Can I limit the Current in a Charging Circuit

[ozhummer]

Hi All,

I have a simple charging circuit for iPhone - basically I take a 12 v supply pass it through a 7805 with a couple of caps and presto my iPhone charges.

Problem is I am using a 7805 and it is rated at 1A - the thing gets really really hot ! - I didn't really want to heat sink it.

My question is - Is it possible to limit the output current somehow ? - This will mean the iPhone doesn't charge as fast, but that is of no concern... I just don't want things to get hot ?

I am also consifering using SM components with an LM1117 which is still rated at about 1A, but really small so the heat dissipated would be too much...

Any help would be appreciated..

 

[dknguyen]

The simplest thing to do with your existing parts is to add a small series resistor to the output of the linear regulator. This will dramaticaly slow down the charging though beyond actual current limiting as it will reduce current over the entire charge cycle whether it is excessive or not.

Lowering your input voltage closer to 5V +Vdropout will dramatically help the heat dissipation at it's source.

 

[ozhummer]

Thanks for this - If I were to use the current components - what value of current limiting resistor would you suggest ? I guess using ohms law R=V/I where v = 5 and I = 250mA would indicate a 20 ohm resistor ????

THe ideal solution would be to get another linear regulator IC that has a lower current limit and has foldback current limting. ALso reducing the input voltage closer to the output voltage will help.

But the simplest thing to do with your existing parts is to add a small series resistor to the output of the linear regulator. This will dramaticaly slow down the charging though beyond actual current limiting as it will reduce current over the entire charge cycle whether it is excessive or not.

 

[dknguyen]

I edited my post while you replied. Foldback isn't going to help you. And I've my own doubts now as to what I said, because Li batteries are pretty finicky to charge which means there's a charging circuit in the phone and its not just feeding whatever voltage straight to the battery. THis charging circuit probably needs 5V to operate so the only choice then would be heatsinking or lowering your input voltage so it is closer to 5V + Vdropout of the regulator.

The resistor is worth a try though since its so simple and straightforard. The equation is actually:

I = (5V - Vbattery) / R

because one end of your resistor is going to the battery, and not straight to ground and the voltage difference across the resistor is what decides the current.

So if we suppose 500mA current which is what most USB plugs can output at maximum anyways, that would make R = 4ohms if Vbattery is 3V (which is around when a Li-Ion battery runs empty which is worst case and draw the most current). Also, because that is worst case, you might even want to use a lower resistance since it's not going to be that hot the entire time its charging. Make sure the resistor can handle the power. 1W should work.

I really recommend you reduce your input voltage from 12V to 6~7V though. THat would solve your heat problem at its source. Because even the resistor isn't going to reduce the heat that much without seriously lowering your charge current.

P = (12V-5V)*500mA = 3.5W
THe 7805 probably heats up by 40C/W so that would be 140C rise. That's the real reason for your problem input voltage too high. SO either reduce the output current or reduce the input voltage.

 

[Ubergeek63]

Hi All,

I have a simple charging circuit for iPhone - basically I take a 12 v supply pass it through a 7805 with a couple of caps and presto my iPhone charges.

Problem is I am using a 7805 and it is rated at 1A - the thing gets really really hot ! - I didn't really want to heat sink it.

My question is - Is it possible to limit the output current somehow ? - This will mean the iPhone doesn't charge as fast, but that is of no concern... I just don't want things to get hot ?

I am also consifering using SM components with an LM1117 which is still rated at about 1A, but really small so the heat dissipated would be too much...

Any help would be appreciated..

it has been said to add a resistor to the output of the regulator. this in not recommended as it will completely disable the charge circuit. it would be better to put a resistor BEFORE the regulator or use a switcher. the total dissipation will not go down but the temperature of the resistor will be related to it's size.

 

[ozhummer]

I want to create a circuit that fits into an ipod connector - I have managed to do this with surface mounted components, so space is at a premium. I could at best a 1w resistor, but cannot change the input voltage 0- in reality it is automotive voltage so could be as high as 14.4v

 

[Nigel Goodwin]

Why not just buy one?, they are cheap and freely available - I recently bought one for my Nokia smartphone, only £2.49 with free postage.

Overwise you need to move to a switch-mode design, a linear regulator will always need to dissipate the same amount of heat - adding aresistor before the regulator just shares the heat out, it's still the same amount.

 

[MrAl]

Hello,


Yes i agree with this last post. If you want to limit the power dissipation in the 7805 put a resistor in series with the input, not the output. That will push some of the power out of the 7805 and into the resistor instead. The limit, however, should be set so it is still high enough to allow proper charging, which could mean a full 500ma. If i remember right, the 7805 needs at least 7v input (it could be 7.5v though) so that means we want to drop 5 volts (12-7) and at 500ma that means we need a 10 ohm resistor. The power in the resistor will be 2.5 watts so the 10 ohm resistor should really be a 5 watt device, 10 watt will stay cooler however, and anything greater than 10 watts all the better.
The 7805 will have about 2 volts across it, at 500ma that will be 1 watt, so the device may be able to take it without a heat sink but im sure a small heat sink would be a good thing to add. 1 square inch on top side means 1 square inch on bottom side so that's not too bad, maybe around 38 degree C temperature rise.

This can also be done with a string of 1N4001 or similar diodes, and the power will be nicely distributed among all the diodes.

To do this in a small space, we have to look at the total power loss required. For a 7v drop at 500ma that means we have to loose 3.5 watts, which is a fair amount of power in a small space. That's with a purely linear circuit, so this means we probably have to go with a switcher. A switching circuit could do this and loose only about 0.5 watt and that would stay much cooler even in a smaller space. This would be a simple buck regulator. Check out the National line of Simple Switchers, of which one comes with a set output of 5v so you dont even need any resistors.

 

[Ubergeek63]

I want to create a circuit that fits into an ipod connector - I have managed to do this with surface mounted components, so space is at a premium. I could at best a 1w resistor, but cannot change the input voltage 0- in reality it is automotive voltage so could be as high as 14.4v

ummm ... no ... https://www.electro-tech-online.com/threads/when-is-12v-so-much-more.110893/#post907504

 

[RCinFLA]

Adjustable step down switching voltage regulator

 

[CafeLogic]

Maybe your iPhone isn't the best place to begin experimenting with electronics? Just a thought.