how to invoke more current from regulator?

ireg lm7805

hi,

i have LM7805 (+5dcV regulator) and ipaq4150(PDA) which charge its own battery through USB adapter.

USB port at PC has 4 pins: (1)DC+5V (2)data+ (3)data- (4)GND

i thought i can charge my pda in my car cigar jack (+12V) which i make 5V regulating circuit for my pda.

but when I connected pda to my circuit powered by +12V car generator, pda was not charged. The boot procedure started and turned off as soon as it finishes.

besides, the LM7805 got hot seriously soon after i connected pda, that even i can't touch it .

but, in the case of connecting to PC USB port power, it works fine.

As I know, PC USB power can invoke upto 600~700mA..? and LM7805 is upto 1A when heat sink attached.

on the back of the Pda, it reads that it needs upto 2A current.

() what if I make several LM7805s to be connected parallelly on the circuit? Can this supply more current without overheat?

thanks in advance~

read data sheet for 7805 there is application example where external transistor is used to increase output current .

From otehr side it could be case that when connected to USB it wont use USB source to charge internal battery .

niceguy77 said:

As I know, PC USB power can invoke upto 600~700mA..?

Click to expand...

Take care: the maximum intial current for a USB device is 100mA. This can be increased up to 500mA if the device shows this in its descriptor. But in this case an USB controller is necessary.


Mik

Attach a heatsink to the 7805. The 7805 can handle up to 1.5A since it doesn´t reach the temperature which turns it off (thermal protection). As you are with 12V to 14V of input voltage and the current is 0.5A, the power that generates heat is 0.5 * (12 - 5) = 3.5W, too much. A heatsink (alluminum plate) should be firmly attached to the regulator with bolt and silicon grease (thermal compound).

Added after 1 minutes:

Attach a heatsink to the 7805. The 7805 can handle up to 1.5A since it doesn´t reach the temperature which turns it off (thermal protection). As you are with 12V to 14V of input voltage and the current is 0.5A, the power that generates heat is 0.5 * (12 - 5) = 3.5W, too much. A heatsink (alluminum plate) should be firmly attached to the regulator with bolt and silicon grease (thermal compound).

artem said:

read data sheet for 7805 there is application example where external transistor is used to increase output current .

Click to expand...

thanks! i found the circuit diagram.

but,

R1 = Vbeo1/(Ireo-Io1Bo1) and
Io = Ireg + Bo1(Ireg-Vbeo1/R1)

i don't what Io1, Bo1 exactly mean. I'm afraid i misunderstood βo1 as Bo1..

what do the above variables mean exactly?

thanks, in advance~!

Io1 is I(Q1) (collector current of Q1)
Bo1 is β(Q1)

In some of your symbols o1 should be replaced with Q1: Vbeo should be Vbe(Q1)...
Then the equasions become more clear..

R1 = Vbeo1/(Ireo-Io1Bo1) and
Io = Ireg + Bo1(Ireg-Vbeo1/R1)

Click to expand...

should be:

R1= Vbe(Q1)/{Ireg-I(Q1)*β(Q1)}
Iout = Io = Ireg + β(Q1) {Ireg-Vbe(Q1)/R1}

it's much better to use switching regulator than a linear one, for high efficiency.

Hi niceguy77
Your circuit works fine with simpler calculation formulae.
You have to deside when the Q1 will turn on, at what current level.
For a resistor of 1 Ohm (not 3 as indicated) and at a currrent of 600 to 700 mA there is a voltage drop across R1 of 0.6 to 0.7 Volts, enough to turn Q1 on.
Since Q1 will go on, the rest of the current should pass through Q1 and not the regulator. Actually with small increase in the regulator's current, Q1 will contact more. That make it to be the main current supplier to the load and the regulator remains responsible for voltage stabilization.
For R1 I personally suggest a value between 1 and 2.2 Ohms. The system will work fine with any value.

Consider the heatsinks needed by bot Q1 and Regulator IC.

Also, if you search the datasheets of the manufacturers you will find other regulators which are rated at higher current than the 7805 (e.g. LM323K & LM323T at 3A, KA78T05 and LT1085CT5 also at 3A)

As suggested by arbalez the best option is to use a switching reg. It produces less heat so smaller headsink is needed and it has higher efficiency.
Examples are: LM2576T50, LM2599T50, LM2676T50 etc. all rated at 3Amps.

If you visit the National Semiconductor's site at www.nsc.com you will find many usefull data and an online design and simulator program to assist you in such a design.

Hi.

I have just come across an article on how to boost current driving capability by paralleling few 7805 ICs together to increase the driving current. Please have a look at the following link. Hope it gives you some ideas. Cheers!

**broken link removed**