What is the purpose of the diode in the schematic of the LM317-based "0-V to 30-V voltage regulator circuit" from section 9.3.1 of TI's LM317 documentation?


Excerpt from LM317's documentation - schematic of discussed circuit

I simulated the above circuit using LTspice XVII.

Simulation of discussed circuit using LTSpiceXVII

If I remove the diode from the schematic, the resulting waveforms are as shown below:

Simulation of discussed circuit using LTSpiceXVII - removed diode

Nothing happened. The output voltage did not change, independently of the presence or the absence of the diode. It seems that this diode does not play any role in setting the output voltage (if R3 & V2 values are equal to those in the datasheet.)

What's interesting, this diode is not carrying any current at all. It is simply reversed biased. What is its purpose?

I suspect this diode has something to do with protection of the LM317 terminals (adjust and output). Maybe because of the V2 source could be turned on before the V1 source (or the V2 source could be turned off after the V1 source,) but I'm not convinced.

I'm also curious why R3 is present in this circuit. What is its purpose? Is R3 responsible for setting the lower value of the output voltage when R2 (variable resistor) is set to 0? It seems that the value of R3 has a significant impact on the diode reverse-bias or forward-bias condition.

If the value of R3 is high enough, the diode is biased forward and this circuit operates like a typical LM317 regulated power supply with the lowest output voltage close to the V_REF. The values of R3 (and V2) also has significant impact on the output voltage. If R3 = 0 (removed from schematic), and R2 (variable resistor) is set to 0, we need only the V2=-V_REF voltage source to null the output voltage.

Unfortunately, the V_REF value is dependent upon multiple factors (e.g. temperature, set output voltage), so it is not constant. V_REF is present on R1, so maybe it is possible to probe that voltage by a couple of opamps and then drive the ADJUST pin by this voltage (negative value) to null the LM317's output voltage, if R2 (variable resistor) is set to 0, but it would be like using a sledgehammer to crack a nut.

I derived equation for the V_output (the same as in the datasheet,) but is only possible if I assume that the diode's current is negligibly small (and this is the case, simulation confirms that this current is nearly 0, because the diode is reversed biased) and if I assume that the adjust terminal current is also zero (it is in fact about 200 times smaller than the R1/(R2+R3) resistor divider current.) It is very interesting how resistor values were chosen to fulfill these requirements.


3 Answers 3


Of course the diode is not useful when you have a stable system with -10V supply up and running.

If you are not convinced about the diode being useful when the -10V supply is starting or shutting down, consider what happens if -10V supply is 0V or there is no power yet to sink any current so it floats. Without diode, output voltage will shoot right up to maximum value, if no current flows through R2.

If there is no R3, and R2 connects directly to -10V, the potentiometer range is unusable. If ADJ pin is below -1.25V, output will be 0V. R3 keeps the R2 biased so that there is about -2.92V at node between R2 and R3. So when turning the pot, there is only a small range of 0V before voltage starts to rise.

  • \$\begingroup\$ "If ADJ pin is below -1.25V, output will be 0V." It is not true, because I just have made this circuit work on my breadboard and if ADJ pin is less than V_REF, output goes negative. Only if ADJ is exactly equal to V_REF, output is zero volts. So simulation results (negative output) are consistent with experiment. And that seems OK, because the LM317 is powered from V_IN and V_OUT, and there is no such limitation that V_OUT must be positive or zero with respect to V1 ground, because from LM317 point of view, ground is V_OUT. The only limitation is that V_IN - V_OUT must be less than 40 V. \$\endgroup\$
    – pstr
    Feb 22, 2023 at 12:56

As indicated by "Justme", if the -10V supply is accidentally disconnected or during initial power supply sequence of the +35V/-10V for a brief time if it is at high impedance when +35V is already up, the output voltage V0 will shoot up to a very high value and that can cause (over-voltage) reliability issues to the circuits which are being supplied by this regulator. Suppose the user has connected a 5V logic circuits on V0, a 30V on V0 will instantly fry those logic circuits

As regards the resistor R3, as indicated by "Justme", if R2=0, the output voltage will theoretically be -8.75V if R3 is not present and will be -1.66V if R3 is present. Having V0=-8.75V might violate the condition that you mentioned i.e., VIN-V0 will become greater 40V. Secondly, consider the circuits connected on V0. If the voltage on V0 becomes too negative, it can draw uncontrolled amounts of current from V0 because it will forward bias the diodes in those circuits. This can happen even if V0=-1.66V but it is better than having -8.75V on V0. Thirdly, if someone connects a +ve voltage instead of -10V (very unlikely event), the resistor R3 can limit the current. This might not be the intent of the resistor R3 but it does help in this case anyway.


The diode ensures that the bottom end of R2 can never go more than 0.7V above ground. Let's draw a bit more of the circuit:


simulate this circuit – Schematic created using CircuitLab

Suppose that C1 is charged to say 10V, and you adjust R2. What happens to A when you adjust R2 one way? What happens to A when you adjust R2 another way? And how does the behavior differ when the diode is there vs. when it isn't :)


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