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TPS61042 with external LED FET?

Other Parts Discussed in Thread: TPS61042, TPS61040

Hi All,

I have an application that has 3 chains of LEDs. Only one chain will ever be on at once, so I was thinking of using external N channel MOSFETs, e.g. BSS138 (common source to the Rs resistor) to select which chain is selected, and use the enable pin to turn the driver on and off. I don't need PWM control.

I notice that the TPS61042 has it's own LED FET. From the block diagram the Vfb signal is directly available on an external pin.  Can I bypass the internal FET (Q2 in the TPS61042 block diagram) and connect the LEDs to my FETs' drains and then connect the sources of my FETs to the FB pin?

Thanks!

Mark.

  • Yes, you can bypass the internal FET as you described.

  • Thanks! (BTW,  you've edited my post, but it's has an error - I think you meant to say 'connect the sources of my FETs to the Rs resistor and also to the FB pin).

    Also, I have a prototype of this chain selection working with a TPS61040. Unfortunately this has a feedback voltage of 1.213V which means the FETs have to have a very low gate threshold to work at low voltages (2V or less). Also that chip requires external over-voltage protection, which is why I'm thinking of using the TPS61042 instead as this both has internal over voltage protection and a feedback voltage of 250mV.

    Couple of things to note, in my circuit I have three chains, 2 of them have 3 green LEDs and the third has 7 green LEDs. When the 7 LEDs are on the output voltage is around 23.5V. This remains high for some time when the enable is off because the forward voltage of the LEDs is insufficient to discharge the capacitor. So when you subsequently turn one the 3 LED chains there is a momentary pulse. I think I've solved that by placing the output capacitor from Vout to Vin rather than Vout to ground, then putting a high value resistor (220k) in parallel with it. You can't just place a resistor in parallel with the capacitor in the normal circuit because that is in the DC path when the enable is off.

    Also another thing to watch is the external FETs going into a linear mode at very low voltages. This can cause the FETs to have a voltage between the drain and the source - it will still get the desired current, but at the expense of power dissipation in the FETs. That problem is solved with an accurate low voltage shutdown on the CTRL pin.

    Mark.