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I have purchased a circuit that operates in an unusual way. This circuit firstly only as a 5V DC input terminal to power the 5V components. in addition, the 5V DC input voltage is also fed to a 12V DC boost converter. This 12V powers the 12V components.

My problem is this. I am currently running the circuit via the mains (the power supply outputs 5V DC which goes to the input terminals of the circuit). I would also like the option of running this circuit via a portal power supply such as an SLA battery. However, I only have the option of using a 12V 7Ahr DC SLA battery. (the entire circuit pulls about 840mA/hr)

First I would use a buck converter to step down from 12V DC to 5V DC then input to the circuit.

My question is, will the battery get drained quicker if use so many voltage converters i.e. I step down the voltage to 5V using a buck converter, then using the boost converter on the circuit to step the voltage back to 12V DC to power the respective components?

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You may not consider this answer entirely helpful as I do not intend so solve your specific problem. However, I would like to point out the universal way of solving such electronic questions.

As you well know, with each conversion cycle there are some losses. Hence, one can safely say that the more conversion cycles you have (boost, buck etc) the more your losses will add up. For the kit that you use, have you checked whether you can access any datasheets and do those datasheets provide any information about the efficiency of the converter? If the answer is yes, then you can figure everything out easily!

Let's assume your buck converter is 80% efficient and that your boost converter is 80% efficient as well. Then, working backwards, your supply drawing 840mA of current would draw 840mA / 0.8 out of the 5V rail. That is 1050mA. Following, this would mean that your 12V supply (the battery) will see 1050mA / 0.8 = 1312.5mA. Do the same analysis for an alternative circuit and compare the power draw.

Just draw your circuit, figure our all the conversion cycles and then calculate your expected draw.

One thing to note, however, is that the efficiency stated in the datasheets for any power converter (big or small) is usually the peak efficiency. That peak efficiency is met at a particular set of input and output voltages and currents. For instance, if you have a 1kW converter and you only used 50W of power you can easily expect 50-60% efficiency with some technologies, even if the datasheet mentions an 90% efficiency. That 90% efficiency is met at around 80% of max power (800W in this case). Hence, in order to avoid turning this small problem into an academic issue, I recommend you first do some estimations using the calculations above and then actually have a go at measuring the currents. Do you have access to measuring equipment?

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  • \$\begingroup\$ Thanks for the example. I do have a digital multimeter \$\endgroup\$
    – JoeyB
    Mar 19, 2019 at 21:58

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