Voltage regulation

**Joe McGivern** · 8th June 2008 09:48 PM

Voltage Regulation post.

The charging voltage for my PDA (hp iPac) is 5 volts DC. I've got an adaptor (with centre +ve) whereby I can connect it to a 230 volt ac charger or to a USB port on my laptop.

I connected a 12v dc supply from my car cigarette lighter to the input to an L4940V5 voltage regulator and fitted a smoothing capacitor across the output. On no load the voltage is stable and is 4.98 volts.

I connected the output to a suitable low volt plug (4mm OD and 1.7mm ID) hoping to charge the battery in my PDA from my car socket.

The regulator very quickly overheats when I connect the regulator to my PDA.

Can anyone advise me what the problem is likely to be?

I Tried to put a small explanatory sketch in with the post but the file size is too big.

**Hero999** · 8th June 2008 09:50 PM

How much current does the PDA take?

Did you use a heatsink?

**Joe McGivern** · 8th June 2008 10:36 PM

The current output from the 230 volt charger is 2.0 amps nominal.(made by Delta ).

From the data sheet on the regulator it should be good for up to 1.5 amps.

No. I dont have a heat sink on it

Don't know the exact current drawn. I thought that 1.5 amps would have been ample. Are there similar regulators on the market with greater current carrying capacity
.
Regards
J McG

**OutToLunch** · 8th June 2008 11:31 PM

A 7V differential while processing 1A is a hefty 7W being dissipated through the pass element. If the current being drawn really is that much, then a large heatsink will be needed or you need to think about a buck regulator

**audioguru** · 8th June 2008 11:32 PM

The heat is made by the voltage across the regulator (7V) times the current through it (1.5A?). Then you have 10.5W of heat to get rid of.

The datasheet shows that the small D2pak will have an internal temperature of 681 degrees C without a heatsink when it tries to dissipate 10.5W but its max allowed temperature is only 150 degrees C.
The regular TO-220 package can dissipate a max power of only 2.5W without a heatsink.

You are using a linear regulator that gets hot when it has such a high voltage across it. A switching regulator circuit is much more complicated but would be cooler.

**Joe McGivern** · 9th June 2008 10:42 PM

Thanks to all who answered.

The package of the regulator I am using is TO 220

I feel that even a large heat sink would be unlikely to dissipate the heat quickly enough.

The heat is generated so quickly that I am concerned damage may occur before I get a stable current measurement.

I am not familiar with the term "buck regulator" as used by "Out To Lunch" and would appreciate it if you could expand a little on that subject.

I would also like to consider a little further the "switching regulotor" as suggested by "AudioGuru".Point me i n that direction if that is possible.

Thanks Again
J McG

**audioguru** · 10th June 2008 12:18 AM

A buck regulator is a switching regulator. I have never used one.

The datasheet for your regulator shows that the TO-220 package can dissipate a max of 41.7W if it has a huge "perfect" heatsink. It can dissipate a max of about 20.8W if it has a pretty big 2.5 degrees C per Watt heatsink and thermal grease but no insulator.

**eblc1388** · 10th June 2008 12:35 AM

An all-in-one 5V switching regulator IC is not cheap or else you have to build one yourself from separate component like regulator, inductor and capacitors.

If you don't mind the noise, you can mount the existing regulator on a small heatsink along with a 12V mini cooling fan, like those used on PC graphic cards.

**Joe McGivern** · 11th June 2008 08:46 PM

I could probably buy a car charger but that would not be educational or of any recreational benefit.

I've gone off to do some reading on switching regulators an dc to dc converters etc.

As Homer said to Marge : Put the kettle on , this may take some time.
Thanks for now.
Catch you later
J McG.

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