I know there are several topics about relays but for me this is kind of a mistery. I have a relay 5DC - 250VAC 10 A that the minimum operating voltage (at 20°C ambient temperature) is 3,6 V and the min. operating current is 71,9 mA.
Thats about 3,6*71.9= 260 mW (aprox).
I protected arduino I/O with a diode to unload the coil. I also bought some TIP139 and a couple of resistors to regulate the current, however I tested it connected to the computer (5V power?) already and it's working without the transistor.. now the math:
260mW/5V = 52 mA... how is this possible if the Arduino board only gives me a 40 mA on the I/O? :-? oh... and I don't live in the tropics its about 18ºC
I always remember to think... the Arduino is the BRAINS and not the MUSCLE. So, you want the BRAIN to move the MUSCLES... so you have it TELL the transistor that you want it to do the heavy lifting.
If you strain the arduino brain... it dies.... if you strain the muscles... well... you might get smoke... but it's not brain dead and the muscles are usually cheaper to replace.
Is it advisable to use the lowest power possible to turn on the coil, or its a bit indifferent ? (only matters in terms of power consumption ?)
Practical matter: With the resistences, I've here (1.2 to 1.4 ohm), I will provide an Ib of 3,58 mA with Beta (hFE) aprox 100.. I'll have 358 mA on Ic when I only need 72mA to switch on the coil. Should I get bigger resistences? like 5K? or it doesnt matter?
I think you are not grasping the basis to electricity here.
You can't have 358 mA on Ic if you only need 72mA. The coil will take what it takes. The fact that you have more current going into the base that you need for full saturation is only an issue when it comes to turning off the transistor. The extra current in the base makes it take slightly longer to turn off, that's all.
Ok ok.. I haven't played with electronics for a while! I thought you could say simply if the transistor is ON when VCB and VBE are directly polarized then Ic=BIb (and the B would vary according to VCE).. I don't have a lot of practical experience tho..
I'll test the circuit today! thank you very much for the patiente!
I'm powering the coil with 5V and the relay datasheet says that gives me about 85 Ohm resistence by the coil so:
5=Vcoil+Vce
Vcoil=85Ic
and VcoilIc>=270mW (to be sure)
85Ic^2>=270mW so Ic=56mA
Ic=BIb so Ib=0,56mA
From arduino onde gets 5v on high so 5=R*0,56+0,7 so R=7,68 KOhm (isn't this huge?)
Back to 5=Vcoil+Vce we get
5=85*56mA+Vce so Vce=0,24V
But this is barely above VceSat (Active Zone is Vce>VceSat=0,2v)... and if I lower Ic go get a higher Vce I wont have enough power to turn on the Coil "safely".
It is Ib that controls Ic all the way up to saturation, after that Ic is controlled by the resistance of the collector load. No amount of extra base current will affect the collector current.
So you need a collector current of 80mA. With an assumed gain of 100 then this means you need a base current of:-
80/100 = 0.8mA
Given you have a 4.2V drop across your base resistor then it needs to be 5K2.
However lets say the gain of the transistor was only 50, it that case you would need a 2K6 base resistor.
In practice you don't know the gain of your transistor, any data sheet will only give you a range, so you take the minimum of that range and use that for your base current and hence base resistor calculation. This gives you the maximum size of resistor you can have to turn on your relay.
Having more base current than this does not give you any more collector current because the transistor is saturated.
In general you will be OK with a 1K base resistor, nothing to fuss about.
Well you could but there is no point in taking Vce off the 5V supply, you simply don't have to be as accurate as that in this application.
Vce=5-6,8=-1.8v
No that is not right, Vce will never get much above the forward diode voltage drop. This is because a diode is a none linear device and so does not obey ohms law.
And I have almost the same question about the relay :
The Omron relay can take 15A max according to the datasheet , But I don't know can the PCB take 15A? :-/ :-/
Same question to the Mosfet Relay : http://www.emartee.com/product/41874/Arduino-4-Route-MOSFET-Button-IRF540
Is one channel capable to be loaded with 33A (FET AND PCB?) or is that total power?
And I find another relay on this website : http://www.emartee.com/product/41894/Wrobot-4-Channel-Relay-Shield
I can't make sure of the difference of them ,and which one is what I need to purchase for my Arduino....... in trouble......... :-/ :-/ :-/ :-/......who guys can help me out from the relay...............
I can't make sure of the difference of them ,and which one is what I need to purchase for my Arduino....... in trouble......... ......who guys can help me out from the relay...............
Maybe you could contact the folks at emartee.com who sell the parts via their Contact Us
You are mixing up the current the contacts can switch with the current through the coil it takes to make the contacts close.
You don't put 15A through any PCB, you use wires to connect to your load which can be up to 15A.
Is it advisable to use the lowest power possible to turn on the coil, or its a bit indifferent ? (only matters in terms of power consumption ?)
No, its rated for 5V, use 5V if you can. If you skimp on the drive current it may not work very quickly, you will have to derate the current it can handle. Worse case it will be unreliable and the contacts will fuse together because there's not enough force to make good enough contact.
The output ratings will definitely assume standard drive conditions.
