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Lm2575 adj feedback resistor 4K ohm

henry schaier87
Prodigy 15 points

I have an existing circuit that regulates down from 24VDC to 9VDC which feeds a LM317 6.8V 1A battery charger and a 7805 reg.  I don't understand how the 4K ohm resistor is calculated since this pin 4 accepts only a resistor where the internal R1 and R2 disabled on the LM2575.  Any one understand how we got 4K?

Our board uses an LM2575-adj regulator to generate 9V @1A.  PArt of the circuit is used for a 78L05 and the is for a LM317 6.8V@1A battery charger.  Pin 4 of the LM2575 has a feedback resistor of 4K which is tied to the output after the 330uH inductor.  I don't understand how to select this feedback resistor since no application is shown for this configuration.  See circuit below.  any ideas on how VDD is created.  Note for the ADJ version R2 is open and R1 is 0 ohms for the internal reference circuit.

I have a 4K resistor tied to the feedback pin 4.  I don't know this got calculated.  This power supply drives a 6.8V 1A bat charger and 5V linear regulator 7805.

  • 0
    TI__Guru 57555 points

    You are correct that using the circuit shown will not work with an -ADJ device; you would only get Vdd=1.23V. 

    However, with the -5V option you will get something like 9.9V or so.

    For the -5V device the bottom FB divider is 1k and the top is 3.1k.

    Adding 4k to the 3.1k you get 7.1k.  That would get you a Vdd=1.23*(1+(7.1k/1k))=9.9V

    This is not a good way to get your non-standard voltage.  I would use an -ADJ with the

    correct external divider. 

    The absolute value accuracy of the internal divider is very poor; only the matching is good.

    Also, the tempco of these resistors will be large.

    In other words, they will not match well with your single external resistor.