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Is there any variable DC-DC converter suitable for pre-regulation?
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T3sl4co1l:

--- Quote from: bloguetronica on September 06, 2018, 10:34:03 pm ---Well, I will have to weight all the options carefully. Probably a self made DC-DC converter will do, although I saw an application based on the LM2576 from Texas Instruments. In my application, I don't require a sophisticated DC-DC converter. However, I was not expecting that the DC-DC converter would be one having a fixed voltage, which is counter intuitive. With that feedback loop, the output voltage is now variable?

--- End quote ---

Easy -- start with an adjustable regulator.  The feedback pin ties to a voltage divider sensing the output.

So, suppose we tie another resistor to the feedback pin.  If we tie the other end to ground, it's in parallel with the lower resistor of the divider, and the ratio drops, so the output voltage rises.  Simple enough? :)

Suppose we tie the other end to a voltage reference.  If that voltage is 0V (GND), we have the above situation.  If VREF = V(FB), nothing happens, because there's no voltage across the resistor, no current flow and no one's the wiser.  If we tie it to a higher voltage, then some of the current into the bottom divider resistor comes from this one, and some comes from the original (top divider) resistor.  The bottom resistor always draws the same current (because V(FB) is kept constant by the internal control), so that current gets divided between the two top resistor.

This is a linear circuit, so there's no distinction between any of these cases, they're all equivalent.  In short, biasing the FB pin down causes the output to rise, and vice versa.

Or, overall, the FB pin is the -in of an inverting opamp, so putting current into that pin (with the feedback divider connected as usual) causes the output to fall proportionally.  A simple geometric leverage problem. :)

So it's very easy to set the output with a pot or opamp or DAC. :)

Tim
coppercone2:
I thought about doing this a few times but I never saw the benefit of not just using a large transformer and heat sink. Only got interested when I thought about running power MMIC circuits. But they are so expensive I am scared of not just using a commercial PSU because I might not anticipate a failure mode. But with a 800 $ IC you can throw together a giant heat sink and the most inefficient regulators imaginable and it will still be worth it........
bloguetronica:

--- Quote from: T3sl4co1l on September 06, 2018, 11:49:32 pm ---
--- Quote from: bloguetronica on September 06, 2018, 10:34:03 pm ---Well, I will have to weight all the options carefully. Probably a self made DC-DC converter will do, although I saw an application based on the LM2576 from Texas Instruments. In my application, I don't require a sophisticated DC-DC converter. However, I was not expecting that the DC-DC converter would be one having a fixed voltage, which is counter intuitive. With that feedback loop, the output voltage is now variable?

--- End quote ---

Easy -- start with an adjustable regulator.  The feedback pin ties to a voltage divider sensing the output.

So, suppose we tie another resistor to the feedback pin.  If we tie the other end to ground, it's in parallel with the lower resistor of the divider, and the ratio drops, so the output voltage rises.  Simple enough? :)

Suppose we tie the other end to a voltage reference.  If that voltage is 0V (GND), we have the above situation.  If VREF = V(FB), nothing happens, because there's no voltage across the resistor, no current flow and no one's the wiser.  If we tie it to a higher voltage, then some of the current into the bottom divider resistor comes from this one, and some comes from the original (top divider) resistor.  The bottom resistor always draws the same current (because V(FB) is kept constant by the internal control), so that current gets divided between the two top resistor.

This is a linear circuit, so there's no distinction between any of these cases, they're all equivalent.  In short, biasing the FB pin down causes the output to rise, and vice versa.

Or, overall, the FB pin is the -in of an inverting opamp, so putting current into that pin (with the feedback divider connected as usual) causes the output to fall proportionally.  A simple geometric leverage problem. :)

So it's very easy to set the output with a pot or opamp or DAC. :)

Tim

--- End quote ---
Oh, now I see! Hence the PNP transistor controlling the FB pin. So, essentially, if the output voltage (after the linear pass element) drops, the base of the PNP is pulled up, then the collector current rises and pulls up the FB pin (as if the output voltage of the DC-DC was too high), and then the voltage of the output of the DC-DC converter drops accordingly. That's actually simple and brilliant. Essentially, the PNP inverts the compensation effect of the FB pin, right? In this case, it shouldn't matter if the DC-DC converter is fixed or not.

Thanks Tim!


--- Quote from: coppercone2 on September 06, 2018, 11:57:04 pm ---I thought about doing this a few times but I never saw the benefit of not just using a large transformer and heat sink. Only got interested when I thought about running power MMIC circuits. But they are so expensive I am scared of not just using a commercial PSU because I might not anticipate a failure mode. But with a 800 $ IC you can throw together a giant heat sink and the most inefficient regulators imaginable and it will still be worth it........

--- End quote ---
It is worth it, IMHO. No relays or transformer taps, no relays, you can use a smaller pass element and much smaller heatsink, and less heat drifting the precision of your voltage reference (but that last reason is debatable, and I'm nitpicking here).

Kind regards, Samuel Lourenço
prasimix:

--- Quote from: coppercone2 on September 06, 2018, 11:57:04 pm ---I thought about doing this a few times but I never saw the benefit of not just using a large transformer and heat sink. Only got interested when I thought about running power MMIC circuits. But they are so expensive I am scared of not just using a commercial PSU because I might not anticipate a failure mode. But with a 800 $ IC you can throw together a giant heat sink and the most inefficient regulators imaginable and it will still be worth it........

--- End quote ---

Using commercial PSU is no warranty that something does not go wrong. @dmg learned that in hard way. Neither large transformer and heat sink (i.e. classical linear regulator) alone is enough that sensible load can be safely powered (just imagine for example a large power up overshoot). You need a properly designed voltage and current control loops combined with various protection mechanisms to decrease chance of failure to the minimum.
T3sl4co1l:

--- Quote from: bloguetronica on September 07, 2018, 12:25:18 am ---Oh, now I see! Hence the PNP transistor controlling the FB pin. So, essentially, if the output voltage (after the linear pass element) drops, the base of the PNP is pulled up, then the collector current rises and pulls up the FB pin (as if the output voltage of the DC-DC was too high), and then the voltage of the output of the DC-DC converter drops accordingly. That's actually simple and brilliant. Essentially, the PNP inverts the compensation effect of the FB pin, right? In this case, it shouldn't matter if the DC-DC converter is fixed or not.
--- End quote ---

Not quite.  I think their intent was to have a current source into the pin, but the current is proportional to output voltage minus control voltage.  It's a high-side current source, and therefore referenced to the output rail.

So the current mirror really doesn't accomplish anything, and it looks like another resistor in parallel with the top divider resistor.

I suppose that's not inherently a bad thing, it just reduces the gain.  The current mirror does invert the control signal, though, so it acts as a follower rather than an inverter.

If the control input is in turn set by the output (postreg) voltage, it should track, give or take gain and offset, which I'd have to write down and figure out to be sure about.

I think I would prefer to use the two-way resistor divider, and drive the control node with an op-amp that computes the required gain and offset.  At least for starters.

I might then decide to optimize it, potentially to as simple as a single transistor -- in which case, it would be roughly equivalent to what's shown, give or take resistor values and offsets and all that.  Again, I'd have to write it down.

The rest of the circuit looks a bit sausage to my eyes, it could be simplified, and linearized better for more stable control and probably lower noise at the same time, or something like that.  That's just more to the point that almost all lab supply circuits on the internet are crap, just to varying degrees.  I've only seen one that's mostly good (unfortunately, the author got very defensive when given constructive criticism).

I mean, I'd design the perfect supply to end all supply design threads -- but I don't need one, and no one's giving me any money to. :P

Tim
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