5v Voltage Regulator Help

[JoeBear99]

This is my first time putting together a small AC to DC board which converts 240V AC to 5V DC via a 9v 2a transformer, a bridge rectifier and an L4940V5 regulator with a fairly large heatsink, terminating in a USB socket. I've also added an LED in parallel immediately before the USB socket.

The original circuit diagram calls for a 7805 regulator however I opted for a low dropout one as it seemed to be more versatile would I have to use it somewhere else. The schematic I've followed is this one below:

The problem I'm having is the USB socket seems to only deliver a maximum of 300 mA (tested with a handset phone and an ammeter). This is somewhat short of the current I was looking to deliver through the USB (800 mA - 1A)

As I understand it, the voltage regulator here drops 4v (9 - 5), which at 300 ma draw, leaves 1.2w to be dissipated as heat. I can confirm the heatsink reaches temps of around 80 degrees c, but I can not tell if this heat is "throttling" the regulator and limiting the maximum current to 300mA. The transformer itself should certainly be capable of handling 2A at 9v AC.

I'm a novice at all this but does anyone have any advice as to what's happening and how I can effectively push closer to 1A through this USB socket?

 

[Arpan Sarkar]

Check the output current with other load like ,5v led or something.

 

[kellys_eye]

You can't 'push' current into a load - it will 'draw' only what it needs and no more. You can fit an additional resistor across the output to deliberately increase the draw but.....why?

 

[JoeBear99]

You can't 'push' current into a load - it will 'draw' only what it needs and no more. You can fit an additional resistor across the output to deliberately increase the draw but.....why?

I realize this, but if my handset can draw 1.5A when connected to a typical wall charger, shouldn't a 2A transformer also be able to deliver the same current to the device? What dictates that a maximum of 300 mA is being drawn in this scenario?

 

[kellys_eye]

if my handset can draw 1.5A when connected to a typical wall charger,

...it won't do this all the time though. The charging current will vary over the charge cycle.

If you want to test your PSU then deliberately load the output using a known value load - a 5Ω resistor will cause 1A to be drawn. The resistor will need to be suitably rated though (5W minimum).

 

[kellys_eye]

Note that the greater the load the more the regulator will dissipate and a heatsink will be required in most circumstances.

 

[Audioguru]

First of all, 9VAC has a peak of 12.7V and the bridge rectifier will drop about 2V when loaded so the input to the regulator will be 10.7V, not 9V. When the circuit is fixed the regulator will heat with 11.5W, overheat, then shutdown. The value of the smoothing capacitor is too low for 1.5A.
2nd, you forgot to read the datasheet for the regulator you are using. ALL low dropout regulators need a fairly high output capacitor (22uF for this one) to avoid oscillation.

 

[dorke]

It would be better for you to use a 6V transformer with that Low drop-out regulator.
That would minimize the power dissipation on the regulator to about 3.5W at 1.5A load.

 

[(*steve*)]

Also the 1A fuse, as shown, will blow when you draw in excess of about 700 mA. A 2A fuse (and by extension a 2A transformer) would give a maximum DC current of about 1.4A.

For use as a USB charger, I would get a switch mode DC-DC buck converter module from eBay. Get one rated for 3 to 5 A. With this magic you'll be able to draw more current with less wasted heat.

If you really want to use that regulator, follow @Audioguru's (and possibly @dorke's) advice and also use a larger heatsink. Something around 5 degrees C per watt would be best unless you use a lower voltage transformer.