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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.

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  • \$\begingroup\$ Note that the three answers below all require that you have a 5 volt power supply capable of delivering the 73 mA that the relay requires. \$\endgroup\$ – Peter Bennett Mar 4 '14 at 20:19
  • \$\begingroup\$ Please take the time to use proper capitalization and punctuation. Readability goes a long way to helping you get your questions answered. I've edited your answer accordingly. \$\endgroup\$ – JYelton Mar 4 '14 at 21:28
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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).

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ This is how I would do it, too. In general, I prefer MOSFET transistors for switches. The 0.8 mA current may cause M1 to be in the transition zone for "a while," but with 80 mA switched current, that shouldn't be so bad. Choose a MOSFET with low gate charge. A BS-170 will probably work great. \$\endgroup\$ – Jon Watte Mar 4 '14 at 22:03
  • \$\begingroup\$ what is v+ and control? \$\endgroup\$ – user38211 Mar 4 '14 at 22:39
  • \$\begingroup\$ I mean v+ is the input or control ?If control is the input then what is the value of V+ \$\endgroup\$ – user38211 Mar 4 '14 at 22:40
  • \$\begingroup\$ @user38211 The schematic shows a low power control signal that switches the mosfet on/off, and a power supply (V+)that can provide enough current to power the load (relay in this case). V+ can be 5v or 9v or 12v, irrelevant of the control input signal level. \$\endgroup\$ – alexan_e Mar 4 '14 at 22:44
  • \$\begingroup\$ will this circuit run 12 v,30 ma relay?my inputs (10v,0.8 ma).?can i use IRF 540 N MOSFET? \$\endgroup\$ – user38211 Mar 9 '14 at 14:12
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Read what it says in the data sheet: -

enter image description here

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: -

enter image description here

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.

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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.

schematic

simulate this circuit – Schematic created using CircuitLab

Any NPN transistor that can handle 100 mA should work.

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  • \$\begingroup\$ will the same circuit run for 12 v,30 ma?my inputs(10v,0.8ma). \$\endgroup\$ – user38211 Mar 9 '14 at 14:10
  • \$\begingroup\$ Yes. The 2N3904 is good for up to 40 volts and 200 mA. Just be sure the transistor you use is rated for sufficinet voltage and current for your circuit. \$\endgroup\$ – Peter Bennett Mar 9 '14 at 16:42
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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:

enter image description here

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:

enter image description here

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.

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  • \$\begingroup\$ Thank you.Are there any direct current amplifiers available in market which will do the above specific tasks? \$\endgroup\$ – user38211 Mar 4 '14 at 22:04
  • \$\begingroup\$ what about current driver? \$\endgroup\$ – user38211 Mar 4 '14 at 22:05
  • \$\begingroup\$ You might google "relay driver IC" and see what you find. The ULN2803, for example, should pop up in your search. A problem with it is that it is a Darlington configuration and this means there will be a significant voltage drop across the "switch." You have 5V power and you nominally want 5V for the relay. But in this case, your relay may work at 3V, so the ULN2803 may almost appear to work for you here. But there's another problem. The max spec on TI's datasheet suggests perhaps \$1.5mA\$ is required to drive them. That's a possible problem. So that should make you search more. \$\endgroup\$ – jonk Mar 4 '14 at 22:18
  • \$\begingroup\$ I can give 10 v dc,then will ULN2803 work? \$\endgroup\$ – user38211 Mar 4 '14 at 22:23
  • \$\begingroup\$ You could take a look at TI's DRV8803 as an example IC using MOSFETs on its output. They aren't cheap. But they are available. \$\endgroup\$ – jonk Mar 4 '14 at 22:24

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