1
\$\begingroup\$

The formula for the output current of a LM317 is:

LM317 CC regulator

In this schematic, what will the output current be?

schematic

simulate this circuit – Schematic created using CircuitLab

MY approach:

I thought R in this formula is the resistance LM317 sees. so I should calculate the whole resistance. it will be:

$$\left( \frac1{R_1}+\frac1{R_2+R_3}\right)^{-1}$$

but how can I neglect the node R2-R3 which current is being sunken through?

So I tried Delta-wye conversion :

$$R_{13}=\frac{R_1R_3}{\sum R_{\Delta}}$$ $$R_{12}=\frac{R_1R_2}{\sum R_{\Delta}}$$

So the resistance LM317 sees will be R13 + R12.

Which approach is correct? Am I right at all?

\$\endgroup\$
  • 4
    \$\begingroup\$ The circuit you made doesn't make sense. There is no load and the maximum input current into the adjust terminal is 100uA. This exercise is not realistic or useful. What is it you are trying to demonstrate or learn ? \$\endgroup\$ – efox29 Jan 2 '16 at 20:06
2
\$\begingroup\$

The key to the circuit is the current going through R1 that gives you the Vref.

Trying to solve this circuit by going the route of using the resistance the LM317 sees will not work, because that would build in a wrong assumption of how the current would flow.

Assuming the current through the ADJ pin is negligible (and the voltages and load are in the regulating range of the LM317), a step by step approach works:

As defined by the operating point of the LM317, \$ V_{R1} = V_{REF} \$: $$ I_{R1} = \frac{V_{REF}}{R_1} $$

Given \$ I_{R1} = I_{R2} \$: $$ V_{R2} = I_{R1} R_2 $$

Given \$ V_{R3} = V_{R1} + V_{R2} \$: $$ I_{R3} = \frac{V_{REF}+V_{R2}}{R_3} $$

$$ I_{OUT} = I_{R1} + I_{R3} $$

It turns out quite straight forward and you can substitute from top to bottom to get a simple formula.

\$\endgroup\$
  • \$\begingroup\$ I understood non of your step by step approach. can you tell the reason why those four MathJax lines are true? \$\endgroup\$ – AHB Jan 3 '16 at 6:29
  • \$\begingroup\$ 1: that's the operating point of the regulator 2: by kirchoffs node law. 3: by kirchoffs ring law 4: by kirchoffs node law. \$\endgroup\$ – Jasen Jan 3 '16 at 7:29
  • \$\begingroup\$ Jasen has given the explanations. I have also edited the post to give additional info. \$\endgroup\$ – rioraxe Jan 3 '16 at 7:55
  • \$\begingroup\$ @rioraxe I got them now. thx. \$\endgroup\$ – AHB Jan 6 '16 at 9:18

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.