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Is this a valid circuit for high current 12 to 5 volt regulator? I need approx 10 amps. The TIP's will have a massive heat sink.

Schematic

The source is a car battery mounted on this huge R2D2 thingy robot. TheraBot

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it wouldn't be valid in my book but what does valid mean to you? –  Andy aka May 10 at 12:09
    
As in, what's wrong with it? –  Pål Thingbø May 10 at 12:18
    
Thermal runaway, that is wrong with it. You wrongly assume the transistors are equal, but in practice they aren't. –  jippie May 10 at 12:21
    
TIP35C is 125W 25A transistor. Dropping 7V at 10A is just 70W. I think one should be enough. –  Kamil May 10 at 12:32
    
If I use 3 TIP's, with emitter resistance added, it'll be easier to cool the system? (as each TIP will yield approx 20W) –  Pål Thingbø May 10 at 12:48

4 Answers 4

That is not the right topology for using transistors to increase the current of a linear regulator. Here is how it's done using a single transistor to provide more current:

This still keeps the output voltage well regulated. In your circuit, the B-E drop of the transistors will make the output voltage lower.

At low currents, there is little voltage accross R1, so Q1 stays off. When the load current increases, the voltage accross R1 increases, which turns on Q1, which dumps more current onto the output. The regulator is still regulating, but the current thru it will stop increasing at around 3/4 Amp in this case, after which the transistor takes over most of the additional load.

One big power transistor with a big heat sink should be able to handle your 10 A output current. However, if you want to spread the heat accross multiple transistors, you can't just add more of them in parallel. The way to add more transistors is to give each its own emitter resistor. This provides a little negative feedback so that if a transistor is passing more than its share of the current, the voltage accross its emitter resistor will be higher, which will take away from its B-E voltage, which will decrease the current thru the resistor.

Here is a example with 3 external transistors that take most of the current load, while the regular is providing the regulation:

This is basically the same idea as before, but each transistor has its own emitter resistor. R1 is also increased a little bit to make sure there is plenty of base drive available for all three transistors, and to account for additional voltage drop accross the emitter resistors. Still, R1 is larger than it needs to be in this example. However, you have plenty of headroom voltage available, so dropping a little more in a resistor is no problem.

Keep the dissipation of the resistors in mind. Let's say to account for a little imbalance and some margin, we want each of the transistor to be able to handle 4 A. That is 400 mV accross the emitter resistor, plus 750 mV or so for the B-E drop, for a total of 1.15 V that needs to be accross R1 at full current. That means it will dissipate 660 mW, so it needs to be at least as "1 W" resistor.

Each emitter resistor must be able to safely dissipate (4 A)2(100 mΩ) = 1.6 W. These should be at least "2 W" resistors.

All this said, I agree with Wouter in that this is the wrong way to address your overall problem. Linearly regulating down 12 V to make 5 V will be more trouble and a lot more wasteful than a switcher. However, the real way to address this is to step back a few levels and re-think at the system level. Running lots of high current stuff at 5 V from a 12 V battery makes little sense. You should be able to find motors that run at 12 V, actually more easily than ones that run at 5 V at this power level. You then only need to provide 5 V for the control logic, which controls switches that enable power to the 12 V devices. Or you can still use 5 V devices with a proper PWM drive so that you are switching the 12 V on and off fast enough so that the devices only see the average of 5 V.

There should be several good options at the system level, none of which include wasting 70 W as heat to run 5 V motors from 12 V.

I described how to make a higher current linear regulator from a existing one and some external transistor to document how to do it right, but this should not really be part of your overall solution.

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Thanks a lot for this. It really helps. –  Pål Thingbø May 10 at 14:10
    
Note that R1 can be much larger as IC1 gets what current it needs from bases and R1 is essentially a turnoff resistor for transistors. (I know that that will be obvious to you once noted - just not necessarily noticed along the way). –  Russell McMahon May 11 at 8:28

A number of remarks, in approximate order of importance:

  • why do you need 10A at 5V? If you want to get a warm feeling, light a candle!
  • if you really need 10A at 5V, why create it from 12V (which now must supply that same 10A)? (Take a PC PSU!)
  • if you really really want to make 5V/10A from 12V, why not build a switching power supply? It will probably cost less than the massive heatsink you need now. (I am all in favor of linears for small currents, but this is ridiculous.)
  • If you really really really want to male a 10A linear 12V->5V regulator, don't use this circuit. Thermal runaway is one problem. It has no current limiting. And what do you think the output voltage will be? (check the Vbe of a TIP35 at a few A's). You tried to compensate with that diode, but I don't think it will be enough. Or stable.

