# Buck converter source current draw

I'm doing some calculations for a system that is going to use a few buck converters. I need to make sure that the battery will be able to feed enough current but I'm not 100% certain how power on the output and power on the input relate.

Using a power resistor in series the battery will see the same current draw as the lower-voltage circuit.

What draw does it see with a buck converter and is it relative to the load?

• Pin = Pout / efficiency. eg if Pout = 100W and efficincy is 85% then Pin = Po/0.85 = 100/0.85 ~= 120 Watts – Russell McMahon Oct 7 '19 at 11:19
• Using a power transistor in series the battery will see the same current draw as the lower-voltage circuit. When the transistor isn't conducting, it won't... In other words, you need to elaborate this way more. Why using a power transistor in the first place? Is it the buck's switching transistor? What schematic are you going to use? etc, etc, etc – Huisman Oct 7 '19 at 11:20
• Efficiency of any SMPS depends on what load it primarily is designed and what load really is applied. A 10A buck converter may have 50% efficiency with a 10mA load. – Huisman Oct 7 '19 at 11:22
• @Huisman I meant to say power resistor, there are a bunch of flies in my head. – php_nub_qq Oct 7 '19 at 11:23
• by example specs you will learn to understand more details like startup surge current and voltage sag with power losses, Source ESR, output ESR vs load vs load regulation error. and variations from 25kHz to 2MHz Buck converters – Tony Stewart Sunnyskyguy EE75 Oct 7 '19 at 15:45

For each converter,

$$\P_{out} = P_{in} \times efficiency\$$

Therefore,

$$\P_{in} = \frac{P_{out} }{ efficiency}\$$

The total battery current is

$$\I_{in} = \frac{\sum P_{in}}{V_{in}}\$$

The conservation of energy and the regulator efficiency are the key to understand your problem. simulate this circuit – Schematic created using CircuitLab

One thing is certain: $$P_{out} = \eta \cdot P_{in} \\ and \\ P=U \cdot I$$ Thus as an example if your load draws 3 A at a regulated voltage of 3 V and your regulator as a 90% efficciency and a voltage input of 5V we have the following: $$P_{out} = 3 \cdot 3 = 9 = 0,9 \cdot P_{in} =0,9 \cdot V_{in} \cdot I_{in} = 0,9 \cdot 5 \cdot I_{in}\implies I_{in} = 2A$$

Thus now you should have all you need to complete your analysis. Also always work your way from one end to the other. For instance if you know the load and the output voltage work your way back from here. Be careful when putting your shunt in series with your supply because now Vin will change with the current your drawing from V1. Here $$V_{in} = V_{1} - R_{shunt} \cdot i_{in}$$

• What is the purpose of a shunt in this schematic? – php_nub_qq Oct 7 '19 at 11:39
• It represents the power resistor in series with your battery, that you were planning on putting. Which will see a current through it proportional to the current through the load by factor of the voltage ratio between the input and output of the regulator and of its efficiency. This shunt/power resistor will induce a voltage drop, but, as long as this voltage does not bring the input voltage of the regulator bellow its minimum it should be fine. – benguru Oct 7 '19 at 13:40

Quoting from Wikipedia

Switching converters (such as buck converters) provide much greater power efficiency as DC-to-DC converters than linear regulators, which are simpler circuits that lower voltages by dissipating power as heat, but do not step up output current. Buck converters can be highly efficient (often higher than 90%)

Most converter designs will be optimised to be at their most efficient around their maximum rated output. So, for example, if the output were 9W and the efficiency 90% you would need 10W input. As you have a number of converters you need to calculate the maximum required input power of each from their specified efficiency and load. Sum these and that is the maximum power the battery will need to supply.

For example, if you have a battery supply of 12V and 3 converters each with an output of 10 W but with peak efficiencies of 70%, 80% and 90% they would require 10/0.7 + 10/0.9 + 10/0.9 = 38W (rounded) so your battery would have to be able to supply 38W or about 3.2A assuming all 3 converters were on full load at the same time.

If by "Using a power transistor in series [with] the battery ... " you mean as a linear regulator the efficiency will be worse. All it can do is dissipate the excess power in the form of heat.

Taking the above example again and say that the 3 DC outputs required are 3.3V, 4V and 5V all at 10W each then the total current draw will now be 10/3.3 + 10/4 + 10/5 = 7.5A . The battery would have to supply this current in full or 12 X 7.5 = 90W for a useful 30W output. 2/3 of your power would be wasted as heat.

Whether using a buck converter or a linear regulator the power required from the battery will be roughly proportional to the load. The efficiency of converters tends to drop off with part load so there will be a proportionally larger power draw from the battery.

I see in a comment you have changed "power transistor" to "power resistor". This is even worse as not only do you still have to dissipate the power, but you've lost any regulation.