I am calculating a power estimate for a circuit board which is a power distribution board.

The main input to the power distribution board is 20 V. This 20 V on the power distribution board is connected to two DC-DC converters having adjustable outputs. There are two identical DC-DC converters on the power distribution board.

Now each DC-DC converter is powering up further three linear voltage regulators. Each linear voltage regulator has an output connected to the load that require 1 A current and the voltage requirement of the load is 2 V.

If we set the output of the DC-DC converter to 2.5 V and feed 2.5 V to three linear voltage regulators then what will be the input current of each linear voltage regulator ? Given that each linear voltage regulator output is set at 2 V and the load current is 1 A. The input to each linear regulator is 2.5 V. The voltage drop across each linear voltage regulator will be 0.5 V which multiplied by the current is the power dissipation which is like 0.5 Watt per linear voltage regulator. In total there are six linear regulators on the power distribution board.

How can I calculate the input power for linear regulators ?

I am doing this

Power = 6 x Voltage X Current
Power = 6 x 2.5 X 1
Power = 15 Watts

How do I calculate the input power at the DC-DC converters which will be the power budget of the power distribution board. There are two DC-DC converters. How can I calculate an estimate for the power distribution board ?

 

Here are my calculations.

The input and output power of the linear regulator can be found as

Pin = V X I = 2.5 X 1 = 2.5 Watt

Pout = V X I = 2 X 1 = 2.0 Watt

The voltage drop in the linear regulator will be 0.5 V which translate in to 0.5 Watt power dissipation.

There are six linear regulators. The input power to the linear regulators will be 6 X 2.5 Watt = 15 Watt.

Because there are two DC-DC converters so each will have to deliver 7.5 Watt. In other words the output of the DC-DC converter has to be 2.5 V and delivering 3 A current.

How about the input current to the DC-DC converter ? Given that the efficiency of the DC-DC converter is 0.8.

Does it mean that the input power of the DC-DC converter will be 7.5 Watt / 0.8 = 9.37 Watt for each DC-Dc converter. Is that correct ?

The input current to each DC-DC converter will be.

Input Current = Input Power / Input Voltage
Input Current = 9.37 Watt / 20 V
Input Current = 0.46 A

The total input current for both DC-DC converters will b than 0.46 x 2 = 0.92 A. Is that correct ?

The power budget to the power distribution board is = 20 V X 0.92 A = 18.4 Watt

 

Correct... though 9.37W @ 20v = 0.47A to 2 SF. = 18.8W

DC-DC converter efficiency is only 0.8 at a specific operating point, it could be a lot lower, but rarely much higher, so your power budget is probably nearer 20W or more to be safe. Beware of spurious accuracy...

 

I also have head about the peak current required by the DC-DC converter at the ON time. Generally how much headroom should be have in the input power supply to the DC-DC converters ?

 

Does the 2.5v input to the linear regulators provide enough margin to getv2.0v out?

 

The .5V margin will require a low dropout regulator. Look for one that has a dropout voltage of less than .5V at 1 A.

 

I understand that 0.5 V drop across linear regulators requires LDO voltage regulator. I am considering a LDO with drop around 400 mV.

 

I also have head about the peak current required by the DC-DC converter at the ON time. Generally how much headroom should be have in the input power supply to the DC-DC converters ?

Start-up current for a buck converter can be significant as the input and output capacitors charge up, but its not easy to say 'how much?' as it depends on their design and whether they have soft-start capability. It could be 2 or 3x or not... Best solution is to test one...