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I just started my journey into electronics and have successfully built a 2 digit 7 segment counter (using IC 40110 and a pushbutton) on a breadboard. I used the 3.3V and ground pins of Arduino to power my breadboard.

I am trying to make a standalone circuit which can be powered by a battery. The goal is to increase the lifetime as long as possible while the circuit serves its function. Once I do this on my breadboard, I plan to design PCB for this circuit. But I feel stuck by not understanding the best way to power my circuit and need your help on this.

Q) How can I determine the power consumption of circuit? Should I measure the current and votlage across each of the 7 LED segments and sum their product?

Q) Is determining the input voltage of my circuit important? If so, how can I determine the minimum input voltage required? (I assume I should go for the minimum operating voltage so that I can reduce power consumption)

I researched online and found few ways of doing this:

A) Use 9V battery connected to this power supply module:

enter image description here

B) Use 9V batter and connect it to a down converter

Method A seems to be the easiest way but it uses linear regulatobrs which is less efficient than the buck converter. Q) Apart from simplicity, when/why would anyone use this?

I think Method B is the way to go because they are efficient than linear regulators but I don't know how much difference it will make. Q) How can I compute and compare these 2 methods?

Thanks a lot in advance!

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  • \$\begingroup\$ Why use linear regulators? Simpler, yes. Efficient, no. Do not create noise like a buck reg, yes. Rarely do you get a circuit that is ‘perfect’ - there’s almost always a tradeoff, so linear regs have their advantages and switch regs have a different set of advantages. You choose the one which gives the least compromise to your circuit. Sometimes you use both to capitalize on both their strengths. \$\endgroup\$
    – Kartman
    Commented Jul 29, 2023 at 8:50

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The power consumption of the circuit is going to be dominated by the LEDs in the 7-segment display unless you have some weird and extremely inefficient other parts of the circuit. Let's say you're powering your LEDs with 10mA each; that's 70mA x 2 displays for 140mA total (the 40110 itself draws nothing in comparison to this). If you have a 9V battery and want to drop the voltage to 5V, you will dissipate 4V*140mA in the regulator and make 560mW of waste heat, which would heat up a standard 7805 in a TO-220 package about 14C, not bad but as you said, you could be more efficient (the 7805 is 5/9 or 56% efficient in this application).

There are many drop-in 78xx regulator replacements for a few dollars that require no external parts with efficiencies in the low-to-mid 90%s. Specific recommendations are not allowed here, but if you search "linear regulator replacement converter", you should find what you need. Just make sure you get one with a minimum input voltage of ~4.5-6.5V; 9-36V is a very common input voltage range that won't work in your application.

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  • \$\begingroup\$ Thank you for the response. It clears things up. But why would anyone use linear regulators? For simplicity of the circuit? \$\endgroup\$ Commented Jul 29, 2023 at 0:22
  • \$\begingroup\$ And how can I determine what is the mind voltage required? Or does it not matter? \$\endgroup\$ Commented Jul 29, 2023 at 0:24
  • \$\begingroup\$ Linear regulators are still fairly commonly used with light loads or when the difference between the input and output voltages isn't large (e.g. making 2.8V from a 3.3V bus). They are generally simpler than switching regulators and do not have switching noise. 9V batteries are "dead" (i.e. the voltage begins to drop very quickly) once they are discharged to about 5 or 5.5 volts so a regulator with a minimum input voltage around or below that will allow you to extract most of the energy from the battery. \$\endgroup\$
    – vir
    Commented Jul 31, 2023 at 18:38

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