I am trying to build a arduino based solar information system.
as a low battery voltage disconnect I have connected the relay (5V) between battery and load with NC config. On low voltage the relay is energized and the relay goes into NO state. flyback diode is connected parallel the relay in reverse polarity.
The 5v supply to relay coil also supplies power to the other digital components like LCD, ACS current/voltage sensing modules and arduino.
Initially I planned using a IRF3205 mosfet for the low voltage disconnect but i skipped it due to the Rds-ON of the mosfet and fear of frequent burnout of the mosfet.
Thus I want to know that if i connect the circuit and allow for the relay to work, will it adversely affect the other digital components.
An amplifier (transistor) is required to drive a relay, see what's contained in a ready made relay module.
The right MOSFET is much more reliable than a mechanical relay with stress and wear. Take care to use a logic level FET (IRL..., not IRF...) for the required current. If electrical isolation is desired, insert an opto coupler. For a high side switch an opto coupler can do the level conversion at the same time.
+1 for a MOSFET if only because relays and batteries don't sound like the best combination, unless it's a big battery normally supplying lots of current in comparison to which the relay coil current is insignificant.
Nonetheless it makes sense to have the relay energised when the battery has sufficient voltage, not energise when voltage is low (which is when you don't have charge left to power the relay in the first place).
wvmarle:
Nonetheless it makes sense to have the relay energised when the battery has sufficient voltage, not energise when voltage is low (which is when you don't have charge left to power the relay in the first place).
However that does not prevent the battery being discharged to harmful levels trying to operate the relay.
You must design to prevent battery over-discharge.
Can you please post a copy of your circuit, in CAD or a picture of a hand drawn circuit in jpg, png?
Thanks.. Tom..
please find the schematic.
Also, the Relay is 5V and the battery is 12V lead acid.
secondly, as I understand with the discussion , whether I put the relay is active low or active high, in both cases there will be some voltage drop and that can damage the battery. So can anyone suggest any other way.
Also, if go for Mosfets, getting logic level mosfets is difficult to procure locally so can anyone help me with N-channel mosfets with low side switching. In my circuit schematic, ground is common for PV battery and load. Please suggest some suitable mosfets also.
I see you're using the battery voltage (how high?) to power your Nano through the built-in regulator, then use the Nano's 5V output to power everything else. That's not going to work, unless the battery voltage is at 6-7V normally your regulator will overheat.
wvmarle:
I see you're using the battery voltage (how high?) to power your Nano through the built-in regulator, then use the Nano's 5V output to power everything else. That's not going to work, unless the battery voltage is at 6-7V normally your regulator will overheat.
Hello
No i am not using arduino's regulator for powering other peripherals. I have included a separate 5V input for all the peripherals. Vin for arduino is 12V and that's for powering arduino and for LEDs (four) only, rest all peripherals ACS module/LCD/relay are on separate 5V input (via buck/boost converter).
4 LEDs, about 40 mA.
Nano itself, about 35 mA.
75 mA * 7V = 525 mW dissipation. That's seriously pushing the limits of that regulator, and if your battery voltage goes up (12V nominal easily goes to 13-14V) you have a problem.
Why don't you power your Arduino off the 5V you have already?
Other than that, the circuit should work, but I still don't understand why you try to power the relay off a near-empty battery, instead of powering it when the battery actually has enough charge in it to do so.
A MOSFET doesn't waste the current a relay does, and if wired properly will easily outlast any mechanical relay.
I second that, at a guess he is powering the nano from 12v then the 5v for everything else is getting it from the Nano.
Also having a quick look at the schematic I can’t see any smoothing/decoupling capacitors ?
To me the LED connected to the relay does not look right, your expecting the resistor to control the LED and the transistor for the relay.If it was me I would move it to the switched contacts of the relay or go with a mosfet
Steveiboy:
I second that, at a guess he is powering the nano from 12v then the 5v for everything else is getting it from the Nano.
Also having a quick look at the schematic I can’t see any smoothing/decoupling capacitors ?
To me the LED connected to the relay does not look right, your expecting the resistor to control the LED and the transistor for the relay.If it was me I would move it to the switched contacts of the relay or go with a mosfet
The decoupling capacitor is placed just after the 5V input (far top right)
Also I have attached the schematic for the Mosfet as relay.Please let me know if you think any changes are required.
wvmarle:
4 LEDs, about 40 mA.
Nano itself, about 35 mA.
75 mA * 7V = 525 mW dissipation. That's seriously pushing the limits of that regulator, and if your battery voltage goes up (12V nominal easily goes to 13-14V) you have a problem.
Why don't you power your Arduino off the 5V you have already?
Other than that, the circuit should work, but I still don't understand why you try to power the relay off a near-empty battery, instead of powering it when the battery actually has enough charge in it to do so.
A MOSFET doesn't waste the current a relay does, and if wired properly will easily outlast any mechanical relay.
I agree and i should have seen it before. will change the schematic to deliver power to arduino via 5V supply.
puneet1984:
Also I have attached the schematic for the Mosfet as relay.Please let me know if you think any changes are required.
A bit less white around it would've been great
Basically it looks OK but as pointed out in reply #10 that LED is indeed not going to light up this way. Use a separate current limiting resistor + LED connected to the pin.
What is the current that MOSFET is going to handle? It's limit is 19A, but with a 100 mΩ on resistance you will need a heat sink at much lower currents. Why did you pick this part? There must be p-MOSFETs out there with much lower on resistance.
Can you elaborate why you think that the LED will not light up?? in theory it should work.
Also if you can suggest few mosfets which you think can do the job.
As far as the current capacity, I am not sure about my total demand but my battery is 150Ah and my solar is 100W and I just want to initially try few connections, definitely not connecting to a inverter at present.
My plan is to use this with Joule thief type circuit (https://www.lynxwind.com/ljl.html) and use 12V as input to run 220V LED/CFL bulbs along with routers and raspberry pi and home automation system.
puneet1984:
Can you elaborate why you think that the LED will not light up?? in theory it should work.
Theory says it doesn't. VBE is normally much less than VF of even a red LED. No current is going to flow through that LED.
Also if you can suggest few mosfets which you think can do the job.
Use e.g. digikey's parametric search.
Check the highest current you want to draw from that battery pack. Best to add a fuse as well, to protect your MOSFET.
My plan is to use this with Joule thief type circuit (https://www.lynxwind.com/ljl.html) and use 12V as input to run 220V LED/CFL bulbs along with routers and raspberry pi and home automation system.
That's just silly. LED bulbs are readily available in 12V, and most routers and so also run on low voltage DC - better use 12V-5V and 12V-3.3V buck converters for those. Completely forget about 220V (and if you really need that, get a real inverter, much safer).
Also while the "joule thief" is an awesome demo circuit it's horribly inefficient and not for daily use, let alone for use with high currents/voltages.
