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.
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.
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.
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.
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.
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.
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.
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.
