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Calculating Power Budget

engr_joni_ee

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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 ?
 
Hi,

Why no skech?
Why no links to the datasheets?

All information you ask for should be given in the datasheets. Did you read them? What do they say? Wat exactly is unclear?

Klaus
 
I am not sure if we need datasheets to calculate power budget. The load requirement is given also with the number of loads.

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
 
Not
I am not sure if we need datasheets to calculate power budget. The load requirement is given also with the number of loads.

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
Not really.

You need to consider that your supply has to be able to deliver the peak currents that your switching converters actually draw during the ON-time. On top of that, you need to provide some headroom.
 
Generally how much is that peak current to be supplied by the power supply that is needed at the ON time of the switching regulator ? and how much the headroom is sufficient.
 

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