If you really ^ 4 want to build something like this: there are standard circuits for this that use a PNP power transistor, or multiple ones with load balancing resistors.

One think you got right is that it will be easier to cool the system with multiple transistors, because their Rth j-c ( 1 C/W each) are in parallel. For TIP35 (with 70 W and 140C temperature difference) you would need a total Rth of 2C/W, hence a heatsink of 1C/W. With 3 in parallel you would need a heatsink of 1.6C/W. Still a big one, but not as big a s a 1C/W. (Note that in practice 140C might be too high, so you'll need a 1C/W anyway).

================================================

With the added info:

  • That will be a pretty 'hot' robot ;)
  • I'd use an 7805 (or more than one) for the intelligence, or a switcher if it uses too much current (what are you carrying, a blade server?)
  • MOST IMPORTANT: for the power stuff, try to get 12V versions, or use PWM.
  • except for the for steppers, for those use the 12V directly and use constant current drives (or a PWM equivalent). This will give you better torque.
  • At yours estimated electronics-knowledge-level I recommend buying a DC-DC module rather than building one (I certainly would not attempted to design and build one)
  • another option: use a 6V battery for the 5V power stuff.
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I need this on a robot with lots of 5V gear; servos, steppers, microprocessors, video, etc, etc. The 12V source is a car battery. I have this circuit working with now at 2A, but the TIP gets very hot, even with a huge heat sink. I'm gonna check out a switching supply. Any pointers? –  Pål Thingbø May 10 at 12:58
    
So could I go for something like this: linear.com/product/LTM4641 –  Pål Thingbø May 10 at 13:09
    
@PålThingbø it's certainly worth having a different power supply for your microprocessors to your actuators, it may even be worth having say 5 2A supplies for your actuators rather than one 10A if none of them take more than 2A. –  Pete Kirkham May 10 at 13:46
    
The 5V things are microprocessors (five of them each drawing approx 500mA), servos (don't know how many yet), about 30 different sensors (ultrasonic, PIR, IR, etc), a HD camera, a Fourier-transform camera (detecting color changes) and some minor other devices. The problem is space, I can't have more things on my robot. –  Pål Thingbø May 10 at 14:09
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@PålThingbø "The problem is space, I can't have more things on my robot." A switching power supply is smaller than the heat sink required. –  jippie May 10 at 22:16

Thermal runaway, that is wrong with it. You wrongly assume the transistors are equal, but in practice they aren't.

The transistor that carries slightly more current will warm up slightly more than the others, resulting in further increasing its current and warming up further. One transistor will end up taking the majority of the load.

To solve this, you can add small emitter resistors which will cause feedback and equalize currents across the branches.

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How small? 1 ohm? –  Pål Thingbø May 10 at 12:45
    
Do some math. At 3A, how much (extra) voltage will those 1 Ohm resistors drop? –  Wouter van Ooijen May 10 at 13:13
    
I agree Olin's circuit is even a better option than trying to fix the original with extra resistors. –  jippie May 10 at 22:17

I prefer a switcher to a linear regulator running so hot you can't touch it, but I couldn't find any buck regulators in a through-hole package with the specs you need (12v to 5v @ 10A). Everything available appears to be surface mount, in packages that are definitely unfriendly to work with (pins hidden on the bottom, QFN and such).

I don't know what your budget is, but I did find this 12v to 5v DC-DC converter which will do 10A. (Input can actually range from 10v to 14v.)

enter image description here

Costs less than $15 at Digi-Key, much better than the earlier one I found ($65).

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The link is not working. –  Pål Thingbø May 10 at 13:42
    
@PålThingbø fixed –  tcrosley May 10 at 13:44
    
Thanks. I went for this: digikey.com/product-search/…. Same price, more options. –  Pål Thingbø May 10 at 13:49
    
The board you found is actually an evaluation board, which is intended for engineers to use to evaluate a particular IC before incorporating it into a product, which is why it has so many options. Either of these (the product I recommended or this one) get around the problem of having to deal with a surface mount switcher IC directly. –  tcrosley May 10 at 14:14
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@PålThingbø -- I found another board that does the same thing for under $15, and revised my answer. Smoking deal. No, actually not, that's what we're trying to avoid. :) –  tcrosley May 10 at 15:08

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