TPS54260 -12V output,

The maximum output current for that configuration is about 1.5 A.  Also you will need to add 10 uF from pin 2 to the circuit ground at R9.  The EN external UVLO will not work with the inverting buck boost.  You would have to level shift iy as at start up -12 is at GND level.  I would just remove it and use the internal UVLO.  You may have other issues as well, but try those first.

Thanks very much. The following table is about my test,there are some things puzzel me.

Hi Cindy,

For the 18-30V to -12V conversion you cannot use switcherpro for component selection. I recommend using the equations in the following application report: http://www.ti.com/lit/an/slva317a/slva317a.pdf

After looking at the layout it does not look optimal for thermal performance. Both IC's appear to not have the recommended vias to inner copper layers to help with thermal dissipation. For the IC providing the 12V output it can simply be via'd to the existing GND copper below the IC. For the IC providing the -12V output, additional copper are should be added below the IC to connect to the -12V for this purpose. This one is worse off and I will guess is why you had issues with it getting too hot.

A quick explanation of why the output currents are different is the difference in topology. In the regular buck topology, the inductor is conducting to the output during both the on time and off time of the internal switch. In the inverting buck-boost topology, the inductor is only conducting to the output during the off time of the internal switch. This increases the peak current on the internal FET causing it to reach its peak current limit at a lower output current.

Did you also make the changes John suggested?

Regards,
Anthony 

Hi Anthony

I recalculated the parameters of the -12V outut circuit referring slva317a.pdf.

Design requirements: Vi=18V-30V,24V nominal, Vout=-12V,Io=2A, fsw=500kHz

I also have some question about thermal metrics. According to the datasheet of TPS54260,θja=57

I have worked out that: if Io=2A,Pdevice=2.5W,if Io=1A,Pdevice=0.97W.

According to SPRA953,I can't use the equation:Tjunction = Tambient + (θja x Power )

(if Io=2A,Tjunction=25+142.5=167.5 celsius degree,if Io=1A,Tjunction=25+57=82 celsius degree)

instead,I should use the equation:Tjunction = Tambient + (θjaeffective x Power ).

I want to know, if my design is accoeding to (3) test board conditions,what's the value of θjaeffective?

Hi Cindy,

I want to be sure, did you use Equation 5 to estimate the maximum output current? 2A seems high.

If your PCB design is the same as (3) test board conditions, the Θjaeffective will be approximately Θja given in the second row of the Thermal Information table on the TPS54260 data sheet. This is because the PCB design is the main contributor to this value.

If needed, one of the many methods to estimate the Θja effective for a different board it would require some measurements and the use of one of the Psi parameters to find the junction temperature. See section 3 for guidance to make the temperature measurement and the junction temperature calculation. Then with your junction temperature, rearranging Equation 1, you can then solve for your Θja effective. Θjaeffective=(Tjunction-Tambient)/Power.

Regards,
Anthony

Hi Anthony,

I used Equation 5,Iomax<(Iclmin-0.5Iripple)X(1-Dmax)=(3.5-0.5x0.25)x(1-0.4)A=2.025A,so I got Iomax=2A. Normally,what value of Iomax should I set up in this situation?

Even if I set up Iomax=1A,Pdevice=0.97W,Tjunction=82 celsius degree,it's also too hot to touch.isn't it?Or this temperature is normal for the device?

Hi Cindy,

This looks correct. I didn't take into account the size of your inductance when I took a look. You should also note in an inverting buck-boost the full load efficiency of your final design may limit your maximum output current further. Equation 5 does not include it.

82 Celsius sounds like a reasonable temp but yes I wouldn't touch it. Any inverting buck-boost does dissipate more power in the package than the same IC in a similar buck design. The rms currents are higher so you can expect it to get hotter. If you look at the equation for the inductor rms current, for a buck the first variable is Io^2 and in the inverting buck boost this is (Io/(1-D))^2. If there isn't enough copper area for thermal dissipation you may see the IC hit the thermal shutdown before reaching your maximum output current.

If the 2A on the negative 12V output is an absolute requirement you may need to consider some other options such as using a wide VIN controller to generate this instead for a more effective solution.

Regard,
Anthony