3
\$\begingroup\$

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.

\$\endgroup\$
  • 1
    \$\begingroup\$ Have you tried the datasheet yet? \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 4 '16 at 22:34
  • 3
    \$\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
7
\$\begingroup\$

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.

schematic

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.

\$\endgroup\$
  • 1
    \$\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
  • 1
    \$\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
6
\$\begingroup\$

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.

\$\endgroup\$
  • 1
    \$\begingroup\$ +1 Good answer, and a better alternative. \$\endgroup\$ – Sparky256 Jul 4 '16 at 22:51
3
\$\begingroup\$

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

\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.