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I have a dc source which gives 5.5V and I want to know how to give only 2.7V to the supercapacitor for charging. I'm thinking about this circuit : enter image description here What are the limitations of the circuit? How can I improve the circuit ?

Will it work ?

Thanks for your help.

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  • \$\begingroup\$ It will take a long long time to charge .Remember that a farad is an Amp second per volt . \$\endgroup\$ – Autistic Jan 19 '17 at 9:51
  • \$\begingroup\$ I see a battery in the schematic, but then the text says capacitor. Huh? What? This is unclear. \$\endgroup\$ – Olin Lathrop Jan 19 '17 at 12:02
  • \$\begingroup\$ Yes sorry I put a battery because there is no supercapacitor on Proteus but I'll do the circuit with a supercapacitor \$\endgroup\$ – heisenberg Jan 19 '17 at 12:42
  • \$\begingroup\$ You'll get better results by using an (adjustable) voltage regulator like for instance the LM317 set to 2.7V. Maybe also note that many super caps are only rated for 2.5V and that those caps generally live longer when not pushed to the max permissible voltage. \$\endgroup\$ – JimmyB Jan 19 '17 at 13:27
  • \$\begingroup\$ I thought of it too, but I have to keep the price of the system the cheapest possible so I want to have the less component possible \$\endgroup\$ – heisenberg Jan 24 '17 at 9:03
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The 1N5223 zener is nominally 2.7 volts and can be as low as 2.565 volts or as high as 2.835 volts so if you can live with this variation then that's fine.

The BAT54 being schottky will look like a small volt drop as the capacitor reaches full charge so the only problem here is what the residual current taken by the capacitor is - the higher the residual, the more the BAT54 drops voltage. It could be about 0.1 to 0.2 volts at 100 uA.

Is this good enough? Only the OP can decide.

If you want something more accurate I'd use a linear voltage regulator to produce 2.7 volts and feed that to the supercap. If you had a couple of more volts available at the input there is this Linear technology solution: -

enter image description here

LT describe it as: -

The LT3663 is a 1.2A Step-Down Switching Regulator with Output Current Limit that is ideal for use as a supercapacitor charger since it provides adjustable output voltage and adjustable charging current limit.

I'm sure there will be switching or linear solutions out there that will do the job rather than rely on the tolerance of a zener diode (+/- 5%) and the unknown leakage current of the supercap.

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  • \$\begingroup\$ I thought of it too, but I have to keep the price of the system the cheapest possible so I want to have the less component possible \$\endgroup\$ – heisenberg Jan 24 '17 at 9:03
  • \$\begingroup\$ @heisenberg then you have to trade performance with price and only you know how to pitch this equation. \$\endgroup\$ – Andy aka Jan 24 '17 at 9:07
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It should work (i didn't look on zener datasheet, but you probably took 2.6V or something). But i would suggest a bit more interesting circuit: a constant current charger (PNP biased with a resistor) and a comparator that would measure supercap voltage and switch the PNP off once 2.8 is reached. enter image description here

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I tested my system IRL and it works, I have the 2.7V I want. Using the LT3663 is a great solution but as I answered to some people, I have to keep the price of system low and therefore keep the number of component as little as possible.

Thanks everyone for your help.

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    \$\begingroup\$ If you are going to answer your own question then answer those questions you have raised. If you just want to comment then leave this as a comment under your question. The questions you asked are: What are the limitations of the circuit? How can I improve the circuit ? Will it work ? \$\endgroup\$ – Andy aka Jan 24 '17 at 9:10
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I believe not, here is why: I don't know how precise your 2.7V has to be, so you're going to have to make a decision here. :)

The zener voltage is not precise. I had a look at Fairchild's datasheet, It says that Vz = 2.565V Min / 2.8V Max @Iz, add to this the temperature drift you're gonna be off.

Moreover, even if you have a precise 2.7V zener diode, you have to take the voltage drop across your D1 into consideration.

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  • \$\begingroup\$ Why does this answer have a downvote? \$\endgroup\$ – Bort Jan 19 '17 at 14:54

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