I have a LM340T15 voltage regulator and want to use it as current limiter (150mA) for a 30V supply line. In the reference design (datasheet) following circuit is suggested

enter image description here

My modified design looks like

enter image description here

Since the LM340T15 has a fixed output voltage of 15V I changed the formula to

\$ I_O = \frac{15V}{R} + I_Q => R = \frac{15V}{(I_O-I_Q)} = \frac{15V}{150mA-1.5mA} \sim 100\Omega\$

Therefore the circuit draws a maximum \$ 30V\times 150mA = 4.5W\$.

Is this design applicable? Who can I calculate the power dissipation over the resistor R9 and the regulator?

Also: applying a supply voltage (input) will not result in the expected output voltage. At around input=13V the output voltage (voltage over the circuit) is capped at 8.5V. Increasing the supply voltage has no effect on the output. The current rises until 45mA and stops there (as expected). Since the LM350T15 is regulated to 15V, are these 15V still the limit when using it as current-limiter?

  • \$\begingroup\$ Worst case is short circuit, so 4.5W is equally shared. They’ll both get a little toasty. \$\endgroup\$
    – Kartman
    Apr 30, 2021 at 9:11
  • \$\begingroup\$ Note that maximum Vout is about Vin - Vreg - 3 (due to ~~~= 3V across regulator.) || Vout = IR = Iput x Rload. || Vreg_drop ~~= Vin - Vregout (across) R9) - Vout = Vin - Vregout - Ireg x Rload. | Preg is max at Rload =0 and decreases as Vout (= R10 x Ireg) increases to the maximum allowed. \$\endgroup\$
    – Russell McMahon
    Apr 30, 2021 at 10:36

1 Answer 1


Note that maximum Vout is about
\$V_{out_{max}} = V_{in} - V_{reg} - 3V\$ (due to ~~~= 3V across regulator.)
The Higher the valuie of \$V_{reg}\$ the lower the available \$V_{out}\$ range.
With \$V_{in} = 30V\$ and \$V_{reg} = 15V \$ max \$V_{out}\$ is about 12V ( = 30-15-3)

\$V_{out} = I\times R = I_{put} \times R_{load}\$.
\$V_{reg_{drop}} \sim Vin - V_{reg_{out}} - Vout\$,(across R9) so
\$V_{reg_{drop}} = V_{in} - V_{reg_{out}} - I_{reg} \times R_{load}\$.
As \$P_{reg} = V_{reg_{drop}} \times I_{reg}\$ and as \$I_{reg}\$ is constant,
then regulator dissipation is maximum when \$V_{out}\$ is 0 (as there is no dissipation in R10.

So - Pregulator is max at Rload =0 and decreases as Vout (= R10 x Ireg) increases to the maximum allowed.
\$P_{reg_{max}} = (V_{in} - V_{reg}) \times I_{reg} = (V_{in} - V_{R10}) \times I_{reg} = (30-15) \times 0.15 = 2.25 W\$
Lower voltage regulators will allow a higher Vout range and incur higher worst case regulator dissipations when the output is shorted, and lower (constant) R9 dissipations. .

Power in R9 is a constant \$P_{R9} = V \times I = V_{reg} \times I_{reg}\$.
Here \$P_{R9} = 15V \times 150 mA = 2.25 Watt\$

  • \$\begingroup\$ Based on your calculations I tried the proposed circuit and the lost energy is unnecessary high. It works but is inefficient. a better choice is the classic lm317, which should arrive next week. The guide above will prove very useful in the future \$\endgroup\$
    – v3xX
    May 2, 2021 at 19:31
  • \$\begingroup\$ @v3xX With 30V in, for a given load at 150 mA , the combined dissipation in the regulator plus R9 will be the same. You can reduce dissipation in the regulator by using a lower Vin and lower Vregulator. If you keep Vin at 30V and use an LM317 at 1.25V then regulator dissipation will RISE as you have to drop (Vin-Vload), V_R9 is now lower so the regulator must dissipate the balance. See my formulae which will confirm this. Do you HAVE to use Vin=30V? If so, using an LM317 plus input "R8") of about R = 12V/150 mA ~ 80 Ohms will redcue R9 and regulator dissipation. R8 ~= >> 12 x 150 mA = 1.5W. \$\endgroup\$
    – Russell McMahon
    May 3, 2021 at 7:09

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