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The LM2575 is a simple switcher regulator for smaller currents/voltages. There is some fixed voltage versions (5,6,12V) and a variable.

The most flexibe stock could obvious be the variable version, but if You have a stock of, say 6 volt and want to use the same trick as seen with the L7805 where a set of voltage-divider resistors on the regulator pin makes it possible to get higher output voltages than the fixed 5V (the trick is that the L7805 seeks 5V on it's reg. pin).

Similar there is a direct feed-back pin on the L2575 (and a 2xR voltage divider on the variable to the feed-back pin).

I could test it myself, but if anyone has more knowledge and allready tested, then I will see..

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    \$\begingroup\$ Not sure I understand what actually your question is. Also unsure how you use an 7805 to change its output voltage, your description doesn't match the trickery in my cookbook (diode in the ground lead). \$\endgroup\$
    – jippie
    Dec 20, 2015 at 16:50
  • \$\begingroup\$ Looks like you could change the output voltage by putting resistors in parallel with the built-in divider network. But I wouldn't do that. I would just order the right part. \$\endgroup\$
    – user57037
    Dec 20, 2015 at 16:58
  • \$\begingroup\$ The way you normally trick a fixed [linear] voltage regulator is by "lifting" its ground pin: i.stack.imgur.com/xEpVN.png \$\endgroup\$ Dec 20, 2015 at 17:03
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    \$\begingroup\$ Is there a 6V version? \$\endgroup\$
    – Andy aka
    Dec 20, 2015 at 17:33

5 Answers 5

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Yes, you could do this, but the results may not be all that satisfactory. Below is the block diagram of the IC:

enter image description here

There are two resistors inside the chip for fixed versions. For the adjustable version, there are no resistors (R1 is open and R2 is short).

You might think you could simply insert a series resistor to increase the voltage (1.23V/K ohms), however the tolerance and temperature coefficient of R1 and R2 may be very poor (the ratio however will be very accurate).

You could insert an LED or a diode or a zener diode in series. You could add a lower resistance network from pin 4 to the output/GND so that resistance of R1/R2 wouldn't matter so much, but that would waste power.

If the increase you want is small (say increase 5V to 5.5V) it won't matter so much. If you don't care much about stability (say you're using a pot for a one time setup) it might be okay. But nobody is guaranteeing the accuracy or the stability of those resistors (only the ratio) so you can't count on much.

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  • \$\begingroup\$ I was thinking that the main stability issue might be that R1 and R2 will heat up simultaneously on the chip, but whatever is added outside won't so that could cause accuracy problems at higher current/power. \$\endgroup\$ Dec 20, 2015 at 17:50
  • \$\begingroup\$ Read the data sheet - it says NO. \$\endgroup\$
    – Andy aka
    Dec 20, 2015 at 18:17
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    \$\begingroup\$ @RespawnedFluff It's worse than that- the resistors on the chip might have a tempco of 1000ppm/°C so when the chip heats up the external resistor (call it perfect) will have more or less voltage drop since the current is changing. Since the on-chip resistors are being heated by the chip it could be pretty bad. But if you want to add a bit of voltage to compensate for IR drop in a cable or get a higher voltage to run some LEDs I bet it works more than well enough. \$\endgroup\$ Dec 20, 2015 at 19:04
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The data sheet for the device says this: -

FEEDBACK CONNECTION

The LM2575 (fixed voltage versions) feedback pin must be wired to the output voltage point of the switching power supply. When using the adjustable version, physically locate both output voltage programming resistors near the LM2575 to avoid picking up unwanted noise. Avoid using resistors greater than 100 kΩ because of the increased chance of noise pickup.

Now I don't want to spoil the party but when a data sheet says "MUST" I tend to think that messing around with this is something you do when all else fails.

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  • \$\begingroup\$ I have to agree. I wouldn't bother. There are approximately 1 million other ways to approach the problem. \$\endgroup\$
    – user57037
    Dec 20, 2015 at 18:04
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    \$\begingroup\$ I saw that paragraph, and gave it 'due consideration'. \$\endgroup\$ Dec 20, 2015 at 19:01
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For the LM2575 I would put a variable resistor in series with the FB pin. That variable resistor will get added to the internal R2.

enter image description here

The output voltage is Vo = Vref (1 + R2/R1). So if you increase R2 (by adding a resistor in series wiht it) you'll get more output voltage. You need to first figure out what value the fixed version uses, which is not possible without trial/measurement since you only know the ratio of the internal R2/R1. EDIT: Luckily they are given in the above schematic.

