I want to design a variable voltage regulator that will get from it`s maximum output (lets say 10V) down to 0V (or very close) considering the input as regulated 12V (from a PC power supply).

All the designs that I found use the 317 IC and can not go under 1.25V and I am pretty sure that there must be a way to do so.

I could not find any tutorials that explain in an easy way the way that the 317 behaves (beside the classical configuration) so any additional explanation is welcomed.

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    \$\begingroup\$ Have you tried the datasheet yet? \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 4 '16 at 22:34
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    \$\begingroup\$ Is the LM317 the only adjustable regulator you can get your paws on? Because there are more modern regulators that can regulate down to 0V without as much work...do you have a minimum load for this supply? \$\endgroup\$ – ThreePhaseEel Jul 4 '16 at 23:41
  • \$\begingroup\$ @Virgil Litan, a useful explanation and demonstration of an LM317 implementation that can output 0V: youtu.be/CIGjActDeoM \$\endgroup\$ – venzen Jan 14 '17 at 5:03

To get the LM317 down to zero volts you need to bring the control pin down to -1.25 V.

enter image description here

Figure 1. This dual LM317 circuit consists of a current limiter based around LM317(1) and a voltage regulator based around LM317(2). The voltage regulator section is relevant to this post as it is adjustable down to zero volts. Source: ON-Semi datasheet.

  • The control pin of LM317(2) is pulled low by Q2 which is wired as a simple constant current sink pulling several milliamps from the adjust pin.
  • D3 and D4 clamp the top of Q2 at two diode drops (2 x 0.7 V) below zero (-1.4 V).
  • The 240 \$\Omega\$ resistor and 5k pot can then adjust from zero up to the supply limit.

The problem with this circuit is that you need to generate a negative supply capable of sinking the few milliamps. My answer to Smartest way to use current limit using LM317? (where I explained the current limiting section) may help in this regard.


simulate this circuit – Schematic created using CircuitLab

Figure 2. If a centre-tapped transformer is used the negative rail can be generated quite easily. In this example a half-wave rectified signal is smoothed by C2 which doesn't have to be very large.

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    \$\begingroup\$ You know this. Others mightn't: The use of D3 + D4 as a 'voltage reference' in the ONSemi LM317 app note fig22 circuit is a terrible choice by the ONSemi 'designer'. The two diodes form an extremely poor reference (even when current driven as they are by the low-accuracy (irrelevant here) JFET current sink. Far better would be to use a TL431 2.5V shunt regulator or TLV431 1.24V version (the latter having a 0.2% accuracy option). ... \$\endgroup\$ – Russell McMahon Jul 6 '16 at 0:04
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    \$\begingroup\$ ... The 1.24V can be used directly as the -1.25V reference or adjusted to 1.25V if desired using two (precision) resistors. | LM317 reference accuracy varies but can be as poor around +/- 5% eg this example \$\endgroup\$ – Russell McMahon Jul 6 '16 at 0:05

A somewhat nasty solution is to place two silicon diodes of suitable current rating in series with the LM317 output (after the voltage-setting resistor divider). This would reduce the output voltage by about 1.4 volts. This will degrade the voltage regulation slightly, as the diode voltage drop will vary with load current.

A better solution is to provide a -1.25 volt (or so) low current supply to the bottom of the voltage setting resistors, rather than connecting that point to ground.

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    \$\begingroup\$ +1 Good answer, and a better alternative. \$\endgroup\$ – Sparky256 Jul 4 '16 at 22:51

A regulator is an amplifier which stabilizes the output at a target voltage. The LM317 does this by comparing the output to a 1.25V reference, and while you can get higher voltages than 1.25 (by voltage dividing the output) you cannot get lower (voltage dividers top out at 1:1).

Instead of an LM317, you can use an operational amplifier that senses near ground, and for a 20V range, divide output by a fixed ratio (20:2.5) and amplify difference of that divided output to a second divider on a 2.5V reference voltage (TL431 being a suitable reference source).

When the reference divider is at 1, output stabilizes at 20V; when it is at 0, the output stabilizes at 0V.

See figure 13 here: http://www.ti.com/lit/an/snoa589c/snoa589c.pdf

  • \$\begingroup\$ "_ The LM317 does this by comparing the output to a 1.25V reference, and while you can get higher voltages than 1.25 (by voltage dividing the output) you cannot get lower ..._" Yes you can get lower than 1.25 V. It's explained in the datasheet referenced in my answer. \$\endgroup\$ – Transistor Jul 5 '16 at 18:17
  • \$\begingroup\$ The answer is correct in principle but incomplete. The OA needs either to be a "high power" one or, more usually, to drive a power stage which is perhaps adumbrated but not made clear. | Transistors point can be addressed by adding eg "... without also needing a negative supply voltage". | The LM317 has current limit, thermal shutdown and is reasonably robust when mistreated. Use of an opamp + LM317 in some manner has advantages. \$\endgroup\$ – Russell McMahon Jul 6 '16 at 0:13
  • \$\begingroup\$ The use of an LM317 has a single disadvantage, as well: it works according to its specs only when it delivers 5 mA or more from the 'output' pin. Unless there's negative voltage, that output pin will misbehave at low V (the 'adjust' output resistor draws the necessary current in 1.25V-and-higher connection). \$\endgroup\$ – Whit3rd Jul 7 '16 at 4:57

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