TPSM53603: TPSM53603RDAR power interruption calculations

Hello,

Do you have a schematic which you can share?

Best regards,

Ridge

Power interruption capacitor calculations, placed after a 12V DC-DC converter - then fed to 2 x TPSM53603 which converts to 3.3V and 5V. the total load on 12V DC-DC is 550mA. Is it right to calculate the capacitor voltage after time t and estimating if the voltage is greater than the required voltage. I would like to share my calculations.

Or, is it right to use the energy calculations for a capacitor.

Which efficiency parameter is to be considered if I am using 2 x TPSM53603. Of TPSM or my 12V DC-DC converter used before TPSM

I am not able to understand how to proceed. Sharing a snip below. I understand that this may not be the best way to deal with interruption. But is there a way to make things calculated and better in this existing design?

a. 12V DC-DC converter output. 11.5V because the schottky diode drops about 0.5V

b. 3.3V using TPSM53603RDAR. Load around 1.5A. Is this 2200uF really helpful and needed? Does power interruption work if capacitors are placed on 3.3V or 5V rails?

c. 5V using TPSM53603RDAR

Hello,

I don't think putting the large capacitors on the output of the buck converters will help if the goal is to maintain the input voltage. 

It looks like you are on the right track for the capacitance calculation, but it is difficult to estimate the fact that the output current of the capacitor will change as the capacitor is discharged. Make sure to follow i=c*dv/dt. Also keep in mind that your efficiencies will likely be different for the different output current and voltages you have setup on the two converters. I recommend using similar data found in the datasheet for the efficiency estimation. 

Have you tried simulating this with 2200uF yet? 

Best regards,

Ridge

Yes. I tried Ridge.

1. With 2200uF

2. Recommended 47uF

Warning before simulation was this

If we do not place 2200uF here, but have one near to one of our high power consuming chips, will it be fine?

Hello,

Having too large output of an capacitance on the buck converter can cause instability. 2200uF of output capacitance is too much for the TPSM53603 devices. I cannot comment on how it will impact devices at the point of load.

Best regards,

Ridge

Ridge Lahti Okay. Thanks Ridge. Will take care of that.

So, if I want my capacitor to discharge above 6V in 20ms considering Vout+1 threshold of TPSM53603, what is the correct way to estimate the Hold up capacitor value? I have 2 x TPSM53603 each giving out 3.3V and 5V. Let's consider the minimum UVLO as 6V for 5V output TPSM.

A. If I calculate according to C hold-up = 2 * P out * t / Eff * (V² cap - UVLO²), and if I calculate voltage of a capacitor with lesser capacitance than C hold-up value, that it discharges to after time t, it still is greater than the UVLO value. What seems the right way to calculate the capacitance value?

B. Also, can you shed some more light on this: It looks like you are on the right track for the capacitance calculation, but it is difficult to estimate the fact that the output current of the capacitor will change as the capacitor is discharged. Make sure to follow i=c*dv/dt. 

I want to know how will the capacitor be able to provide the load current if it is decaying?

Hello,

Your equation looks correct. You would integrate the the capacitor equation for your timing and input, which you've already done. It looks like you are sizing the capacitor correctly. 

The capacitor provides the current by i=cdv/dt. Again, your equation looks right.

Best regards,

Ridge

Okay Ridge. 

May seem basic, but is there any justification on how capacitor discharges but acts as hold-up, and how is the capacitor current not affecting the connected load, as discharged current is always less than the load connected? Should it be calculated in terms of energy?

Hello,

For all buck converters, the input current can be calculated from the total power in and the total power out: Vin*Iin*efficiency= Vout*iout

Since buck converters are step-down converters, the input current will be less than the load current. Your capacitor equation looks right where you are calculating with the total input current of the buck converters in mind with respect to the change in time and voltage. 

Best regards,

Ridge