0
\$\begingroup\$

Here is my circuit design. Are there any problems here? LM317

Here are my questions:

1, I notice that R1 is either 120 or 240 ohm in the datasheet. Is that because of the 1.25v reference voltage, and the LM317 requires a minimum 10ma/5ma to operate? (1.25/240=5ma; 1.25/120=10ma)

2, Which resistor should I use? 120 or 240? Is 240 the maximum value for R1?

3, With 240 for R1 and 5k for R2, I get 25v with this calculation, 1.25/240*5000=25v; but with the Vout = Vref (1 + RL/RH), I get 27v. which one is correct?

4, For the pot, or R2 here, Is its value limited to 5k and lower? ( I don't have low value pots. 10K and 50K are all I got, and my test showed very bad voltage range with 10k.)

5, From my circuit above, is there a 6v drop across 2 LM317? so to get 10v output, I need at least 16v.

\$\endgroup\$
2
  • \$\begingroup\$ The advantage of using R1 to achieve the minimum current is that it results in a constant current being drawn regardless of output voltage. If you load the output directly so that it draws (say) 7mA at 1.25V (178\$\Omega\$ it will draw 140mA and dissipate more than 3W at 25V. As R1 it will only dissipate 9mW at any output voltage. \$\endgroup\$
    – Spehro Pefhany
    May 21, 2017 at 13:51
  • \$\begingroup\$ Your first lm317 is not going to do current foldback to 10 mA/1.25A \$\endgroup\$
    – sstobbe
    May 21, 2017 at 14:43

1 Answer 1

Reset to default
2
\$\begingroup\$

  1. This value is given to maintain the minimum load. It isn't a magical value, pick whatever satisfies the minimum load in your application.

  2. Using 120R guarantees 10mA but uses more power. 240R is above the absolute minimum of 3.5mA and is probably but not certainly fine.

  3. The latter is the correct datasheet equation.

  4. R1 and R2 form a divider to present the required voltage to the adjust pin. If you want to use a 50k pot instead of a 5k pot for R2, you can multiply R1 by 10 and satisfy the minimum load requirement with another resistor to ground.

  5. The drop depends on the load, see the National Semi version of the datasheet for the graph of input-output differential versus current. If you design for a smaller value than 3V you had better know your load very well.

In your design, when SW1 is connected to R4, the first LM317 has its current limited to about 10mA. This will all be consumed by the second LM317 and R1, leaving nothing available for the load.

You seem to be going for a large output voltage range. You still need to do the thermal design, this project is going to be bulky.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.