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The classical current booster for LM78xx/LM317 series is the use of a BJT in this way:

LM78xx series current booster Source

I understand the principle of how it works. However, I have a doubt: can this work with any linear regulator? How this BJT interacts with the internal of the linear regulator?

In my particular application, I need to use the TPS7A45xx series (long story, let's consider it a design constraints). In the datasheet of TPS7A45xx there's this circuit to boost current:

TPS7A45xx series boost circuit Source

In this way the current is just doubled, while in the LM78xx case is almost 10x. I need to increase it close to 10x rather than 2x and, of course, the solution of parallelize 10 TPS7A45xx doesn't seem a good solution... So, can I use the BJT for TPS7A45xx?

(NOTE: I know that I'm dissipating enormous power but my application needs a linear regulator and, in particular, a TPS7A45xx. I'm aware of the thermal limits.)

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    \$\begingroup\$ Why must you use a TPS7A45xx device. I mean, if a much cheaper device is available why can't you use that or, if you are insistent on using it why not make the cheaper device do the action end of regulating and have the TPS7A45xx doing absolutely nothing? I mean, that seems to match your strange criteria. Plus then there's the problem of ruining the low-drop-out capability of the TPS7A45xx by adding a series resistor that drops maybe 0.7 volts to 0.9 volts hence, the LDO bit kind of gets ruined in the process. So, your question needs more information to get an answer off the ground. \$\endgroup\$ Commented Oct 30 at 17:24
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    \$\begingroup\$ So, this circuit needs to be used in a radiation environment and TPS7A45xx, while not a rad-hard IC, has radiation test data provided by CERN. That's in short the reason. So I cannot use another linear regulator because I don't have that radiation test data (unless I use rad-hard devices that cost 1000x more). In any case, we can leave this question more general: can the BJT be used with ANY linear regulator or it depends on the internals of the regulator? \$\endgroup\$ Commented Oct 30 at 17:59
  • \$\begingroup\$ @ocirocir "any" is a strong word. Any counterexample and the answer is simply "no". As always, it depends. And don't you need a rad-hard bypass transistor as well? Is there one big load or multiple separate smaller loads? \$\endgroup\$ Commented Oct 30 at 18:02
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    \$\begingroup\$ Ok, you're right, but my question is that I don't know how to evaluate it. How do I know if this works or not with the TPS7A45xx (or with regulator X)? What should I look into? Of course, there are other problems to consider (including finding a BJT with radiation data - that should be easier), but I'm stuck on the first problem. \$\endgroup\$ Commented Oct 30 at 18:13
  • \$\begingroup\$ It's less likely an.LDO regulator will work compared to a standard emitter or source follower regulator. I'm talking about random instabilities that may not show themselves under standard testing so, in answering your revised comment, I can only advise that the answer is no. \$\endgroup\$ Commented Oct 30 at 18:24

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I only just read this question and the comments. I understand that radiation data on the TPSA7A45xx series is a determining factor. I've also found at this link the following comment about making a Spice model:

We would possibly make one in the future. Issue is as TPS7A4501-SP is targeted for space customers, Pspice model for TPS7A4501-SP needs to be accurate - thus the reason for delay.

So it appears TI has definitely got you in their sights.

I'm not familiar with the TPSA7A45xx series and I've also not seen a schematic quite like the one they suggest. But from a glanced high level view I think the TPS7A4533 must be a fixed \$3.3 \:\text{V}\$ regulator and has its sense pin tied to the output to guarantee that voltage, plus a TPS7A4501 that allows the use of a voltage divider to set its output, plus an integrating opamp used to integrate the current (turned to a voltage via a small resistance) so that, on average, the two regulators share the load, equally. If I got that right, the whole idea of it makes sense and I kind of like it. More to their point, though, is that it maintains an LDO reality when looking for more current compliance. I think they did the right thing adding that schematic because it keeps the LDO reality while adding current compliance.

The problem gets way more complicated if you add more, though. A system of multiple integrators all working magically together to keep the currents though all the regulators balanced will be a monster of a project. It's non-trivial. Two is fine. But you immediately starting getting into trouble the moment you move off the reservation, so to speak.

Getting back to your question, though.

You do not specify your input source voltage. These regulators are called LDOs, with all that entails. Looking at the typical chart, these are LDO (PNP-style control for positive-side.)

The first schematic you have shown adds a PNP to an NPN-style voltage regulator. Using the same technique on these will add another \$V_{\small{BE}}\$ of voltage overhead, which in the of the TPSA7A45xx series is a significant add:

enter image description here

Are you willing to accept that?

I usually only consider an LDO in cases where it is actually needed. If my voltage overhead is high enough, I go for linear regulators of the NPN-style, not PNP-style. What's the point of an LDO if doesn't, in fact, matter as the input voltage is several volts higher?? It's kind of silly and it adds complications that may be avoided if you allow the regulator some overhead with which to do its magic.

