The purpose of a voltage regulator is to regulate voltage. In layman's terms, it outputs a fixed voltage as long as the minimum input voltage requirements are met. The input voltage may change drastically, but as long as it's above 7V, the output should be around 5V for a 7805.
1. Ensure that the correct terminals of your 8.4V batt are connected to the 7805.
2. Ensure that the batt is giving out the correct voltage.
3. Ensure that your 5V and ground are not shorted.
4. Change a 7805 if all the above are satisfied.

Anyway, go to http://www.national.com/pf/LM/LM1086.html and download the datasheet for the LM1086. Ignore all the crap but the diagram on very 1st page. You can see that the pinouts are different from a standard 7805. Modify your circuit and it should work.
Go ahead and dip your feet into some beginner electronics. I believe that's what got all of us started.

 

Ah, right, I think the different wiring set-up is where I went so massively wrong- didn't even realise the top of a voltage regulator could be one of the inputs/outputs! In terms of my previous circuit diagram would I therefore connect the following bits to the following pins:

Vout= This is where I take the +'ve regulated voltage from
INPUT=This is where I put the +'ve unregulated voltage IN
'OUTPUT' and 'ADJ/GND' = Not sure what to do with either of these- is GND short for 'Ground'? If so is this where I connect both the input and output ground terminals to? sorry, I'm all confused because what do I then do with the OUTPUT terminal? It may be explained on that sheet you suggested, but I feel that I may not be able to understand the way it's presented and some of the terms used

 

i'm a bit supprised that you have an output terminal... I shall have a look at the data sheet when i have a free moment

'ADJ/GND' GND = Ground (as you suggested), ADJ = Adjust.

Normally, with a fixed regulator (as you had originally), the 'gnd' pin is connected to ground. Ground (in this instance) is the 0v (or negative) connection on the battery. This should also be connected to the 0v rail in the circuit you are powering.

However, with an adjustable regulator, this pin is connected to a pot (usually), which is used to adjust the output voltage - hence the pin is renamed 'Adjust'

Any help?

Tim

 

Yeh, thanks Tim, I'm clearer on the ground terminal now and how it might also be called 'adjust'. You've also hit the nail on the head with the confusion between the 'Vout' pin and the 'OUTPUT' pin- I'm not sure which one to take the regulated +'ve voltage from, although it sound like the 'Vout' but in that case what does the 'OUTPUT' pin do?? Any help on this distinction would be most appreciated.

Also, would it still be suitable to use 10uF capacitors with a 3.3v regulator as I plan to do, or would another capacitor value be more appropriate?

I thought I had it all sorted and then this 3.3v regulator seems to be set up all different...nothing's simple! :roll:

 

it if was simple it woudn't be so much fun... :lol:

10uF should be fine still

err, had a look at the data sheet, and it says that 'Vout' is the tab, and the 'output' is the middle pin. Someone may say that this is wrong, but personnally, I'd be tempted to try it using the 'output' pin as the regulated power, and leave the 'Vout' tab unconected. If it doesn't work, I was wrong!! Then if you heatsink it, make sure it is isolated from any other regulators.

I am assuming that your regulator is a T0220 (looks a bit like a flat transistor)? Because in my experince, having the tab as one of the conenctions on this style of case is quite rare. (You see it all the time on TO3 style cases, but not so much on T0220)

 

I guess you're right, but at the moment I just want an incredibly simple solution!

Ok, I'll give that a go Tim with the middle Pin as suggested, I'll try anything, and that may be one combo I didn't attempt in my frustration the other day..

not sure what you mean by 'any other regulators' - do you mean the other connections on the regulator?

You're right, the 3.3 v regulator I have does look like a flat transistor

 

lol, sorry I was just confusing things. What I meant was that if you built more than one regulating circuit (dunno why you would, but anyway), then you have another regulator, in this second circuit. And there is every chance that you would try and heat sink them on the same heatsink. Normally, this isn't too much of a problem, as the tab is connected to ground, however, if it is connected to 'Vout' (as it is on yours), then the heatsink may connect the tabs on both regulators together, which would cause some problems - this is why you use a lil' plastic thing, to stop this happenening.

don't worry about it, i was just trying to cover all angles.

 

Phew...

One things that got me thinking that things might be a bit different with the 3.3v regulator I bought was that in the name of it was "-3.3"
as in LM1086CT-3·3

Does the -3.3 mean anything crucial, like it will output a negative voltage or something equally weird?

Also, as a very general point, are heat sinks hard to fit because with the voltage drop from 8.4 to 3.3 volts my bad boy's bound to get a little hot under the collar...how much difference will a heat sink make?

 

http://www.national.com/pf/LM/LM317.html

There's the datasheet for the LM317. You set it up using the ADJ pin.

 

Good call Tim, you were right- tried it last night and the output pin does appear to give the regulated voltage output, so thanks a lot. Supposedly then if I want to attatch a heat sink I can attatch it via the Vout tab at the top which should conduct the heat into the heat sink?

Thanks also checkmate- that information sheet saved the day! Much appreciated

 

pretty much... although I wouldn't say 'via' the tab, as connect it TO the tab - you need a reasonable contact between IC and heatsink

 

OK, cheers, I presume I just bolt the heat sink on with a small bolt.

Minor question, was does IC stand for Tim?

 

IC = Integrated Circuit - usually used to refer to 'chips', but technicaly a regulator is an IC aswell

yeah, I think you need an 'M3' sized bolt (although as long as it fits, I don't see what difference it makes...) Personnaly, I use an insulator (something like this http://www.maplin.co.uk/products/mod...889&Products=2) so that the regulator is not connected (in an electrical way) to the heat sink, so if I use the same heat sink for other devices, they won't interfere with each other, but as you only have one IC, its entirely up to you.

So (to summarise) in this instance just bolt the heat sink on with a small bolt.

Tim

 

Pallen33,

If you want something real simple........

A doide drops voltage by 0.6 volts so if you stuck 7 of them end to end you would drop (7 x 0.6) 4.2V giving you 4.8v supply.

Join the doides so all the stripes face the same way connect one end to the battery and the other to what you need to power. If it doesn't work swap it around as it will only work one way.

Doides come in differrent current ratings
IN4001 is 1 amp
IN4004 is 4 amp etc

Just make sure you buy ones rated higher than what you need, it is ok to use 30 amp doides to run 1 amp stuff but not the other way round.

 

How about a zener diode.

If the circuit does not draw more current ( few mili amps)
use a resistor to drop 9V by few volts and use zener (5V zener ) to produce 5V constant.

Please correct me if I am wrong.