I've this schematic, where I'm blinking output bulbs (21W @ 12VDC). Rate of blinking is 85 counts/min. The wave comes to be of 700ms. On time 350ms.

In this 350ms of On time, initially I skip 50~100ms for inrush, next 50msec for load sensing, next 50msec for short circuit detection and remaining time to check for other parameters. In Off time of 350msec controller does nothing. Again in next on cycle,all the stuff is checked. If the short circuit occurs before those 100msec, my 100mohm shunt burns out. The shunt is of 5W. is there any way in software or hardware to stop burning of shunt, and detect short circuit immediately? enter image description here

  • \$\begingroup\$ What is the maximum current that your voltage supply can source? Is it a battery? \$\endgroup\$
    – BeB00
    Aug 26, 2020 at 6:45
  • 1
    \$\begingroup\$ Also, check the specs on your shunt resistor. This (vishay.com/resistors/power-metal-strip-calculator) suggests that their 5W 100mohm resistor can take 100W for 100ms \$\endgroup\$
    – BeB00
    Aug 26, 2020 at 6:49
  • \$\begingroup\$ Yes battery of vehicle 7Ah \$\endgroup\$ Aug 26, 2020 at 7:15
  • \$\begingroup\$ @winny I didn't get you about club. \$\endgroup\$ Aug 26, 2020 at 7:49
  • \$\begingroup\$ What are all the different terminals on your circuit connected to? Some of them are a bit mysterious. \$\endgroup\$
    – Simon B
    Aug 26, 2020 at 8:55

2 Answers 2


Here is a way to protect against a short circuit:


simulate this circuit – Schematic created using CircuitLab

The current limiter drops about 300 mV for the lamp load, but limits maximum current to about 5 A. The capacitors provide a "soft start", but for the OP's application of a flasher, they may be changed or removed.




Assuming a maximum car battery voltage of about 14 V.

On a short circuit from L to battery minus, the power dissipation on your shunt is about
\$P_{shunt} = \frac{U_{max}^2}{R_{shunt}} = \frac{(14 V)^2}{0.1 \Omega} = 1960 W\$
So it is clear that your shunt quite immediately burns...

One possible solution is to provide an additional relais to feed a smaller current to L and to probe its voltage before switching to full power.

Your bulb takes nominally 21 W at 12 V, therefore it has a resistive load of

\$R = \frac{U_{nominal}^2}{P_{bulb}} = \frac{(12 V)^2}{21 W} \approx 6.8 \Omega\$

With a shunt of 10 times this resistance you should be able to detect a short circuit or overload. Its assumed maximum power dissipation is
\$P_{shunt\_overload} = \frac{U_{max}^2}{R_{shunt\_overload}} = \frac{(14 V)^2}{68 \Omega} \approx 2.9 W\$

Given that you load this power only on a fraction of time to this testing shunt, you can get away with a component of lower permanent allowed power dissipation. Read the data sheets!

  • \$\begingroup\$ A 68 ohm "shunt" will not provide the 1.8 A current for the bulb with a 12 V supply. The OP's schematic does not look right, and it has been more than 2 years since the last response. \$\endgroup\$
    – PStechPaul
    May 4, 2023 at 6:22
  • \$\begingroup\$ @PStechPaul As I write, this is an additional current path just to safely check for overload or short circuit. Its purpose is not to drive the bulb. \$\endgroup\$ May 5, 2023 at 6:06

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