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.
To get the LM317 down to zero volts you need to bring the control pin down to -1.25 V.
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 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.
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.
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
if there is sufficient loading, try one LED to drop a fairly constant voltage. You an always bury it where it can't be seen.
The way I have done this many decades ago: Use two LM317: one in a classic regulator w. a 5k pot, and the other in simple 1.24V constant non-adjusted supply. They have to be supplied from separate windings of the same transformer, or from two separate transformers. The 1.24V only provides negative -1.24V reference to the main regulator, which makes it regulate perfectly from 0V instead of 1.24V. Just make sure that the filtering cap in the 1.24V regulator is large enough, so when you turn it off, the output voltage doesn't jump up because the reference voltage capacitor discharged sooner. It's super-simple and it works great!
Here's the lTspice sim of a circuit similar to that suggested by JohnnyB:
The sim shows the output voltage (green trace) versus the R4 pot resistance (red trace), which goes from 0V to 10V for a 0 to 1kΩ pot resistance change.
If an isolated negative voltage is not available, it can be supplied by a small wall-wort 5V (USB) supply, as it only requires about 10mA of current.
A cheap and dirty option would be to use a NiMH or alkaline cell to provide a negative voltage. An alkaline D-cell should last for at least 1200 hours of operation. To prevent draining when the power is off, the battery is isolated using a small N-MOSFET (e.g. 2N7000). (Below)
