Voltage Regulator Design

Too late at night to think about this one, so I'll just point you to the HP AN90B DC Power Supply Handbook, you might find something in there that will point you in the right direction perhaps. If anything, it's great reading.

Dave :)

Doing some further research, I'm wondering if I might have better luck with a Linear voltage regulator for this type of application? Something like the LM138/LM338?

Have you measured the input source voltage under load, this must stay above 8V. The regulator cannot create power, it merely meters it, and if your input source is a wimp, the circuit fails.

Yes, I have measured the input source, which is a 14V NiMH battery. The input voltage is barely affected.

And the whole is breadboarded with 0.5mm^2 wires ? Apparently the feedpack input of the regulator has the input it should.

Rene

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Is your circuit similar to figure 2? Could you post to alt.binaries.schematics.electronic (or email me, or post somewhere on the web) a photo of your layout?

Yes, John, the schematic is exactly the same as Figure 2 on page 11. There isn't anything different on my schematic, but if you still need it, I can find a way to put it up.

The design right now is on a PCB with adequate trace widths (100mil).

I've seen these used with great success, so I doubt that it's the chip design. Here are some thoughts about troubleshooting, however:

1. Double check your layout -- are you _sure_ you have it right? You don't have it mirror-image from what it should be? You don't have two critical pins swapped?
2. Is the chip good? Not the design -- have you blasted that particular part to kingdom come?
3. Look at the input to the inductor (the junction of the inductor and the zener. It should have a two-step or three-step waveform where it goes to near supply levels, down to -0.5V or so, then perhaps to the regulated output. This PWM duty cycle should increase as you increase load to compensate.
4. Look at the input to the feedback pin of the chip. If the chip is working correctly this should be held constant. If it is and you're getting crappy results you have trouble in your part of the circuit.
5. Look at the power input to the chip -- is it smooth, or does it have big negative spikes when the internal FET switches on?

This isn't the _whole_ set of things that I'd look at to debug this circuit, but it's where I'd start.

--

Tim Wescott
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"Applied Control Theory for Embedded Systems" came out in April.
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Good. Then I will refer to parts and connections, based on that schematic.

I did not ask for a photo of your schematic, asked for a photo of your layout (the actual physical parts and wires of your particular build of the schematic).

What about lengths and routes? When you switch multi amperes on and off, with fast edges, hundreds of thousands of times a second, a half inch of trace in the wrong place can cause trouble. So a photo of each side of your board would be very useful.

I got it!

Sorry to post such a stupid question, but after reading through Tim's suggestions, the answer to the problem dawned on me, and of course, was a design problem.

More like a paper to computer entry problem :P

I connected R2 to the chip side of the inductor instead of the output side. I was reading suggestion 2 from Tim and thinking how the hell would the PWM know what the load was, then I thought, the feedback! Then I thought, oh crap, I know what I did... and sure enough, I did. Right on paper, wrong in the design, oops!

It works beautifully now!

Thanks again for your quick help with this!

Glad to be of help. While I'm usually quite capable of synthesizing fresh new stupid mistakes totally from scratch, in a pinch I can always screw up while copying a perfectly good design.

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Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google?  See http://cfaj.freeshell.org/google/

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html

A linear reg will certainly reduce the component count and circuit complexity, but at the expense of efficiency. What is the mimimum input voltage at 5A?

You'll definitely have better luck with a linear. They are much simpler, at the expense of being less efficient. What is your *minimum* input voltage?

Dave :)

You really should include context when you post.

You're drawing from a 14V source to power a 6V rail, at 5A. A linear regulator just flows the current through and drops the voltage, so it will require greater than 5A at 14V, or more than 70 watts, while supplying 6V at 5A = 30 watts. The other > 40W gets burned up as heat.

A switcher would theoretically only take 30W from the 14V battery, or around 2.14A. Assuming that it's only 85% efficient you'll really need (30W)/(0.85) = 35.3W, or about 2.52A -- practically that's only 1/2 the power draw than with a linear, _and_ you only have to heat sink for 5 watts.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google?  See http://cfaj.freeshell.org/google/

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html

Ok, so, to bring this thread up once again, I'm having a new problem :(

Once I realized that my feedback was connected to the switched output instead of the filtered output, I cut the trace and ran the wire to the output instead. I tried the circuit and it worked perfectly. Under load, the voltage did not drop like a rock, it stayed very steady, just like I wanted.

Well, I modified every other board I had with this simple modification, and low and behold, none of them work properly! Their voltages drop very very quickly under load. By approximately 1 amp average DC load, the output voltage has dropped 3 volts or so! The feedback is connected to the output, as it should be, and this still happens.

The data sheet indicates that the feedback voltage likes to sit at

1.21V. When checking the feedback voltage at the pin, it is indeed 1.21V. Under load, it drops according to the load. By 1A or so it is 800mV or less. I tried to check the duty cycle of the switching output to see if it increases as the feedback voltage decreases, which it should, right? I cannot get a good look at the duty cycle on my scope here. It is just sooo noisy, I can't really tell much of anything from it. From the way the output voltage dives so quickly, I can't imagine that the chip is paying any attention to the feedback voltage, but why?

