TPS63060DSCT Followed by LMR10520YSDE

How much output current are you trying to get from the 5V and 3.3V supplies? The best way to verify this design would be first make sure all parameters are well within spec to the datasheet, then get an EVM, TPS63060EVM-619, and modify it to test the circuit. It is not recommended to reduce the input or output capacitance since the converter is internally compensated and both the input and output loops depend on sufficient low ESR ceramic capacitors placed immediately next to the VIN/VOUT and PGND pins. As you design please follow the design and layout guidelines of the datasheet and match that of the EVM. 

It's going to be a universal board so some applications will draw more on the 3.3V rail while others will pull from the 5V. I don't see any notes in the datasheet about specific ESR values like some other switchers which need the ESR to be within a certain range. My goal is try to minimize the total size of the design as much as possible so perhaps a tantalum in place of the 3 x 22uF on the output?

When chaining 2 switchers together and placing them right next to each other could the output caps of the TPS63060 be used as the input caps for the LMR10520 leaving only the bypass cap for the LMR10520? I still need to look at the different options and see which take up more room and then balance the cost but I'm thinking if I could place just 1 100uF between the 2 switchers that would probably eliminate a lot of board space but I've never tried this before.

Hello George, 

I have not tried this either, but I will just mention a few things to consider.

The output capacitor of the buck-boost converter has certain RMS current capability requirements that you need to satisfy:

Buck boost output cap I RMS = IOUT * SQRT( D / (1-D)), where D is the buck-boost duty cycle, and IOUT is the output current of the buck-boost.

The input capacitor of the buck converter also has RMS current capability requirement:

Buck input capacitor I RMS = IOUT * SQRT( D*(1-D)), where D is the buck duty cycle and IOUT is the output current of the buck. 

The capacitor will have to be able to handle both I RMS current requirements which may require you to use a physically larger capacitor anyway and defeat the area savings you are trying to achieve. 

Another thought is that the Buck switching current (normally supplied from the input capacitor) will now load the buck-boost output cap and appear as a load transient on the buck boost regulator at the switching frequency of the buck converter. 

Regards, 

Denislav

OK, well then worst case for the boost case would be with an input of near 5V bring the duty cycle down to almost 0% (call it D=.01). Maximum current out is 2.5A so this yields an Irms of ~250mA (if I did my math right). I've been told that Imax for tantalums can be calculated loosely by Imax=[Vrated}/[(0.65 + ESR)] so almost any tantalum with a low ESR should have plenty of headroom in this application (a 100uF/10V/0.9ohm == 6.45A).

It may be all for not since I don't think I could get a 100uF tantalum in less than a Case C size and that's bigger than 2 0805 ceramics.

I'd be curious to see some results if you took these 2 eval boards and chained them together and then removed the input caps on the LMR10520 and measured again. I would think that with the right selection of cap(s) you should be fine but I don't have those eval boards to test this theory out. 

Hi George, 

Please double check the RMS current numbers. 

Here is what I got:

Buck-boost topology case:

Duty(buckboost) ideal = Vout / (Vin + Vout) 

For VIN = 2.5V and VOUT = 5V, Duty=0.67

Cout IRMS = IOUT * SQRT ( D / (1-D)) 

IOUT = 2.5A

Cout IRMS = 3.54A

Boost topology case:

Duty(boost) ideal = (VOUT - VIN) / VOUT

for VIN = 2.5V, VOUT=5V, Duty = 0.5

Cout IRMS = IOUT * SQRT (D / (1-D)) = 2.5A

Buck topology case:

Duty (Buck) ideal = VOUT / VIN

VIN = 5V, VOUT = 3.3V, DUTY = 0.66

IOUT = 2A (?)

Cin IRMS = IOUT * SQRT(D*(1-D)) = 0.95A

In terms of the capacitor current capability, I would suggest checking the manufacturer's datasheet for the part you have in mind. The formula you presented results in 6.45A which I find hard to believe. It seems too large. I would expect it to be in the sub 1A range for a 100uF 10V tantalum.

Regards, 

Denislav