I've been shopping for a voltage regulator that works in a range of .5-1.5 (+/-.5) Volts, but they are not easily found.
I'm wondering if anyone knows whether they even exist, or possibly how to make one. Thank you,
L.
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1\$\begingroup\$ They are very easily found. Did you not check TI? More than 350 matching parts. \$\endgroup\$ – Samuel Jun 27 '14 at 6:58
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1\$\begingroup\$ 0.5 Volts is pushing the limits. Why do you want such a regulator? What is your source and what is your output? How much current do you need? \$\endgroup\$ – abdullah kahraman Jun 27 '14 at 6:59
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1\$\begingroup\$ Do you really need to power something or do you just need a particular voltage to do stuff with? In that case, a voltage reference might be what you are looking for. \$\endgroup\$ – Tom Jun 27 '14 at 8:40
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\$\begingroup\$ What is your 0.5-1.5V range ? Input or output range ? Do you want to cover this range, or do you just need 1 voltage ? There are 2 very common voltages in this range : 1.2V and 0.8V If the provided range is your input range, what is your expected output voltage ? \$\endgroup\$ – Jacen Jun 27 '14 at 9:08
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2\$\begingroup\$ "Works in a range" is a useless spec. Does that mean input, output, something else? We do engineering here, not handwaving. \$\endgroup\$ – Olin Lathrop Jun 27 '14 at 16:46
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The problem is that your 500mV output requirement is rare, so that very few regulators use a reference that is lower than ~1.25V (a few use 800mV).
One way to get this to work is to amplify the feedback signal using an op-amp. A gain of 3 would allow you to use any regulator with a reference voltage up to 1.5V.
Edit:
Say your regulator has a 1.25V reference, so you'd normally use a voltage divider RA/RB on the output such that Vout = 1.25V (1+ \$\frac {R_A}{R_B})\$.
Add a suitable op-amp to amplify the output voltage:
Vin comes from your output voltage. Say it is 500mV and your regulator has a reference voltage of 1.25V. Vout goes to the feedback input of the regulator (where the tap on the voltage divider goes).
In general your output voltage will be Vout = Vref (\$\frac {R_G}{R_F+R_G})\$, so in this particular example we might pick Rg = 10K and thence calculate Rf = 15K.
answered Jun 27 '14 at 9:52
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\$\begingroup\$ Do you mean produce the reference 500mV and amplifiy it to 1.5V with an amp? I think I am just short of understanding what you're trying to describe here. \$\endgroup\$ – sherrellbc Jun 27 '14 at 16:02
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\$\begingroup\$ No, amplify the feedback voltage. Normally it's a voltage divider from the output to take the output down to the reference. If you amplify the output voltage by a factor of +k to (say) 1.25V (or whatever your regulator uses as a reference) then it will control at Vout = 1.25V/k (or oscillate, but that's a solvable problem). \$\endgroup\$ – Spehro Pefhany Jun 27 '14 at 16:17
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\$\begingroup\$ Adding an amplifier in the feedback loop of a regulator that is already near stability limits is a recipe for creating an oscillator. You might get away with it, but lots of compensation might be needed. \$\endgroup\$ – Edgar Brown Dec 1 '18 at 18:30
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This chip manufacturer shows you how to do that: Maxim. The link shows a very clear circuit diagram and explanation.
This way, you can realize arbitrarily low regulated output voltages.
Looks simpler than the amplifier method of Spehro Pefhany. Just use a second voltage regulator with an output voltage V2 that is higher than the reference voltage Vref of the primary regulator V1. Normally, you'd have a voltage divider between the regulated output and ground. Now you should put that voltage divider between the outputs of both voltage regulators. The middle voltage of the divider is connected to the feedback input of the main regulator in both scenarios. V2 only has the voltage divider as a load.
It is fun to realize that for the working of a regulator, it does not matter at what voltage you put the "ground" end of the voltage divider in the feedback loop, because that is a static voltage. The feedback loop is about the dynamic changes of the output voltage. If that goes up, the regulator should try to get to lower that, and vice versa.
It also does not matter if you have a fancy switching regulator (buck) or a traditional dissipative type of regulator. Of course, Maxim shows Maxim regulators, but you can utilize other regulators.
Here is how the classic voltage regulator circuit uses a voltage divider between output and ground. The regulator will adjust the output voltage in such way that the voltages on both amplifier inputs become equal. Clearly, the output voltage cannot be lower than the Vref.
(drawn with inkscape)
Here is how you can get output voltages below Vref:
It is important to note that the above is about basic circuits where the voltage reference (1.25 V) is connected to the positive input of the regulator, and the feedback voltage from the voltage divider to the negative input. However, popular 3-pin regulators like the LMvv series for a fixed vv volt output, and the LM317 adjustable regulator work slightly different. Here, the "adjust" pin is NOT the negative input, but the positive input with the bandgap voltage reference in series. The regulated output voltage is therefor 1.25V higher than the voltage on the adjust pin. So here the trick is to connect the voltage divider between the regulated output pin and a NEGATIVE voltage, provided by a second, small-scale, voltage regulator. You'll find the schematic for a LM317 as a 0-30 V regulator on this Texas Instruments link, see page 11.
Here is the circuit diagram of the low-voltage regulator using a LM317:
The LM317 is a bit special, as the "adjust" pin is not a high-impedance input, as it has a constant current of about 50 - 100 uA, which you have to take into account as a load of the voltage divider. But the LM317 is cheap, abundant, has simply only 3 pins, and I had a few in my junk box, so I actually tested this circuit. Because it has only 3 pins, it needs a minimum load current of about 5 - 10 mA in resistor RL, for its own internal working.
Note also that the 4.00 V and -5.00 V indicate that these voltages should be stable, as any drift and noise will directly appear in full in the output voltage.
Of course we build such circuits for fun, but if you have any real need for low voltage, any 50-100 euro hobby bench power supply is adjustable down to zero volt, often with steps of 10 mV. How that works? Probably a arduino type control unit with digital to analog converter (DAC) output connected to the positive input in the diagrams above, instead of the fixed zener diode, and an additional negative voltage input supply . . .
Direct link to the Maxim application note as a PDF file.
Example schematic of using a second reference voltage to generate a low voltage output from a standard regulator (from the Maxim application note:)