An interesting thing that I tried, was to remove the inductor (15uH) and try different combinations of these inductors in series and parallel. Even though the inductor is far away from the feedback and any other wires I thought it would affect, the gator clips I used to connect the inductor(s) allowed me to move the inductor(s) far away from the circuit to ensure that it's EMI was not messing with anything. The combinations of inductance values here did not seem to make any difference, the output voltage was pretty smooth all the time, but the voltage drop was not effected by any different values.

Just to try it, I removed the inductor(s) entirely and shorted where the inductor connects with an alligator clip. Surprisingly (to me) it worked great. The output was a *little* noiser than with the inductor, BUT, there was almost no voltage drop (500mV at 2.1A load). Overall, an acceptable result. Funny enough, though, if I just used two pieces of wire (18 gauge I believe), and twisted them together, no luck, it did not work at all, very noisy output and it voltage dropped just like with the inductor. What gives with this?! What are the properties of your typical alligator clip connector (I confirmed this with a few different alligator clip connectors, all about 1 foot in length) that would cause this and make them so much different than a typical wire? Increased resistance? But is it really that much?!

So, to summarize, for some reason, the LM2678 seems to be ignoring it's feedback voltage level and does not seem to be increasing the duty cycle as the load increases. I should also note, that, on the same circuit board, I have a 9V regulator. It is the exact same chip and designed the exact same way (selecting recommended components per the data sheet) and it works GREAT. It is designed for ~3A load instead of

5A like the 6V, and therefore uses a different schottky, different inductor, and less capacitance, but it does not have voltage drop problems, it works great.

Any ideas? I'm very stumped. I'm going to spend a little more time trying to determine the duty cycle and see if it is in fact increasing as the load increases, but I suspect it is not, or I would not be having the problem I am having... right?

Just another note, with the inductor removed and replaced with the alligator clips (described above), the feedback voltage remains within a couple mV of 1.21V, like it should, it does not dive to 800mV as it does with the inductor in place. Why does the inductor have such a serious effect on the feedback?

There is no way I can read the duty cycle of the switched output. It is tough to lock onto, but there is a bit of a square wave, and alot of a sinusoid in between the square wave, but the square wave "on" time is not consistent even under a relatively steady load, so I don't know how to read the duty cycle. When the load changes, the switched output does seem to change a little bit, but still, nothing that can give me a good reading, it just turns more into a square wave and it seems that the sinusoidal part in between disappears a bit. With the alligator clips and no inductor (semi-working config), there is almost no square wave to speak of (it looks something like a (1/x)*(sin x) repeating) on the switched output, until load is applied then, it *appears* to look similar to that of the *with the inductor* config, while that config is under slight load (before it drops so much voltage that the circuit it is supplying voltage to resets).

I know the way I am describing this may be confusing, but it truly seems that with the inductor, the chip ignores the feedback voltage level!

Another update...

So, as I mentioned, there is 2 circuits on the one board, the same design, with the exception of components specified for different current levels. The 6V at 5A side, and a 9V at 3A side. The 6V side is and has been the problem.

Well, since the 9V side is working perfectly, I decided to change the feedback resistor to make it another 6V side. Hooked it up to a load, and it is still perfect, works great.

I tried making the 6V side into a 9V side, and it still drops voltage. At 2.1 amps or so, it drops to baout 4.5V!!

I then decided to make the 6V at 5A side into a 6V at 3A side, and use the EXACT same design that the 9V at 3A side uses, since, even at 6V /

3A, that side is perfect.

So, I used the 22uH 3A inductor, the lower amperage schottky, and changed the input/output capacitance to mimic the working 9V side. Unless I'm just missing something obvious, which, after looking at it this long I doubt, both sides are identical, yet the non-working side (was 6V/5A, now 6V/3A) still does not work.

Barring me missing some obvious difference between the two sides, which like I mentioned, I doubt is the case, we are down to the traces, I guess...

The trace width of the 9V/3A side is 15mils, and the width of the 6V side is 100mils. They have a common ground and power (battery +) which are both 120mils. No traces are longer than 3 inches total. There are almost no traces near the inductors, except those actually connecting to the inductors. Both the working 9V side and non-working 6V side have their associated components equally close and the trace lengths are roughly the same to their associated components. Other than the trace width difference, I can't see any reason for the circuit not working!

The board is a two layer board (top and bottom) that was auto routed using SPECCTRA. I have verified that all connections are going where they should be (according to the design), so I guess it is down to the actual layout, as John mentioned? Is this possibly the problem or? Should I attempt to post my layout somehow? I'm not sure what format would be helpful to you all?

So what are the possible problems when switching multi amps at high frequencies with sharp edges? Is this probably my problem, or is it very unlikely and more likely that I am missing some kind of design difference?

Thanks for any help with this, it is driving me nuts!!

that is never a good idea with analogue circuitry. auto-routers should really be called auto-rooters, cos thats what they do :)

I'm really new to this anyway, I wish I knew the difference. I can figure out how to route manually, but it would take alot longer, and I still wouldn't know what to / not to do with the traces/planes.