# Can simple DC/DC linear regulators be used in parallel?

Can linear regulators such as the L7805CV (Rated at 5V output, 1A max) be used in parallel? I need to create 5V from 12V, but I need more than 1A, so a single regulator won't do. Can they just be wired up in parallel to achieve a higher maximum current?

simulate this circuit – Schematic created using CircuitLab

I know there are more efficient alternatives than linear regulators, but this shall be a cheap backup solution if the primary DC/DC converter fails.

• Back in the day I had success with a 7805 with the output connected to the base of a 2N3055 and a diode between the GND pin and ground itself. The regulation won’t be quite as good but maybe adequate for your application
– Frog
Apr 24, 2022 at 19:49
• There are much simpler ways to boost the output current of a linear regulator. Datasheets usually show at least one configuration; you just need to add a power BJT and a resistor or two. Apr 24, 2022 at 19:52
• The simple answer is Yes. But... it is not considered an elegant way of doing such. 7805 regulator is a "Source Only". Putting 3 in parallel will increase the output current but only one unit will supply its maximum current while the two other will be OFF, until more current is needed, then the second unit will start turning ON, and so on. The best solution is from devnull ( below). Cost wise and complexity wise adding one single PNP power transistor is much better. Apr 24, 2022 at 23:04
• Just adding a related post. And by no means an endorsement, I did find this site discussing a variety of ways that come to mind. It's useful in showing what people may come up with.
– jonk
Apr 25, 2022 at 0:54
• Just use a more powerful regulator? From memory (nearly 40 years ago) the higher power (5A) version of the 7805 was the 78H05. I am sure there are better ones around these days. Apr 25, 2022 at 6:17

Can simple DC/DC linear regulators be used in parallel?

No, they cannot (at least not directly, as detailed in the question). To increase the maximum current an external transistor can be used, as shown in the datasheets of different manufacturers (which recommend the same solution):

ST:

TI:

Trying to improve, as requested in the comments:

• Note 1: disregard the actual values as they depend on models and resistor values (the purpose is just to show the operation of Q1 an Q2).

• Note 2: this is not what will happen in a real circuit as the output current increases with time because the simulation doesn't take the IC thermal protection into account (this is merely an operating point analysis).

To be clear without requiring the reading of all the comments: this is not a recommended solution for many situations. It is very unlikely that this circuit (at least the power transistor) would survive a prolonged short-circuit.

• I now decided that I'd rather go with a professional DC/DC converter, costs a bit more, but will be reliable and allows to select the precise output voltage. In the long term, I will anyway need to redesign the casing to make more room as the project grows.
– PMF
Apr 25, 2022 at 18:43
• @PMF Good to know. It is a reasonable decision given the problem description. Apr 25, 2022 at 18:47
• Could you add an explanation regarding how all this works, rather than just the diagrams? I can figure it out for myself, but answers should be reasonably accessible. Apr 26, 2022 at 8:31
• I have edited the answer to make clear that while this does solve the problem with the maximum current, it's a completely different solution than using the regulators in parallel. Please check if you agree. Apr 26, 2022 at 13:08
• @AI0867 I was not expecting this to be taken as a recommended solution, and the person asking the question arrived at the same conclusion after a good discussion in the comments. I hope the added simulation provides some clarifications regarding the action of the external transistors. Apr 26, 2022 at 15:16

This is an XY problem.

Real problem:

I'm powering a Raspberry Pi with quite some peripherals (TFT display, many sensors). It's got an intelligent HAT as power source that converts the 12V to 5V, but I fear it's going to fail soon. It doesn't properly switch on/off any more, has significant voltage fluctuations etc. I don't want to just replace that HAT because I lost confidence in the manufacturer (long story) and I can't find something similar on the market. So I need something that can replace the missing output. But apparently this here was a bad idea.

Solution:

