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I am busy thinking of how to test a power supply i am designing and came across an article which discussed using an electronic load to test the supply. I will not have access to an electronic load and so thought of using a dummy load to mimic the load the supply will see. I'd like to clarify if I have the right idea of how this works?

The primary requirement of this supply will be to provide a steady 5v and 3.3v through the use of two buck converters with a 12v input voltage and to meet the current requirements of circuits connected to the 5v and 3.3v lines. The 12v input voltage is derived from 230v 50Hz AC through means of a step-down transformer and full bridge rectifier. I have a list of components and max currents they draw. Am I correct in saying I will need to calculate the sum of current draw from the individual components and then use Ohms law to pick a dummy resistance to simulate this condition? Similar to the procedure described here: Determining Power Supply Requirements

For example. If components which draw from the 5v line will draw a peak current of 2.4A I would design the buck converter to output a regulated 5v 3A and then calculate the load resistance as \$R=\frac{V}{I} = \frac{5}{2.4} = 2.083 \Omega\$. I'd then connect this load across the output of the 5v buck converter and check that it is supplying a steady 5v for this load? I assume I would also need to look at the power rating of the resistor so using \$P=VI = 5\times2.4 =12W\$ making sure that the resistor chosen is able to handle this power amount of power dissipation?

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I will not have access to an electronic load and so thought of using a dummy load to mimic the load the supply will see. I'd like to clarify if I have the right idea of how this works?

This is the way most people would approach this.

Am I correct in saying I will need to calculate the sum of current draw from the individual components and then use Ohms law to pick a dummy resistance to simulate this condition?

The dummy loads are chosen with this method and for your worked example the duumy load would indeed be 2.083 ohms but,, if you are designing the converter to supply 3 amps then test it at a 3 amp load is my advice. It's always a good idea to know you have the spare overhead capacity just in case you need it. I'd go for a 1.67 ohm load and this will produce a power of 15 watts and to prevent the resistor from exceeding its maximum limits I'd consider a power rating that is 1.5 times this value i.e. 22 watts.

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