Let's say that I have a 37v DC source that I would like to reduce to 12v so it can run a 12v light bar (a string of 6 leds and I assume a resistor to limit the current).

I measure the resistance of the light bar at 35 Ohms, so using Ohm's law I calculate that putting a 70 ohm resistor before the light bar will safely reduce the voltage from 37v to 12v and run through the light bar at the expected 351 mA.

Will this really protect the components of the light bar considering the fact that the inital voltage is 37v until the current starts flowing. If you measure the voltage of the 70 ohm resistor before the current starts flowing: it reads 37v. It's only after the current starts flowing: that the voltage is reduced to 12v. Is it possible for the 37v to damage a component that's not rated or such a high voltage? Initially I tried applying 29v directly to it (no initial resistor) and it instantly smoked and died without heating up first despite being supposedly rated for 12-30v so I'm wondering if some arcing will occur internally with 37v and a resistor.

Diagram / Simulation

  • \$\begingroup\$ How did you measure the resistance of the light bar? Just an Ohmmeter? (Because I don't think that's likely to be a good way to go.) Is it possible that you actually power the light bar with a stack of batteries long enough to get a good measurement of the current drawn using an ammeter, for example? Or is there a set of technical specifications on the light bar that you could quote out for us; other than the 12 V specification? \$\endgroup\$ – jonk Sep 17 '19 at 20:06
  • \$\begingroup\$ @jonk I used an multimeter to measure 351 mA and then used Ohms law to calculate the resistance. \$\endgroup\$ – SurpriseDog Sep 17 '19 at 20:08
  • \$\begingroup\$ Thanks. That helps a lot to understand how you did it. May I assume that this was measured with 12 V applied to the light bar? \$\endgroup\$ – jonk Sep 17 '19 at 20:10
  • \$\begingroup\$ Yes, that's correct. \$\endgroup\$ – SurpriseDog Sep 17 '19 at 20:11
  • \$\begingroup\$ Then aside from the dissipation costs of using a resistor, your regulation will be $$\frac{\%\,I_\text{LED}}{\%\,V_\text{LED}}=\frac{\frac{\text{d}\,I_\text{LED}}{I_\text{LED}}}{\frac{\text{d}\,V_\text{LED}}{V_\text{LED}}}=-\frac{V_\text{LED}}{I_\text{LED}\,R}=-\frac{1}{\frac{V_\text{CC}}{V_\text{LED}}-1}$$ The regulation will be okay. (For every 1% variation in the light bar voltage -- it will change with current variations -- the current in the light bar will vary about 1/2%.) You will waste a lot of power. But you've enough over-voltage headroom that the resistor will actually "regulate." \$\endgroup\$ – jonk Sep 17 '19 at 20:18

You can use a resistor to limit current, this is probably the worst way to do that. Why? Because the current will be consumed as power, in this case it will be roughly 9W that's a lot of power, and a large resistor will be needed.

It's not the voltage that damages resistors it's power. (unless arcing occurs, you don't usually worry about arcing until after 60V)

The other problem is the voltage will vary if the current varies, this could have consequences for your circuit.

Use a linear regulator instead, or a few of them or a different power supply.

  • \$\begingroup\$ It's the arcing that interests me. Are you saying that a resistor only limits current, but the high voltage is still dangerous to any component not rated for it? \$\endgroup\$ – SurpriseDog Sep 17 '19 at 20:07
  • \$\begingroup\$ Arcing usually needs to be considered after +60V, your supply does not fit that range. Arcing is the primary concern, for high voltage. Arcing happens when conductors get to close together. \$\endgroup\$ – Voltage Spike Sep 17 '19 at 20:10
  • \$\begingroup\$ Interesting. I had tried applied 29v directly to it without a resistor. (It's rated for 12-30v supposedly) and it instantly smoked and died without having a chance to heat up first. Any idea what happened? \$\endgroup\$ – SurpriseDog Sep 17 '19 at 20:14
  • 1
    \$\begingroup\$ @Benjamin polarity reversed? \$\endgroup\$ – Solar Mike Sep 17 '19 at 20:40
  • 1
    \$\begingroup\$ @brhans Perhaps a buck converter instead? \$\endgroup\$ – SurpriseDog Sep 17 '19 at 23:17

If the light bar has only 6 LEDs @ 351 ? mA. That’s roughly a 3S2P white array (3x2) with 175mA per string at 3x 3.2V approx = 9.6V. Thus the series R would be from 12-9.6V = 2.4V/175mA = 14 Ohms per string or 7 Ohms shared.

At this current the power is 12V* 351mA= 4.2W and if you dropped 24V from 36V you waste 8.4W into a 14 Ohm 10W resistor.

Something doesn’t smell right here.

The equivalent linear load of 12V/.35A=34 Ohms but the incremental resistance could only be 7 Ohms above 9 to 9.6V unless your light bar is a different colour.

Thus applying 29V directly to this incremental load results in ~ 20V/7 Ohms =2.9 Amps

Smell that!
Next time include component links & specs and read them 1st.

  • \$\begingroup\$ Can you explain this idea of "incremental resistance"? How does 34 ohms turn into 7 ohms? Also the specs did say: 12-30v so I don't understand where I went wrong. \$\endgroup\$ – SurpriseDog Sep 17 '19 at 22:58
  • \$\begingroup\$ What specs? Zeners have a knee resistance which as in LEDs is a lower incremental value . That results in a rapid rise in current. It can be added to the fixed Rs above some threshold Voltage. Below this threshold the resistance rises rapidly. \$\endgroup\$ – Tony Stewart EE75 Sep 17 '19 at 23:05
  • \$\begingroup\$ Okay, but about the original question: Does a correctly sized resistor protect low voltage components from a high voltage source? Can I just put a 70ohm resistor in line with a 36v source to make it safe for the leds? \$\endgroup\$ – SurpriseDog Sep 17 '19 at 23:10
  • \$\begingroup\$ Correct is a loaded assumption when the specs are uncertain.Show PDF file, but ok if nothings gets burning hot. This requires computing Rth thermal resistance ‘C/W * power dissipation of each part using specs \$\endgroup\$ – Tony Stewart EE75 Sep 17 '19 at 23:33

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