Anyone good with circuits?

megakid

megakid

megakid

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Hey,

Im doing a tiny project here, basically I've got 12V DC source (my car - so huge current potential) and I'm trying to charge an iPod via a USB port.

Now, I have a lm7805 5v (1A max load) voltage regulator and have wired this up to the two pins inside the female USB port. Before the 5v regulator I added a 2A fuse for protection - I realise this is probably too big as 1) the USB 2.0 "High load" spec says 5v 500mA is the max a device will need and 2) the lm7805 can only take 1A @ 5V which is far less than 2A @ 12V inc any inefficiencies/drop off.

Anyway, It has now come to my attention that the iPod needs it's data pins grounded to "charge", so of my 4 USB pins, 3 are going to common ground and 1 is the 5V feed. I checked all these pins and connected my ipod up - and bingo, the charge icon popped up! perfect!

However, my joy was shortlived as when I went to wrap up the lm7805 and realised it was very hot indeed. Soon after it cooled and the 2A fuse I had fitted in the 12V feed had blown!

Now as you may have realised, I'm not exactly an expert and I've probably done many things wrong but can someone point me in the right direction, am I right in thinking that if the 2A fuse blew, the current heading into the iPod must have been mamoth, around the 4A mark (2A*12V = 24W, 4A*5V = 20W)? Im not too worried bout the iPod, it's still working but what did I do wrong, why would the iPod pull that much current? Does it just suck up as much as it can take? and therefore do I need a current regulator @ say, 600-700mA? Just a simple resistor prehaps? Or am I completely off track?

Any help greatly apprieciated!

James
 
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If you want to make something yourself, have a look at ST's L200 IC which will do voltage regulation down to around 3V and any necessary current limiting up to 2A but as suggested, it's cheaper and easier to buy a ready-made solution.
 
Car electrics are about 13.8V and you're dropping them to 5V with the LM7805. Assuming the Ipod is taking 0.5A the 7805 is handling a volt drop of 8.8v giving a power dissipation of 4.4W , that's a hot 7805 :)

I would agree with matblack , buy one , it's safer & could be cheaper in the end (and if it pops your Ipod you've got somebody else to blame ;))
 
Regulators do run hot, use a heatsink and all your worries will be over.

K.
 
1. Get a multimeter.

2. Measure the amount of current passing between the supply and the regulator.

Current is measured by placing the measuring device in series with the circuit to be tested.

3. Do the same for the supply from the regulator to the device (remembering that the internal resistance of the device being charged changes during charging).

4. Check the specs on the iPod or its charging device to see what its maximum allowable charging rate is.

5. Protect the device being charged with the appropriate fuse.

Compare value 3 with 4. If 3 > 4 you risk damaging the iPod and even if the damage isn't immediately apparent, you could be damaging the battery. It's the same as when you're charging a car battery, you can't just put a huge truck charger on a Mini's battery, it has to be able to charge it within the specifications of the battery. Battery damage in the form of warped plates and reduced life may result and there's even a risk of explosion.

6. Ignore all this and buy the proper device to do the job.
 
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Ive been looking at that L200 IC and that looks like it could do the job but might be a bit much for what I need?

I've seen another regulator IC - the L78M05 (here) and that looks like it has 0.5A current regulator built in. Im struggling to understand the data sheet. To those in the know, will this do the trick? (Ie limit current available to the USB port - this is what I want to do right?).
Im not too sure as it says in the datasheet "If adequate heat sinking is provided, they can deliver over 0.5A output current"

I have noted that I could do the job far tidier and just as cheap with that £7 (once u factor in postage) unit but I fancy doing it myself, improving my understanding of electronics and soldering skills :D
 
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Feek said:
Try what I suggested, just use another 7805 with a decent heatsink.

K.
Click to expand...

I agree with Feek. I used a 7812 voltage reg to power my 12 volt TV from the trucks 24 volt system without any problems. Linear voltage regs do produce a lot of heat. A large heatsink is essential but, you can reduce the power dissipated by the reg by putting a dropper resistor in between the vehicle supply source (12-14 volts) and the input of the regulator. You really need to know the amount of current that the ipod requires which I can almost certainly say is not 500mA.
 
iPod's only use a little over 500mA of 5V power - A USB 2.0 socket is not spec'd to output anymore than this...

This is also backed up by this.

Im not amazingly worried about the heat the regulator got to...
I'm more worried bout the fact that the the 12V 2A fuse blew...why would that be (presumably not just because the regulator got too hot?)?! I'm looking for an understanding more than hit&hope approuch.
If I fitted a 5A fuse, wouldn't that just allow even more current to flow to the iPod, hence screwing the battery over as phykell explained? Or does the iPod have some kind of built in current regulator to stop the battery from receiving too much charge?
 
Fuse would have popped because the regulator overheated, failed, shorted and therefore the current drain went high. The fuse did exactly what it was supposed to do.

K.
 
ok it's making a bit more sense now. I think im going to go for that 0.5A 5V regulator (with current limiter) to see how I get on.

Cheers for your help guys!
 
Now as you may have realised, I'm not exactly an expert and I've probably done many things wrong but can someone point me in the right direction, am I right in thinking that if the 2A fuse blew, the current heading into the iPod must have been mamoth, around the 4A mark (2A*12V = 24W, 4A*5V = 20W)?
Click to expand...

No, it doesn't work like a transformer, the current is the same on input and output, the excess power is disapated as heat from the regulator.

If I was drawing 1A @ 5v from a car battery through a 7805, then the input power would be 13.8v x 1a = 13.8W, the output power would be 5v x 1A = 5W, and the power disapated as heat would be 13.8v-5v = 8.8v ... 8.8v * 1A = 8.8W
 
You should just buy the pre made one, whilst it is commendable that you want to increase your knowledge and skills, making accesories for an iPod isn't the best place to start as it has a large potential for loss compared with other small projects.
 
I am actually building something similar to this at the moment, though it is not for charging a battery.

My mini-project is a little more involved but to all intents and purposes the same.

The thing is, I am not going to ruin an ipod testing it out! :p


Use a heatsink, use a regulator with current protection and use a fuse. Remember the wattages of the other components in the circuit too...
 
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