Relay with that 2.8 volts will make contacts close

I’m working on a lighting circuit for an electric bike. The lights are all 12V LED arrays. The stop and tail lights are automotive, the headlight is an MR16 track light spot light. The battery pack is 48V and there are two places on a circuit board I have available that are switched. One is 53V and the other is 2.8V. I assume this larger one is whatever the battery voltage is, although the wires to the circuit board are small and I don’t think I can draw much power from there. I’d like to use the 2.8V and use that with a relay to send 12V to the head and tailights. The brake light will work similarly and use 5V from the brake lever switches to trigger a relay for that.

I figure the head and tail lights themselves will use 0.8A . There is also a DCDC step down converter (the kind used for golf cart lighting) and I don’t know how much it might add. If I figure if should be below 0.5A. that would be load of 1.3A. So I guess I should be able to use contacts that can handle 2A max OK.

I know they coils need more power to close than remain closed and its not clear just how low the voltage they will close at. I’m thinking that some rated3V might work. I need some help finding something that will work.

Also I’d like to just mount this in an enclosure and don’t really want to make a board to mount the relay. I know there are a lot of Arduiono type relay board assemblies than I could just mount or I could just mount them with a bracket pins up.

Any suggestions on what might work?

Hi Tom,

Digikey does not do any type of design help, so I’m unable to offer much advise.

It looks like I shouldn’t have mentioned what I’m up to.

I’m asking for help locating a relay where 2.9V or maybe 2.5V to be safe will close the contacts which will be able to handle 12VDC at around 2A.

It sounds like a logic-level N-channel power MOSFET would work, if you’re switching the low side of the load.

I’m using to low side to energize the coil if it is an electro-mechanical relay and the 12V 2A would be the high side

I wasn’t referring to your 2v5 drive signal when I said “low side”. I meant, if you can connect 12V to one side of your load and “switch” the other side (the low side) between “open circuit” and “ground”, then an N-channel MOSFET would work. If on the other hand you need to connect one side of the load to ground and switch the other side (the high side) between open circuit and 12V, then a P-channel MOSFET (with an additional transistor to drive its gate) would work. I can post an example schematic if you can tell us which scenario you have.

You didn’t say how much current your 2v5 signal can source. A 3v relay is likely going to pull a significant amount of coil current (100 mA or more). That’s one reason a MOSFET might be a better choice.

OK, that helps a lot. I’ve used mechanical relays, but really aren’t familiar with transistors and their terminology.

I’ll have to do a little reading, but it looks Iike I can just mount this to a heat sink and mount that to the enclosure that would also serve as a heat sink. They are also cheap.

You shouldn’t need a heat sink if you’re driving it properly, as it will be saturated and therefore not dissipating any appreciable power. But if you want to mount it to something metal for mechanical support reasons, just be careful to insulate the metal tab (or buy a MOSFET with an insulated package) so there’s no electrical connection.

You’ll need to select a MOSFET that has a low enough gate voltage that it will saturate while conducting a current level of your load. 2.5 volts is pretty low for driving an N-channel MOSFET into saturation, but you might get by. With such a low drive voltage, it might make sense to use a tiny logic-level N-channel MOSFET as a buffer to drive a P-channel MOSFET on the high side.

Here’s is a picture showing low-side and high-side driving, but since your high-side voltage is 12V you won’t be able to use the high-side circuit as-is, you’ll need another transistor in between. Holler if that ends up being the case.


Hi TomM,

kd0gls’s schematic is a viable option and can work. If you go that way, you might want to add some protection for both MOSFET and the 2.8V signal using a couple of TVS’s to limit voltage spikes.

Another solution to consider would be to use a small solid state relay (SSR) rather than a MOSFET to switch the load. The advantage of this solution is that, like a mechanical relay, you gain galvanic isolation between your signal and the load, yet it can be switched with as little as 5mA and as low as 1.5V. This galvanic isolation protects that signal much better than a direct path via resistors to the potentially noisy load.

On the down side, SSR’s are typically more expensive than the circuit described above. However, for relatively small loads such as yours, some of them are quite reasonable.

Here are a few options to consider:

The input to these devices is an infrared LED, so you need to place a current limiting resistor in series with the input to set the current appropriately. The LED will have a forward voltage that can range from about 1.2V to 1.4V and should be driven with about 10mA, so you would need a resistor of around 150 Ohms to scale the current for roughly 10mA with your 2.8V driver signal.


Thanks, I think the CPC1706Y-ND looks like it will work. I’ll look around for some sockets to fit the pins, etc. I only need a couple so price is fine.

I see these are opto switched using LEDs. I always thought LEDs use 3V, but I can see in the data sheets that they will switch in a range from I think it was 1.4 to 3 volts, so that would work fine for what I’m up to.

Hi TomM,

It might be difficult to find sockets for those pins. However, I’ll take a look to see what I can find.

Regarding driving the IR LED, I would recommend driving it to at least 10mA to guarantee turn-on, but not much higher to maximize life (higher current reduces life expectancy, esp. at higher temperatures). The graph on page 3 of the datasheet, shown below, shows that for a 10mA current, the LED will drop between 1.4V and 1.2V, depending on temperature.

You will need to limit the current with a resistor to prevent burning out the LED. Assuming your 2.8V signal can provide at least 10mA, you can place a 130 Ohm resistor in series with the LED and you will get about the right drive current of 11-12mA.

That part has 0.1" pin spacing with four missing pins and the leads look like normal SIP/DIP leads, so a standard 8 pin SIP socket should work.

To prevent inserting it wrong way round I’d put epoxy in the four unused positions. Since this is on a bike, you’ll need to add a clamp or other device to keep it from vibrating loose in the socket.

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