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LM3409 sense resistor power

Paul Bar
Intellectual 375 points
Other Parts Discussed in Thread: LM3409

Hi,

I designed a nice circuit with the LM3409 that works very well with Imax=1A for one LED with Rd=0.7 and Vf=2.9V. Now I would like to use the same circuit, but with the option of setting 2 maximum led current with a relay.

So, using a switch to control the relay, I would have the Imax=1A option and the Imax=2A option. I do not want to use the dimming function because I want to keep the dimming range intact at Imax=1A.

However, looking at the data sheet, I cannot find the current that flow in the resistor. How to calculate that? In my case the resistor is 0.24ohm, so there would be two 0.24ohm resistor in parallel in the higher current mode. That would result in a certain amount of current that need to flow in the relay. I want to choose the smallest relay to help noise rejection, but I need the current first.

Thanks for your help.

  • 0
    Intellectual 375 points

    Ok.. I found it. In the datasheet, 0.1ohm 1W resistor are use, but the equation to calculate the value is R= 1.24V/(5*2.51A). So according to that equation, there is 12.55Amp of current flowing through that resistor. However, that is not possible since P=I^2*R=15.7W. This is what made me post here.

    However, looking more closely in the theory of application, we have Imax=Vcst/R where Vcst is 248mV at max current. So the current is simply the LED max current which makes sense looking at the schematic.The power of the Resistor need to be at least 372mW.

    Thanks