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I have a solar panel rated at 5V 80mA and want to increase the output current significantly higher to around 500mA and decrease the voltage to either 3V or minimum of 2.7V, how can I go about this?

I understand physics, but I need to know can a buck converter increase current this high from a step down of 2V?

I am trying to power a Module that needs 2.6-3.6V and a minimum of 400mA to turn on. But then I only need an average of 50mAh to stay on during the day.

The solar Panel below: https://store.sundancesolar.com/5-0v-80ma-solar-cell/

Also I have been thinking about using a Supercapacitor because I only need 500mA for 1 second to startup my module. And then an average less than of 5mAh to keep it on.

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  • \$\begingroup\$ you would have to charge a battery ... then you could get 500 mA, but that would stop when the battery discharged, then you would have to wait for it to charge again \$\endgroup\$
    – jsotola
    Commented Jun 30, 2020 at 20:36
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    \$\begingroup\$ The laws of thermodynamics forbids this. 5 V and 80 mA going in and 3 V and 500 mA going out. Do you see the issue here? \$\endgroup\$
    – winny
    Commented Jun 30, 2020 at 20:48
  • \$\begingroup\$ Do you really mean 50mAh or 50mA? The two are completely different things. \$\endgroup\$
    – Simon B
    Commented Jun 30, 2020 at 21:07
  • \$\begingroup\$ @Simon B I need an average of 50mA per hour to keep it on. I put my module into deep sleep mode twice a minute(20 second intervals). It wakes up for 3-6 seconds per minute. \$\endgroup\$
    – Sherman
    Commented Jun 30, 2020 at 21:15
  • \$\begingroup\$ You are right. There is 2 current spikes of 50mA per hour only for 2ms each, but overall current consumption is less than 5mAh. \$\endgroup\$
    – Sherman
    Commented Jun 30, 2020 at 21:58

2 Answers 2

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You need to understand that input and output power from buck converter are the same, minus converter losses. Say on input you have 10 V and 10 A, with 50% duty cycle you will get on the output 5 V and 20 A. Now you say that you want to go from 5 V to 3 V, which is 60% duty cycle, in which case current goes from 80 mA to 130 mA.

Can you provide a datasheet for that PV panel? And please describe your application, there are many possible ways to get what you need. For example you could avoid buck converter by connecting a 3-V battery directly to the PV panel, in which case you will get surge current capability and voltage you need.

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  • \$\begingroup\$ I honestly forgot about that. Engery of conservation... but I cannot use a battery for this. I seen other ways like using a supercapacitor. I only need 500mA for less than 1 second to start my module and after that a consistent average of 50mAh at 2.7-3.6V. \$\endgroup\$
    – Sherman
    Commented Jun 30, 2020 at 20:59
  • \$\begingroup\$ Calculate what would the needed capacitance. What is the allowed voltage deviation? Exactly how long do you need surge current? You calculate the energy needed, and then capacitance. Try to simulate stuff... \$\endgroup\$
    – Marko Gulin
    Commented Jun 30, 2020 at 21:02
  • \$\begingroup\$ I will start to simulate soon! I think I need to test circuits out to understand what I have so far. \$\endgroup\$
    – Sherman
    Commented Jun 30, 2020 at 21:36
  • \$\begingroup\$ How can I simulate a solar panel in Multisim or any circuits platform? \$\endgroup\$
    – Sherman
    Commented Jun 30, 2020 at 22:22
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A buck converter by itself can't do what you want..

5V at 80mA is 0.4 watts. 2.7V at 500mA is 1.35W. You want to get more power out than you're putting in.

The only way it can work is to use your solar panel to charge an energy storage device. That could be a supercapacitor if your power on surge is very short, or a rechargeable battery to provide power for longer.

Bear in mind that if the solar panels gives a claimed 80mA at 5V, then that's in full sun. The moment a cloud goes over, the power will drop a lot. If you need to handle that, it will need a battery, not a capacitor.

As you discharge a capacitor, the voltage across it will drop. A simple case is that if you take 1A from a 1F capacitor for 1s, it will drop by 1V. Or take 500mA for 1s, from a 1F capacitor, it will drop by 0.5V.

So if you took a 1F capacitor bank, rated not less than 6V, and put it across the input of your buck converter, then that should be enough to provide 500mA for 1S, only dropping from 5V to 4.5V in the process.

You would need a buck converter capable of delivering at least 400mA to the load.

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  • \$\begingroup\$ Yes. I have an understanding of my wattage output is too small, but I have seen applications where I can use a supercapacitor to provide a charge to get 500mA. How can I calculate the capacity of the supercapacitor? And what if I want to use only 500mA for less than a second then just have a steady flow of 50mAh right aftwrdas. Is this possible? \$\endgroup\$
    – Sherman
    Commented Jun 30, 2020 at 21:02
  • \$\begingroup\$ Okay that makes some sense! And after this one second my module spikes to 50mA once or twice per hour for 2ms each, but uses a power consumption of less than 5mAh. I didn't mean 50mAh. Would I be able to consistently deliver power? \$\endgroup\$
    – Sherman
    Commented Jun 30, 2020 at 21:56
  • \$\begingroup\$ @Sherman It should work all the time that the solar panel is in full sun. But it won't run for very long without the sun. \$\endgroup\$
    – Simon B
    Commented Jul 1, 2020 at 8:04

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