0
\$\begingroup\$

Let's say I want output voltage from LM317 in range from 8V to 15V. What circuit do I need to built?

I understand perfectly how to set upper voltage range (1 x resistor + 1 x pot), but how do I set the lower point at the same time?

\$\endgroup\$
  • \$\begingroup\$ Explain what you know how to do then explain how you can expect the lower point at the same time? \$\endgroup\$ – Andy aka Dec 16 '15 at 21:34
4
\$\begingroup\$

Assuming you want to set the minimum/maximum voltage range of the external potentiometer, you just use two extra resistors. If you mean something else you'll need to clarify what it is you really want.

schematic

simulate this circuit – Schematic created using CircuitLab

The LM317 will change \$V_{out}\$ such that \$V_{adj} = 1.25V\$. This configuration has the relationship (ignoring adjust pin current) \begin{gather} V_{out} = \frac{V_{adj}}{R_{1}^{-} + R_3} (R_2 + R_1 + R_3) \end{gather} where \$R_{1}^-\$ denotes the "lower" portion of the potentiometer. At the maximum stop, \$R_{1}^- = 0\$, so \begin{gather} V_{out,max} = \frac{V_{adj}}{R_3} (R_2 + R_1 + R_3) \end{gather} At the minimum stop, \$R_{1}^- = R_1\$, so \begin{gather} V_{out,min} = \frac{V_{adj}}{R_3+R_1} (R_2 + R_1 + R_3) \end{gather} This gives 2 equations with three unknowns, and the easiest way to solve this is to pick an arbitrary value for one of the resistors. For example, say you fix \$R_1\$. Then you get (assuming my algebra is correct) \begin{gather} R_3 = \frac{R_1 V_{out,min}}{V_{out,max}-V_{out,min}}\\ R_2 = \left(\frac{V_{out,max}}{V_{adj}}-1\right) R_3 -R_1 \end{gather} As long as \$R_1\$ isn't too big (say, less than 10k) or you have some leeway in the min/max limits, you can ignore the adjust pin current (as I've done for my analysis). If \$R_1\$ is very big or you need very precise bounds (not sure why you would need this), then you'll need to account for it in calculating the resistor values but the same design should work.

Note that I'm glossing over a few details of how to properly use the LM317 (for example, I'm not including decoupling caps). See the LM317's datasheet for more information.

\$\endgroup\$
2
\$\begingroup\$

Okay, the first thing to recognize is that pots come in discrete 1-2-5 values, so we have to accommodate that for a practical circuit. You want a 7V range, and the minimum current should be 5.2mA to correspond to the 240 ohm resistor typically used (so the regulator stays in regulation with no load). So the maximum pot element resistance is 1.34K ohms, so let's pick 1K as the next lowest standard value.

Now the current has to be 7mA to get 0-7V across the pot for your adjustment span. That means the 240 ohm resistor in the classic circuit should be reduced to 1.25V/0.007 ~= 180 ohms.

To get 8V as the minimum, we then need 960 ohms in series with the rheostat-connected pot since the regulator adds Vref = 1.25V, so the resistor can be calculated as Rs = (8V - Vref)/0.007A ~= 960 ohms.

schematic

simulate this circuit – Schematic created using CircuitLab

Note that this circuit supplies a constant 7mA through the resistor chain, thus meeting the minimum output current requirement of the LM317. The voltage output increases linearly with the resistance of R4, and the total impedance is low so the adjust current will have little effect on the output voltage.

Of course you'll want to add input and output capacitors and possibly a protection diode as per the datasheet.

\$\endgroup\$
  • \$\begingroup\$ I think R2 should connect to the tap of R4, otherwise adjusting the pot R4 makes no difference. \$\endgroup\$ – helloworld922 Dec 16 '15 at 22:02
  • \$\begingroup\$ @helloworld922 R4 varies from 0 ohms to 1K whether you flip it vertically or not. In this case, you'd probably want CCW to be the 'top' side of the element in my schematic (0 ohms) so that CW is maximum resistance and thus maximum output voltage. \$\endgroup\$ – Spehro Pefhany Dec 16 '15 at 22:05
  • 1
    \$\begingroup\$ Oh wait, haha. I missed that you're shorting the lower half of the pot. Nevermind, that works just fine. \$\endgroup\$ – helloworld922 Dec 16 '15 at 22:07
0
\$\begingroup\$

It is all wonderfully explained in datasheet made for your convenience by the wise engineers at TI. Please look at page 10.

If R2 = 0 ohm you get 1.25 V at the output. The voltage across R1 (240 ohms) will always be 1.25 V So if you make R2 also 240 ohms you will get 2 x 1.25 = 2.5 V at the output. There's a formula there if you prefer to use that. Now for your minimum and maximum output voltage calculate the value of R2. You will get an Rmin and Rmax. Now implement R2 by a series circuit of a resistor of value Rmin and a potmeter of value (Rmax - Rmin). There you go.

\$\endgroup\$
  • \$\begingroup\$ Sorry, but the link is broken. \$\endgroup\$ – jurij Dec 16 '15 at 21:38
  • \$\begingroup\$ @jurij - Try it now. \$\endgroup\$ – WhatRoughBeast Dec 16 '15 at 21:45
  • \$\begingroup\$ Thanks, I've read it ... but I still don't have a clue. Looks like my problem is too basic to be described there. :D Thanks anyway! \$\endgroup\$ – jurij Dec 16 '15 at 21:47

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.