Problem with relay current requirement and how to increase current in DC circuit?

Question

What is the minimum current required to run a relay of 5 V (part: RWH-SH-105D). Will the relay run with 5V DC and 0.8mA current (my inputs)? If it does not work, then please suggest methods to run the relay by giving suggestions on increasing current.

Answer 1

You seem to suggest that your control source has limited current drive ability (5v/0.8mA).
If this is the case then using the suggested transistor circuits that require a few mA will be a problem.

Consider using a Nmosfet connected as a low side switch. It requires a small current to charge the gate capacitance on transition (on/off) and no current for static operation (steady state).

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

Read what it says in the data sheet: -

It says the 5V coil version has a resistance of 70 ohms and this implies a nominal current of 71mA (they say 73mA but I used my calculator).

They also say that they will all activate at 75% of 5V and this usually means 75% of current so, to guarantee it operates it needs 54mA. However that could be just while the relay pull-in the contact because they hint that the relay (if you are lucky) may stay pulled-in when the voltage/current drops to 5% - most won't of course but an educated guess would be that once the relay has activated you can probably relax the current down to maybe 30% but don't bank on it.

So, by the looks of it you need to use a transistor to turn the relay on. Below is an idea for a 12V relay but using 5V and a 5V coil will be OK: -

Any NPN BJT would virtually fit the bill. PS don't forget the diode across the relay coil - when the transistor open circuits, there can be a substantial kick-back voltage from the magnetic energy stored in the relay coil.

Answer 3

The datasheet states that the nominal coil current for the 5 volt version is 73 mA, so it definitely won't work with 0.8 mA.

You will have to add a transistor as a relay driver.

^{simulate this circuit – Schematic created using CircuitLab}

Any NPN transistor that can handle 100 mA should work.

Answer 4

Andy includes a lot of good information, but the final schematic does need some additional thinking before you can use it, directly. Same with Peter's example circuit, which requires too much base drive given what appears to be your I/O pin limits.

You mentioned that your outputs are \$5V\$ and have a compliance of \$800\mu A\$. (Not much, really.) I think you do need an NPN BJT to support the relay's current requirements and to protect your I/O pin from back EMF, as well.

EDIT: (Or use an NMOSFET, as Alex suggests. I prefer BJTs, broadly speaking. They cost me near zero, less than half a cent each, or about 20 times less than the cheapest NFET I get. And they have more general uses. But they do use more resistors.)

A small signal NPN BJT, like the PN2222A, can handle the required collector current just fine. But it does need to be solidly saturated. You can probably get away with a current gain (\$\beta\$) of 30 (though you will see most people wanting to assume less, perhaps 10 or 20.) With only \$800\mu A\$ to work with, this means you can only get to about \$24mA\$ with any assurances. Since you need more, I don't think the broad sweeping approach Andy gave you will do the job you want, safely or well.

Let's assume you want to try, though. A PN2222A has a rough \$\beta = 200\$ and it's worth experimenting to see if something useful might happen. In that case, the schematic might look like this:

An I/O pin output of "1" is needed here to activate it.

To compute a safe value for your I/O pin, I'd recommend \$R > \frac{5V - 1V}{800\mu A}\$. That's more than \$5k \Omega\$. So use \$6.8k \Omega\$. This provides about \$590\mu A\$ and just barely might do something useful. Probably not. And you should watch the temperature of your NPN BJT when trying this. If it starts getting pretty hot, you know this isn't a good idea after all.

If you need something better/safer to use, try:

In this case, \$Q_1\$ only pulls perhaps \$400\mu A\$, which is safely drawn from your I/O pins. But when active, the \$Q_1\$ collector sinks slightly less than \$4mA\$, which is \$\beta = 10\$ (commonly used "saturated" value.) This causes the pair of resistors, \$R_2\$ and \$R_3\$, to form a divider who's Thevenin voltage is about \$2.5V\$ and Thevenin resistance is about \$340 \Omega\$. That supplies as much as \$4.7mA\$ to the base of \$Q_2\$, which is more than enough to get where you need to be for the relay.