Alas neither TI nor ON give a spice model for LM2575/NCV2575, so I can't easily simulate this. Given how ICs are manufactured, it's also possible for the internal R2/R1 ratio to be well controlled but the individual values of R2 and R1 to vary a fair bit from part to part (in the same submodel).

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    \$\begingroup\$ The R1 and R2 values are listed in the text right under figure 1. And to the OP, if you do something like this there is always a chance that you will upset the stability of the regulator. That is a reason to test the stability, not a reason to avoid trying it. \$\endgroup\$
    – user57037
    Dec 20, 2015 at 17:30
  • \$\begingroup\$ D'oh yes, R1 is 1K on the schematic. \$\endgroup\$ Dec 20, 2015 at 17:43
  • \$\begingroup\$ Read the data sheet - it says NO. \$\endgroup\$
    – Andy aka
    Dec 20, 2015 at 18:16
  • \$\begingroup\$ @Andyaka: I don't see how adding a resistor (comparable with R2) in series with it is all that different from using the adjustable version in terms of noise pickup. YMMV since neither of us is probably willing to breadboard this just to answer a question here. \$\endgroup\$ Dec 20, 2015 at 18:29
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    \$\begingroup\$ So you gave it "due consideration" like @Spehro eh? \$\endgroup\$
    – Andy aka
    Dec 20, 2015 at 19:07
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First ! Thanks to all for the answers so far..

I did'nt notice the mfg. blok diagram with the internal R1/R2 divider. But that says alot.

About: The "must" wiring of the feedback should state that this is not to mess with. I don't agree, the word "must" is just a way to denote where this wire goes.

And as shown the FB goes directly to an internal voltage divider (R1/R2). The mgf. even - as some noticed - list the values for the fixed voltages.

This is to me quite clear an option to add an external resistor to get higher voltages from a fixed version.

What i don't understand though, is the fuzz about the power problems by addding an external resistor. The current, which should be in the microAmp. level on a feedback is'nt critical. No - the most critcal with the Switcher is the overall design and wireing (this is the words from the sheet).

Below is a pic from the sheet. It shows where the mfg. stresses where the lines should (must) be as short (AND good) as possible with thick lines. Notice here, that the FB line it not marked as thick (!?).

enter image description here

I tried to put the mfg. voltage/resistor value into a Excel sheet (below) and found it's a quite straight line aka it can be interpolated.

enter image description here

The interpolation gives the values shown below (up to 35V) and neither exceeds the 100K that the mfg. sheet warn about (!?). So in that context it is far below. Note that the R2 values is the sum of R2 + external (so if You want 24V => R2 = 18,9 and using a 5V version with R2= 3.1K the external should (must) be 18,9-3,1 = 15,8Kohm). Btw sorry, but i'm using comma as decimal point.

enter image description here

Conclusion: Test if You want to use it - and put a scope on to check the noice. And finally - As someone said: Why bother !! (maybe if You dont have one with the right voltage (or an var.)

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  • \$\begingroup\$ Be very careful here. That paragraph in the data sheet is very explicit. Just because you or I cannot see a technical reason why this could not be done doesn't mean you or I have the right to ignore it. BTW the feedback line is not marked in bold because it carries very little current. \$\endgroup\$
    – Andy aka
    Dec 21, 2015 at 11:22
  • \$\begingroup\$ I have just tested a 12V fixed with a 50K trim pot, making it deliver 20Volt. I works very well with that reservation that I did'nt put a scope on and did'nt test it's power margin, but so far it works ! \$\endgroup\$
    – Gearlos
    Dec 21, 2015 at 11:27
  • \$\begingroup\$ Thanks Andy for You critical view. But consider that this voltage divider is setup by one of the most simple components - A resistor. Even the fact, that the mfg. has removed them in the variable version i favor of external should say alot aka. the trick is just to increase a resistor value that consist of an internal plus an external. \$\endgroup\$
    – Gearlos
    Dec 21, 2015 at 11:33
  • \$\begingroup\$ btw: I think I mail and ask the manufacturer. Be back with that ! \$\endgroup\$
    – Gearlos
    Dec 21, 2015 at 11:38
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    \$\begingroup\$ Temperature stability and accuracy is the most likely problem with the resistors. You have not tested that at all. For 20V from a 12V regulator it should be a 6.504K series resistor 1.00K*(20V-12V)/1.23V. If you put a precision resistor in there measure the voltage accurately, and then load the regulator so that it gets hot and see the change you will get an idea of the problem, if any. I'd stick it in an environmental chamber and run the temperature up to 80°C for example. \$\endgroup\$ Dec 21, 2015 at 16:41
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The best answer here would be to use the adjustable version of the LM2575.

