UL approved.

• thank you for sharing this, it helped a lot, i will look in to this. – Niveth Dec 28 '16 at 15:55

Solid-State Relays (SSRs) will be your best option for what you are trying to achieve here. With a small external voltage source (3.3VDC - 5VDC), you will be able to switch a load with higher power consumption such as what you will be running at 230VAC.

Below is a representation of how it will work and how you are able to switch to a load with high power consumption:

This works like an opto-isolator to keep your 230VAC away from your control circuitry.

In terms of current (ampere), you will have to look at the current being drawn from your appliances and make a clear decision which relay would work best. Light bulbs will have a power rating (30, 40, 60 Watts for example). You must calculate the current drawn by each load by dividing the power by its operating voltage. Simple for direct current. A little more complex for alternating current.

• so basically a 230V ac 100W bulb will draw 1/2300 A? something like that? – Niveth Dec 27 '16 at 16:24
• No it would draw 100W/230V.. So 0.5A approximately. Pavg = VIcos(theta) – 12Lappie Dec 27 '16 at 18:03