Buy some switching regulator module for like $10 or less. 5V or adjustable voltage. Either solder it to a micro-USB cord to power the Pi, or connect it to the 5V/ground pins (which is presumably where the hat inputs the power). If it's adjustable, then set the voltage to 5V before connecting the Pi, and maybe put tape over the adjust knob to prevent accidental changes. Glue or screw the module wherever it fits. Probably in the same place you were planning to attach your breadboard with the linear regulators. • by the way, I have done this myself, with no problems other than a loose USB connection Apr 26, 2022 at 8:12 No they cannot. Due to device tolerances their output will not be the same and they will fight each other. You can small so-called "current-balancing" resistors in series with each of their outputs which will take up the differences in voltage drop so they will not fight each other, however. Smaller resistor values result in improved efficiency but less capability to absorb the "slack" due to different output voltages. I know there are more efficient alternatives than linear regulators, but this shall be a cheap backup solution if the primary DC/DC converter fails. What is the application? This does not sound like the proper approach. What are the consequences of the switching converter failing? And why are you expecting the DC converter to fail to begin with? They are quite reliable. • I'm powering a Raspberry Pi with quite some peripherals (TFT display, many sensors). It's got an intelligent HAT as power source that converts the 12V to 5V, but I fear it's going to fail soon. It doesn't properly switch on/off any more, has significant voltage fluctuations etc. I don't want to just replace that HAT because I lost confidence in the manufacturer (long story) and I can't find something similar on the market. So I need something that can replace the missing output. But apparently this here was a bad idea. – PMF Apr 24, 2022 at 20:12 • @PMF Why not a switched mode power supply? A linear regulator from 12V to 5V @ 3A will dissipate 21W. Apr 24, 2022 at 20:14 • @devnull Thanks, I might try that. I asked this question because the question bugged me for a long time. It looked like a very cheap solution. – PMF Apr 24, 2022 at 20:31 • Almost all linear regulators only source current but do not sink current. Therefore they won't fight each other. But you are correct that the current will not be spread evenly. – jpa Apr 25, 2022 at 10:26 • A historical example of a device that worked with this limitation is the BBC Micro. Early board revisions were designed to be operated by a linear power supply, which needed three 7805 linear regulators to provide enough current. As a result, the mainboard had three isolated 5V regions with different chips powered by different regulators. Later board revisions (after the Beeb was shipped with switching power supplies) merged these regions for a single global 5V supply, but the power supplies still provided three (identical) sets of 5V/GND terminals for connecting to the board. – Kaz Apr 26, 2022 at 15:11 There is no problem tying together in parallel all linear regulators because all linear regulators are pull-ups only. This means the regulator with the highest voltage supplies more current, but not necessarily all current. If you have known Voltage tolerances and output impedance and thermal protection, they tend to self-regulate equally, if on the same heatsink. Consider a Load Regulation error of 1% at 1A and a 1% voltage error at no load. This means for Vout say of 10V the Rout=1% of 10V/1A= 0.1 Ohm then for a tandem 2A load = 5 ohms with 2 regulators with a 2% voltage difference @ 10V / 2A, what will the current of each regulator be? (Depends on ratio of Rout * current unshared = Vout offset error) It is wise to choose an adequate margin to allow for temp rise with rated heatsink and current sharing depending on voltage tolerance errors. Derate power and use an adequate heatsink, reduce demand power by at least 1/3rd as every 10'C rise doubles the failure rate. There are some transient situations which can be improved with filters. Above the regulator voltage error is simulated by a +1%, -1%= 2% resistor tolerance error, for the LM317. Your situation requires a more details spec of regulator and load impedance. • Your answer seems to contradict the one from @Justme. Are you saying that it will work if only the heat sink is large enough? – PMF Apr 24, 2022 at 20:34 • The regulation factors do include temperature sensitivity and current sharing depends on voltage error/ Rout Apr 24, 2022 at 21:13 • @PMF The real answer is, as usual, "it depends". The regulators must be current and/or temperature limited to keep the dominant one from quickly going up in smoke. But then, the question becomes "how reliable will this be with one regulator operating continuously at its limit?" Voltage regulators are not normally designed to do this sort of regulation long-term. Apr 25, 2022 at 14:41 No, unless the datasheet says yes for some reason, and it does not. Only the regulator with highest output voltage would try to push all the current which it can't do. And in your case, not even one regulator can provide 1A you assume. To get 5V 1A out, there must be 12V 1A going in to the regulator, so the regulator must drop 7V at 1A, so 7 watts must be dissipated by a single regulator, which it simply cannot do. It would overheat quickly and hopefully shut itself down when it reaches the thermal limit. • There is no logic in saying the highest voltage takes all the power as load reg. Error and ESR determine sharing with Vout error. It's an analog issue. Apr 24, 2022 at 20:57 Here's an EDN Design Idea which parallels standard 3-terminal regulators using diodes: https://www.edn.com/high-current-supply-uses-standard-three-terminal-regulator/ and below is the article from the 2004 Design Ideas archive. The article claims there is no limit to the number of regulators you can parallel in this way. Yes, some linear regulators could be connected in parallel, for example LM317 with bunch of other parts. On the other hand one LM317 can handle 1,5A with only two external resistors. TexasInstruments LM317 It won't be a cheap backup if you calculate how much of a heatsink you'll need for this. You'll be dissipating 21W in those regulators. You'll probably want a 2C/W heatsink at most. That's an easy$25 right there + machining to get the threaded holes to mount the regulators. Are you telling us that you can't get a second 12V->5V 3A converter in parallel as a backup for less than that? I don't believe it for a second.

One idea I've come up with is to have a current-equalizing "current mirror". It auto-selects the highest of a variety of currents to be the reference current. If the outputs of the mirror all drive the same current sink, then the currents will equally distribute among the branches. It's not a high-precision circuit, but it doesn't need to be: there should be plenty of derating applied to any high-current regulator circuit.

This works with any number of regulators, and it also splits the heat dissipation between the regulators and the pass transistors. It has the drawback of requiring about 0.5V-0.7V of additional dropout voltage.

The partitioning of dissipation of each regulator channel between the pass transistor and the regulator is uneven. For the regulator with lowest output voltage, the regulator has most of the dissipation vs. the pass transistor. For other channels, the pass transistors take most of the dissipation.

This circuit is fail-safe: if any of the regulators fails open, the output will largely shut-down.

I expect the transient performance on increasing load current to be worse than when the load current drops off, since all but one regulator are working at the drop-out threshold, and their regulation in that operating regime is not as great for rising loads.

simulate this circuit – Schematic created using CircuitLab

The Schottky diodes select the regulator with the lowest output voltage to be the current reference. R1's value is chosen to provide enough base drive for the lowest input voltage and highest output current. C1 provides a soft-start. The pass element transistor type can be chosen for low Vce at the desired operating current.

This circuit also largely doesn't require the voltage regulators. The regulators, if present, act as per-channel protection devices, with an inherent "wire-or" shutdown for the whole supply if any regulator's protection trips.

If the per-channel protection afforded by the regulators is deemed unnecessary, the regulators can be removed and replaced with a piece of wire from IN to OUT. Only one of them has to be actually a voltage regulator. The voltage regulator, if mounted among the pass transistors, will act as a temperature limiter for all of them, and due to the current mirror action, will also current-limit all the channels in parallel.

Since this then becomes a single-input mirror, the Schottky diodes and R1 can be removed. The reference channel - with the regulator - needs to have the usual collector-to-base connection, perhaps with a base current compensation resistor thrown in.