I use a 7805 for a project where the circuit needs a higher current (~2.8 A) at 5 V. So I assume that if I use both ICs in parallel I can increase the maximum current capacity. Would it work?
Electronic – use two 7805 ICs in parallel to get double current capacity
7805linear-regulator
Related Solutions
Can I use 7805 as current regulator?
Yes, but it's a very inefficient solution. A regulator with only 1.2 or 1.3 V feedback voltage can be used as a current regulator with much less power lost in the sense resistor.
My input is 12-14 vdc (battery) and I require output 5v and 500mA or adjustable. ... Is there any thing that can control both voltage and current?
You can control the voltage or the current, but you can't control both --- the load will determine whichever one you don't control. See this old question.
Can zener sove the problem because it can limit the voltage.
A zener can be used as a shunt regulator, which will control the voltage applied to the load (but then the load itself will determine the current drawn). It can also be set up as a voltage source with a maximum current limit (but if the load draws more than the limit, the voltage will drop in response).
design requirement for emergency (disaster struck areas) is Low cost.
The price difference between a TI and a LT regulator will be much much much lower than the cost of delivering the parts to a disaster zone at short notice. I wouldn't worry about it.
Yes, you could add an entire extra linear voltage regulator and capacitors for a voltage rail you don't even need, just to spread out power dissipation.
Or you could just use a resistor.
You say your problem is reliability? Then let me tell you about the most reliable electrical component ever made: the resistor. They're the only component that can, albeit unreliably, continue to function even after catching fire. Heck, I've seen those beefy sandstone power resistors explode a little and still work, at least until they explode a lot.
But I digress.
Just put some resistors between the 12V buck-boost converter and your 7805. You can parallel several if you like to really spread out the dissipation and give yourself lots of head room. This will be the most reliable, simple, and inexpensive solution.
And, as a bonus, you will almost certainly get better ripple reduction than cascading an additional 780x (7809 in your case).
The 780x series regulators are fairly slow. They have ~60-70dBV of ripple rejection but only if that ripple is 120Hz. I don't know what frequency your dc/dc converter is operating at, but it is almost certainly 100kHz or more, and will have switching harmonics into the MHz. Those will shoot through a slow linear regulator like the 7809 like a bullet shoots through air.
Ok, it's not quite that bad, but I would be surprised if you got more than 40dBV of rejection at 100kHz, and if your buck-boost converter is faster, it could be 20dBV or even worse.
But do you know what you get if you have a resistor in series with a capacitor? A low pass filter! YAY! And if you placed a voltage dropping resistor before your LM7805, you very conveniently have just such a capacitor - the input capacitor for the linear regulator. Sure, the 7805 needs a nice low impedance input, but that's what the capacitor is for. And since you know your maximum load current, you its a trivial matter to size your voltage dropping resistor. Its really just a matter of how much heat you want to dissipate in the resistor(s) vs. in the linear regulator. The nice thing about, say, 1W resistors, is that they don't mind getting hot and will dump that heat with natural convection alone, so I would recommend favoring the resistors when it comes to shedding watts.
I'll just do a quick and dirty example. At 500mA load current, let's drop the voltage to exactly what you'd get with a 7809. To do that, we need a 6Ω resistor. If you want to be really safe, perhaps use 3 18Ω 1W resistors in parallel, then each one will only have to dissipate 500mW. They will drop the voltage to 9V before it even reaches the 7805. If you put a nice fat ceramic directly on the input of the 7805, perhaps a 47µF one or even a fancy 100µF one, in all its 0805 sized glory, you'll solve your problem in a very reliable way, AND have a beefy low pass filter that will give you better ripple rejection than an extra linear regulator ever could, especially at higher frequencies.
And if you really want to clean up the voltage even more, toss in a ferrite bead in series between the resistors and the input capacitor. Ferrite beads are wonderful little critters. Unlike inductors, which might reduce ripple but also radiate some of that energy out as EMI and almost certainly make the situation worse, ferrite beads take high frequency ripple and dissipate it as heat due to core losses. They're best thought of as frequency dependent resistors that only have resistance above certain frequencies. They will aid you a great deal when it comes to cleaning up the analog sections of circuits - now is the best time to start using them!
If you still really have your heart set on adding a 7809, then you can still solve your problem this way. Add a smaller series resistor between the 7809's output and the 7805's input. This will dissipate a bit of power and spread things out further, and will decouple the 7809's output capacitor from the 7805's input capacitor. Just use the values you would if the regulators were being used by themselves. The input capacitor is what provides the low impedance power to the voltage regulator, so it is perfectly fine to have some resistance in series with the input, as long as it is before the input capacitor. Never put it between the input capacitor and the regulators actual input, obviously.
Best Answer
As others have already said, paralleling multiple linear voltage regulators is a bad idea.
However here is a way to effectively increase the current capability of a single linear regulator:
At low currents, there is little voltage across R1. This keeps Q1 off, and things work as before. When the current builds up to around 700 mA, there will be enough voltage across R1 to start turning on Q1. This dumps some current onto the output. The regulator now needs to pass less current itself. Most additional current demand will be taken up by the transistor, not the regulator. The regulator still provides the regulation and acts as the voltage reference for the circuit to work.
The drawback of this is the extra voltage drop across R1. This might be 750 mV or so at full output current of the combined regulator circuit. If IC1 has a minimum input voltage of 7.5 V, then IN must now be at 8.3 V or so minimum.
A Better Way
Use a buck regulator already!
Consider the power dissipated by this circuit, even in the best case scenario. Let's say the input voltage is only 8.5 V. That means the total linear regulator drops 3.5 V. That times the 2.8 A output current is 9.8 W.
Getting rid of 10 W of heat is going to be more expensive and take more space than a buck switcher that makes 5 V from the input voltage directly.
Let's say the buck switcher is 90% efficient. It is putting out (2.8 A)(5 V) = 14 W. That means it requires 15.6 W as input, and will dissipate 1.6 W as heat. That can probably be handled just by good part choice and placement without explicit heat sinking or forced air cooling.