But as an alternative, just leave the set point alone, and let the chip do what it was designed to do. The situation isn't much different than that of needing a remote sense line to make a slight correction of the output voltage. Except in this case it's a local correction, and instead of the usual few hundred milli-volts, there would be 2 or 3 volts of correction.

To add output correction use an OpAmp.

schematic

simulate this circuit – Schematic created using CircuitLab

In practice a diffamp would not be required. Resistors R3 through R6 would be optionally populated as either a divider with unity gain buffer, or as a noninverting amplifier:

  • If Vout less than the set point of the LM2575 were needed a noninverting gain amp would be used. Say the LM2575 was a 5V unit, but Vout of 2.5 volts was needed. Then R3 = 0 Ohms, R4 = DNI, and R5 = R6 for a gain of 2.

  • For Vout greater than the set point, attenuation would be needed. So, populate R3 and R4 as a divider, R6 as DNI, and R5 as 0 Ohm for unity gain. If the LM2575 were a 5 volt unit, but 7V was needed, the input divider could be set as R3 = 4kOhm and R4 = 10kOhm.

This takes care of any concerns about the absolute value and tempco of R1 and R2 inside the LM2575.

What about risk of doing this?

It is generally a bad idea to use any part in an unspecified manner. Such use means you are on your own, and don't get the benefit of having had those who designed the part to have verified that it will work under altered conditions. It's almost always better to find a part that matches the application.

Consider the Simple Switcher. It's simple because it has a pre-compensated loop. Pre-compensated means the poles and zeros were set in advance, and there is no way to change them external to the chip. The datasheet for Simple Switchers go to great length to constrain choice of output filter components so that the pre-compensation will turn out to be what's needed for stability. It is not transparent what the gain or phase margin of such a loop is, or what the designers chose as targets for these parameters.

Specifically here, adding some gain to end up with a lower output voltage will reduce gain margin to some extent, and could cause stability to be insufficient.

Lowering loop gain to end up with output voltage higher than intended is a little more benign, but would still have an effect on loop stability.

This would all have to exhaustively tested to show if it was worthwhile.


Finally, just to be really clear, using parts outside the specification always a risk, and is rarely worth it.

Would I do any of this? I might try it as an interesting lab experiment. For a product, no, I wouldn't do any of this (including adding an external resistor pad detailed in the other answers).

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  • \$\begingroup\$ It may take care of the stability of R1 and R2 but it introduces an active element that will degrade the phase margin of the power supply and, under certain conditions, cause the LM2575 (or any other switcher) to oscillate at worst or have overshoot when the load changes value. Be very careful about suggesting this. \$\endgroup\$
    – Andy aka
    Dec 21, 2015 at 11:17
  • \$\begingroup\$ @Andy, I agree that there is risk in doing this. All of this off spec use is risky. But not every one understands that, so I'll edit to spell out the risk involved. Thanks for your comment. \$\endgroup\$
    – gsills
    Dec 21, 2015 at 13:54
  • \$\begingroup\$ Agree with the 'Danger Will Robinson' comments wrt this circuit, however there are reasons to attempt this, even in a product. Suppose you need a 200mV @1A output and no regulator has a reference that low... \$\endgroup\$ Dec 21, 2015 at 16:37
  • \$\begingroup\$ @Spehro , a lot of the concern centers around the limitations of the simple switcher itself, with the fixed compensation that's hard to do much about outside the IC. For something like 200mV output, LM2575 wouldn't be a good choice anyway. There are other regulators/controllers that have accessible error amps, where adding an OpAmp and custom compensation is commonly done. Usually for a product, it's a better use of time to use a more flexible part to begin with. \$\endgroup\$
    – gsills
    Dec 21, 2015 at 17:31
  • \$\begingroup\$ @gsills I say this because I've actually done it (different, much cheaper part). The more flexible part was 5x the price, so it was worth it. Sometimes it isn't worth it, yes. \$\endgroup\$ Dec 21, 2015 at 21:45

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