So there are two possibilities:

  1. You need an LDO.
  2. You don't need an LDO.

Which is it?

If you need an LDO then the approach they give you is the right choice. You can't go gallivanting around and playing loose and easy with external bipolars, which will only add significant overhead voltage you can't afford being added.

If you do not need an LDO, but are using these LDOs simply because they qualify due to the CERN radiation data available on them and TI's focus on targeting space customers with them, then that's another thing altogether. In that case, then yes you can use external transistors.

(You will want to use large-feature-size transistors as they are more radiation resistant, almost by definition. Sadly, these take up space on a wafer, reducing the number of parts they get, increasing the cost of each, and therefore they aren't making nearly so many of them anymore. Only for high-current devices are they forced into somewhat larger feature sizes. I hate what this has done for me in creating low-noise, low-current front end designs. They've pretty much stopped making them and there's nothing today to replace what was once available to me. But in your case it may be just fine since higher power translates into larger feature size, by definition.)

But if you do that, I would strongly recommend that you add some added circuitry to foldback the current. To make that point, I will illustrate with a circuit:

enter image description here

This actively curtails the current if the load demands more than the design limit. You can see the foldback action in the curve shown.

You will need at least \$3\:\text{V}\$ of overhead. More will be better. Also, more parts to check for radiation data. Granted. But at least consider the idea.

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  • \$\begingroup\$ Thanks, by the way, I'm the second case, I do not need an LDO. However I have a question, you stated "you can use external transistors", but, why? How you know that internals of TPSA7A45xx will not cause issues? \$\endgroup\$ Commented Oct 31 at 11:07
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    \$\begingroup\$ @ocirocir I don't actually know. Because I've not spent the time you should spend working out those details. PNP-style LDO regulators tend to have more significant stability problems with respect to their loads. So added time must be spent there to make sure or to add some extra external components to deal with those issues. So I'm not telling you there aren't any. But ... probably not. The designs -- especially the specific IC you have selected where they are targeting cases where it is difficult to get a technician out there to make a repair -- will be pretty bullet-proof. \$\endgroup\$ Commented Oct 31 at 11:14
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    \$\begingroup\$ @ocirocir Work with the FAE (field applications engineer) to make sure you get all the details right and don't miss something important. But I think you will be fine. (And yes, I've worked together with a scientist specializing in space-based instrumentation design. He's retired now on the Oregon coast. But I learned a lot from him as we worked together. All kinds of crazy things I had no idea about before. But now I think about broken molecules, charged, and sticking to surfaces where they aren't wanted. And heat dissipation. And more. It's a very difficult environment. My heart goes out!) \$\endgroup\$ Commented Oct 31 at 11:20
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    \$\begingroup\$ @ocirocir I toyed with the idea of answering, but in the end periblesis' point about the PNP common-emitter output stage of low-dropout designs requires very careful frequency compensation to keep the things stable, and while my initial thought was "yes, you can use an outboard transistor with an LDO", I realised that I just don't know. It's fine with emitter-follower (non-LDO), because loop gain is constant, not so for an LDO with variable load, and who knows what would happen if you mess with that loop. \$\endgroup\$ Commented Oct 31 at 12:02
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    \$\begingroup\$ @ocirocir I also feel that you can't just triple or quadruple (or ×10) the design you presented, because the adjustable regulators won't be able to balance their own loads between them in the same way. That design is very clever, keeping feedback very small at high frequency (by integrating the current imbalance over time, and re-balancing very slowly), and I think this is for the same reason - don't mess with an already precarious loop at higher frequencies. I only write this as a comment, because I am not at all certain that I am correct. \$\endgroup\$ Commented Oct 31 at 12:07
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It's pretty much valid.

For 79xx you would instead use an NPN transistor because it is a negative regulator.

for 78LXX you would use a larger resistance, like 30 ohms, because the regulator has less current capability.

But yeah it will also work with AMS1117 etc.

How this BJT interacts with the internal of the linear regulator?

The BJT turns on when the regulator takes a large input current ( as V_be forms across the resistor) and assists the regulator by providing current at the output.

This requires some excess voltage it the input . eg instead of 2V you now need 2.9V excess voltage on the input.

While you now have more waste heat it is now on a two devices that can (and should) have separate heatsinks.

The classical current booster for LM78xx/LM317 series is the use of a BJT in this way:

LM78xx series current booster Source

I understand the principle of how it works. However, I have a doubt: can this work with any linear regulator? How this BJT interacts with the internal of the linear regulator?

In my particular application, I need to use the TPS7A45xx

Looks tricky. you have only 2.7V headroom assuming that "6V" is actually 6.0V and not something variable like batteries.

But if you want to go 10x on the TPSA7A45 circuit you can just duplicate the TPSA7A4501 and op-amp parts 8 more times.

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