ATX/PIC battery charger

Hello everyone,

This is my first post on this forum, and I'm not a big genius with electronics or hardware design.
I am looking for people who could give me advise on my design!

The software is based on microchip's AN1015, and the circuit has been 'extensively' tested, apart from the 'anti reverse polarity' relay.

I'd like to know if I am doing something horribly wrong and/or should make some smaller changes. Minding, when this project is 100%, I'd like to make several of these, so want to keep the components as cheap and the disign as simple as possible!
Also I have few questions:

1. Would I be able to draw 15 amps from the modified ATX(normally 12V/16A) regulated at 14.7V ?
2. I am assuming the wire-wound shunt resistors need a fly-back diode. What is the right diode for this(now using SF16 and/or for the relay?
3. Can I be A 100% sure a 12V relay will work with an empty battery of +/-11V ?
4. Would there be a better(cheap) solution to reduce the voltage on the fan M1, from 13.6/14.7 to +/-12V ?

PS: Do not mind the max. voltage of 15.6V stated in the circuit2.jpg, that was intended for a future 'battery equalizing' stage which is not even implemented into the software yet. For now ATX =max. 15.07V and the battery =max. 14.70V.

Thanx in advance.

Mark Sikkema

 

Why bother? The fan life may be minimally affected by the extra voltage. You could use a series resistor if you must.

 

Thanx for the quick reply!

Ok I think I stick with the double diodes bringing the voltage down by 1.4V.
As I wouldn't know how to calculate the resistor to use, not able to find a proper datasheet for this fan and will be using a different fan in any next produce anyway.

Next is the relay, I didn't expain my self well enough in the initial post. The relay needs to take voltages from +/-11V to 14.7V, is this ok for the RAS-12-15 or RAS-12-20M?

As I understand from the datasheet, it says:
Coil rating for 12V at 20 Celsius:
Pull in voltage (VDC) = 75% max.
Drop out voltage (VDC) = 10% min.
max. allowable voltage (VDC) = 130%

does this mean:
12V relay can take minimal 90% of 12V = 10.8V and maximal 130% of 12V = 15.6V at 20 Celcius.
On top of that, the rating for the M-type is even worse(pull in=65% - drop out=8% - max.=110%), which is the special automobile-type???

We will be using this in the winter mainly (-20 Celsius), maybe I should think about a 5V relay with a transistor circuit?
Or would there be a better way, instead of using a relay, to prevent reverse polarity?

thanx, Mark

 

Aai,

Freshly woke-up this Sunday morning, and I realize I forgot to connect an extra diode to the relay circuit (the relay is the only thing I didn't add yet to my developement board). The relay is used as a protection for reverse connecting the battery charger. I'm not sure what will happen when somebody does reverse connect it (do not really want to test it), but I am quit confident it's not to healthy for at least the capacitor in the ATX.

So I have to options:

1. connect a skotky diode to the 12v relay for a low voltage drop and/or maybe somekind of zenerdiode circuit to keep the voltage down too. I do not have these parts lying about, and has to wait for tomorrow.
2. use a 5v relay with a transistor circuit, which I have breadboarded here and seems to work.

I'm very bad for calculating values for the components especially when transistors get involved. Tried to read into it, but end up even more confused.

Could anybody at least give me some indication if it's fine what I'm doing or what resistor values I would need instead and why ?

Thanx, Mark

 

Ok, after some intensive googling I found some schematics which seem to use a MOSFET to deal with the reverse polarity.
I will have to study those schematics and will post it when ready.

 

12V*16A = 192 Watts. You should not exceed the rated wattage for the power supply; for long life you should really limit the maximum power out to 80% or less of the rated output, or ~153.6 Watts. So, for 14.7v out you should limit the maximum current to 10.45 Amperes.

Dale makes non-inductive power resistors.

Why would you be powering a relay coil from the battery? If you're determined to use a relay, you should only energize the coil from a line-powered supply.

A simple fixed resistor of sufficient wattage rating would do the trick. Look at the rated current of the fan, and then use Ohm's Law to determine the resistance required to allow approximately that amount of current to flow across the difference between the fan's rated voltage and the charger voltage.

 

@SgtWookie:
Thanx for the reply, I'm really a bit of a noob with this hardware stuff.
Anyway:

I understand that much, so you are saying I should really count the Watts considering the rectifier and copper windings within the ATX, instead of the Amps. Minding the ATX at a whole(+12V/16A +5V/25A) is rated 300 Watt. Admitting I've only done all my charging tests up to 10A for now. But your not saying that the available Amps go down because I increased the voltage?

I know I read about them, but they aren't available here. I would like to be able to deal with the components readily available.

The idea was, to get the relay to switch on only when a battery is connected the right way around, before the power-supply is even switch on. But as from now I am considering using a MOSFET for this purpose. Less wear-and-tear and maybe even cheaper.

Understood, I was doubting I needed to know the resistance of the coil in the fan to calculate the resistor needed(which I some how could not measure and didn't have a datasheet). From the Amps I will be able to calculate the resistor I suppose. I will have a look at it. I just wasn't sure of the limits of a PC-fan, as they seem to be quite sophisticated little controlled motors.

 

I came across this ATX battery charger schematic which uses a IRF3205 for the reverse polarity protection, as shown in the picture below.

But do I really need an optocoupler to drive a MOSFET?
Wouldn't I be able to turn on the MOSFET directly from the battery either with a resistor or so?
As the charger will not be supplying power till the PIC knows the battery is connected properly.

I can get a IRF3205PBF, which is an expensive MOSFET.
Would I be able to use IRFZ44NPBF for a 15A (max. 20A) load as I will never use a load of 80A?
Probably need to connected to a heat sink though.

the main differences are:
-------------- IRF3205PBF ---- IRFZ44NPBF
Cont. Drain curr.: ---- 80A ------------ 35A
Power dissipation: -- 200W ----------- 94W
On resistance : ---- 8.0mΩ -------- 17.5mΩ

regards, Mark

 

@SgtWookie

Looking at your reply again I realize this was exactly what you were saying, duh...
I'll stick to the 10Amps from now on.

I've been collecting info on mosfets and realize I shouldn't look so much at their rated amps, but more at the resistance. So if I would be using the IRF3205PBF with a Rds of 8.0mΩ, that would calculate to:
10.45A * 10.45A * 0.008Ω = 0.87 watt
I could connect it to the heat sink with in the ATX. As I'm not using the 3.3v and 5v output, there should be some space left on the internal heat sinks.

Am I on the right track here?