Can I use two 7805 ICs in parallel to get double current capacity?

Question

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?

Answer 1

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.

Answer 2

With two voltage regulators in parallel, one might want to naturally produce 4.99 volts whilst the other will want to produce maybe 5.01 volts. The "winning" regulator will be the one that produces 5.01 volts and the losing regulator will basically switch itself off in an attempt to lower the output voltage but, the output voltage won't lower because the 5.01 volt regulator has "won" and will provide all the current to the load until it overheats. Then the "cold" regulator will take over and then it will overheat and really it ends in a bit of a power struggle (no pun intended).

Short story is that you can't reliably or cleanly get twice the current from two paralleled voltage regulators that ostensibly produce the same output.

Here's a decent looking circuit that adds two transistors around a 7805 to give significantly more current and short circuit protection: -

Normally, as current approaches the limit for the 7805, the presence of the 6R8 resistor drops enough voltage for the MJ2955 PNP BJT to turn on and start supplying more output current. If that current reaches about 3 amps, the NPN BJT will shunt the 6R8 thus turning the PNP off.

Circuit taken from here and there appear to be several variants of this on the web such as this: -

Taken from here. Or just build a little 5A switching regulator like this but make sure your target application doesn't require a particularly low noise and low ripple voltage supply: -

Answer 3

If you need that kind of current, linear regulators are usually not the answer, as they will dissipate quite a lot of heat. A ready-made, integrated switcher will stay cool and use less space.

Here is a selection of switching converters for 5V, 3-5A output.

Another one

And another...

Answer 4

If you want a 7805 that can handle more current, use a STS LD1085V50, 5V, 3A

On Semi KA378R05TU
TI LM1085IT-5.0
Exar SPX29300T-L-5-0

Answer 5

Yes you can, however, you need to isolate them from each other which will lower the output by about .707 volts each, the voltage drop of the silicon blocking diode that you would need to install on the output of each one, before putting the output of the diodes in parallel. It is simpler to use a bypass transistor or even a higher output regulator. Just keep in mind that the filtered but unregulated input current to the regulator circuit must be greater in amperage than your desired output, in order to maintain regulation, if the input current drops below the set output current, there is no telling what can happen, from circuit damage, to oscillation of the output, which would be the equivalent of feeding 5v AC to your circuit being powered. And Yes I have seen this happen when this exact same thing was tried, in lab, when I was a student, and another student tried this exact same setup, feeding the regulator circuit from a 12 volt 1 amp regulated power supply. The input voltage was monitored on an Oscope and never changed more than a handful of millivolt downward, but the output of his circuit was a flat topped high frequency pulse in the 1000 Hz range

Answer 6

Paralleling those parts is a bit of a silly idea, but all it really takes is driving them with equal currents.

It's a tradeoff, though: when the currents are equal, the dissipation on the regulators won't be equal. The regulator with lowest output voltage will run "close to" the input voltage, while the other ones will run at a lower voltage, necessary only to drive the same current as the other regulator.

In the example below, one regulator is set for 4.75V, another for 5.25V. Those are the extremes of nominal voltage tolerance for 7805/LM340-5.0.

Q1 and Q2 need heatsinking and should be on the same heatsink as U1 and U2, since these devices will all share the overall dissipation of the regulator system. I.e. the heatsink has to be designed as-if a single 7805 was running at full current, and then Q1,Q2,U1,U2 have to be mounted to it. Q1 and Q2 ideally should be mounted "facing" each other on the opposite sides of the same "thin" section of the heatsink, so that they'd be reasonably thermally bonded, same goes for U1 and U2.

More devices can be paralleled this way too - it's a relatively cheap thing to do, if all you need is a brute-force solution.

If any of the regulators turn off and stop drawing input current, the others will shut down as well. This is the desired behavior when full output current capacity and thermal headroom should be retained: it's a fail-safe.

On the other hand, if the desire is to have multiple regulators as some sort of a fail-safe against fail-open state of a regulator, then this approach won't work. I consider it a fair tradeoff.

^{simulate this circuit – Schematic created using CircuitLab}

Taking this to the logical conclusion, note that you only need one voltage regulator, and a bunch of current sources to help it out:

^{simulate this circuit}

The current mirror Q4-Q5 monitors the bias current of the power current mirror Q1-Q3. For proper operation, it is necessary that a small current (a couple microamps) flows from Q1's collector through Q4 to Q1's base. When the current mirror has insufficient base bias, the monitored current is 0.

The bias monitor current is then subtracted from startup current I1 by the current mirror Q6-Q7. When the bias is insufficient, the Darlington pair Q8-Q9 has all of I1 flowing into the base, and begins to bias the power current mirror Q1-Q3's bases into conduction.

Once the power current mirror starts to conduct, the output voltage rises towards the regulator's setpoint. Concurrently, the current source D1-R1 provides additional operating current to the Darlington's base, ensuring that it can bias the power current mirror bases over the whole range of operating currents.

Once the bias monitor current, flowing through Q4, as replicated on Q6, is sufficient, Q6 starts diverting the Darlington base current, providing the negative feedback needed to close the power current mirror bias regulation loop.

Under normal range of loads, the bias monitor current is a couple of microamps, and the bases of the power current mirror transistors have enough current flowing through them to ensure the mirroring action.

This setup diverts the power current mirror's base currents from the output of the current mirror, and provides the ultimate in base current compensation: the current mirror output does not carry any of the bias current, only a small monitor bias of microamps.

Since there's no base load on the current mirror output, usable accuracy is retained over 2 orders of magnitude of load current, assuming somewhat matched power transistors. The minimum load to assure regulation is about 20mA.

The additional dropout introduced by the power current mirror is about 0.2V at full load of 1A per channel, and less at lighter loads.

The current mirror operates gracefully as the input voltage falls below what the 7805 needs to stay in regulation. Output drops out of regulation smoothly, following what a sole 7805 would do under 1/3rd of the load current.

The decoupling capacitor is absolutely necessary in front of the linear regulator. The current source, configured as shown, synthesizes high differential impedance in front of the regulator. It will get many modern regulators unstable if there isn't enough decoupling. And with decoupling, the current source won't be able to react to fastest load current transients.

Answer 7

So I assume that if I use both ICs in parallel I can increase the maximum current capacity. Would it work?

Yes, if you can find two "identical" ICs.

if you cannot find it, you can nearly double it by using two adjustable or lower output voltage versions of the IC, put a small power resistor in their serial, and then a divider to raise the output voltage to 5v.