3
\$\begingroup\$

Could I use two L7812 voltage regulator ICs in parallel to get 12 volt and more current that with just one?

Schematic of what I want to do

\$\endgroup\$
  • \$\begingroup\$ Not very well. Is a higher current regulator completely unavailable to you? You'll get far better performance and protection behaviour from a single device rated for the current capability you need. \$\endgroup\$ – TonyM Mar 5 '18 at 22:37
  • \$\begingroup\$ You should NOT parallel them! They are general purpose voltage regulators with loose precision at best. One may put out 11.995 volts, the next one 12.105 volts. \$\endgroup\$ – Sparky256 Mar 5 '18 at 22:46
  • 4
    \$\begingroup\$ You really shouldn't accept a answer to fast. Now you'll never know what others might have said. It's best to wait about 24 hours to give everyone around the world a chance to chime in. \$\endgroup\$ – Olin Lathrop Mar 5 '18 at 22:55
  • \$\begingroup\$ See also 2 Voltage regulator wired together to produce higher amperage possible? \$\endgroup\$ – Greenonline Oct 13 '19 at 16:55
14
\$\begingroup\$

No, that is a bad idea. The two regulators won't have exactly the same output voltage. The one with the higher output voltage will take more current. You can't guarantee that both regulators will put out their maximum current when you try to draw 2x the output current.

Another drawback to your approach is that you have a diode drop between the regulated voltage and the output voltage.

Here is how to get more current from a single regulator:

At low currents, R1 only drops a little voltage, and the regulator works as intended. When the current gets to about 700 mA, the 700 mV that causes across R1 starts to turn on Q1. Q1 then shunts input current around the regulator.

In this case, the current thru the regulator is limited to about ¾ A. As more current is demanded, it goes thru Q1.

One drawback of this approach is that the overall regulator has about 750 mV higher dropout than just the bare regulator without the transistor around it.


However, if you're going to go thru all that trouble, you apparently need a decent amount of regulated current. A linear regulator will dissipate a lot in heat. Having to get rid of the heat is large and expensive.

You really should look at some buck switchers. You haven't said much about your application, but it sounds like a switcher would be more appropriate here.

Let's look at the power dissipation more closely. It seems your AC input is 18 V. I'll assume that means 18 V RMS sine. That means the peaks of the waveform is 25.5 V. That is going thru a full wave bridge, so there are two diode drops. You are apparently expecting a few amps, so lets say 750 mV per diode drop. That makes the peaks the caps get chared to 24.0 V.

You don't show the values of the caps, so we can't compute droop between the peaks. Just to pick something for sake of example, let's say the droop is 4 V. We can approximate the input waveform to the regulator as a sawtooth from 20 to 24 V, which averages to 22 V.

Let's say the output current is 1.5 A. I'll assume you wouldn't be asking to put multiple 7812 in parallel if you only needed 1 A.

So now we have 22 V in, and 12 V at 1.5 A out. Any linear regulator, whether a single chip or something more complicated will dissipate the current times the voltage drop as heat. In this case, that is 10 V times 1.5 A, which comes out to 15 W. That's quite a lot of heat to get rid of. You would probably end up with a heat sink the size of your fist at least.

Now compare that to a buck converter. Nowadays you can get buck switchers that are 90% efficient. Let's work thru the numbers assuming 85%. That is certainly attainable. The output power is (12 V)(1.5 A) = 18 W. The input power is therefore (18 W)/85% = 21.2 W. That means the switcher will dissipate (21.2 W)-(18 W) = 3.2 W. That's much more manageable.

Even better, that 3.2 W won't be dissipated by a single component. The switch will dissipate some, so will the inductor, and so will the diode, or the transistor working as a synchronous rectifier. Actually, if the switcher used synchronous rectification, it would likely be more than 85% efficient. But still, dissipating a watt or two here and there is a lot easier than dissipating 15 W.

Use a switcher.

| improve this answer | |
\$\endgroup\$
  • 2
    \$\begingroup\$ OKI-78SR-12 is pin-compatible with 7812 and will give you 1.5A at a claimed efficiency of 95%. But while it's technically correct that the heat dissipation isn't a single component, they're encapsulated together so it doesn't make much difference. It's not the cheapest switching solution but it's simpler even the the OP's original though of paralelling 7812s. As you say: use a switcher. \$\endgroup\$ – Chris H Mar 6 '18 at 9:52
  • \$\begingroup\$ @Chris: I was thinking of a more discrete design using a buck chip and external inductor. \$\endgroup\$ – Olin Lathrop Mar 6 '18 at 12:51
  • \$\begingroup\$ I thought you probably were, but I suspect the OP might not have your skill level, and (like me) would want a simple component \$\endgroup\$ – Chris H Mar 6 '18 at 13:06
5
\$\begingroup\$

7812 datasheet page 25, figures 12 and 13. You can use an additional pass transistor to increase the regulator's current capability. You cannot parallel them without risking unequal current sharing and oscillation. Make sure to cool the pass transistor.

| improve this answer | |
\$\endgroup\$
  • 2
    \$\begingroup\$ HINT: You may get more up votes if you paste the diagram in your answer, so we do not have to download and search for the data or schematic. \$\endgroup\$ – Sparky256 Mar 5 '18 at 22:50
  • \$\begingroup\$ @Sparky256 True, but I thought it might help the OP to make him actually look at the datasheet and its typical application circuits. \$\endgroup\$ – Jonathan S. Mar 5 '18 at 23:05
  • 1
    \$\begingroup\$ The current limit may not be shared but it will regulate to one being at limit while the other picking up the slack with push current only. ( no pull with NPN common collector drivers only) Due to PTC on voltage this is why they should be thermally coupled. (0.6'C) otherwise the equilibrium will shift between regulators. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 5 '18 at 23:07
0
\$\begingroup\$

These are effectively 3 stage Darlingtons PNP>NPN>NPN , so they only have pull-up current with active current limiting and thus 2V drop @1A.

This means you do not need diode or output isolators, but you do may reverse diode protection in case of abrupt input short circuit to prevent reverse voltage stress. (optional)

When one slightly mismatched regulator goes into current limiting the current shifts to the other regulator. They should both be heatsunk together to handle the (V_drop*I=Pd)... Pd * Rth ['C/W]= temp rise to 50'C max rise. suggested.

Since the loop gain in this shutdown is relatively low (<30dB), the proportional current sharing ought to be stable if each voltage with no load is within 2%. However worst case spec is ~4%. But if your input voltage is too high and current limited triggered by overtemp, then your input is too high.

You are best to define the real and reactive loads and input voltage vs current curve and dynamic loads to choose a better supply. But this may work in a pinch.

| improve this answer | |
\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.