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I want to design a charger circuit in my project.

At first I have an IC (AD694) to produce 4-20 mA current, then must charge one 100F, 2.7V supercap.

How can do it?

I want to use a breadboard for this circuit.

I know some ICs such LTC3326 and some LTM can be used. But I want ICs I can use on breadboard.

Maybe some one can offer other way.

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  • \$\begingroup\$ Please use space after each comma and period. As long as you limit the compliance voltage to 2.7 V or less, you should have everything you need. Draw a schematic and we’ll take a look at it. \$\endgroup\$
    – winny
    Mar 29, 2018 at 7:20
  • \$\begingroup\$ Why do you want to do this? how much load is on the supercap? \$\endgroup\$
    – Jasen Слава Україні
    Mar 29, 2018 at 8:18
  • \$\begingroup\$ I want use this supercapacitor as power supply to operate a sensor wich need 200ma input current.I shude charge and discharge the supercap fast as possible. \$\endgroup\$
    – isa.b
    Mar 29, 2018 at 9:26
  • \$\begingroup\$ Why not make your sensor 3-wire instead? \$\endgroup\$
    – MrGerber
    Mar 29, 2018 at 21:47
  • \$\begingroup\$ I have to use one 12V DC power supply at first. After than 4-20mA I shuld charge and discharge the supercap in about 8 minutes. Because I want discharge it in to the sensor wich need 200mA to operate. I should work with 4-20mA ,this is my first part of project. \$\endgroup\$
    – isa.b
    Mar 30, 2018 at 5:50

1 Answer 1

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In theory, it is simple. You connect your 4-20 to the supercap with the correct polarity, and then the circuit will drive a constant current in to charge it. BUT it will raise the voltage to whatever is necessary to drive that current, which will almost certainly be greater than 2V7, and will damage the supercap. So, protect it with a zener across the terminals so voltage cannot exceed 2V7. BTW, shorting out a 4-20 current loop will not damage it.

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    \$\begingroup\$ After than AD694 we have 10 volt constant output and I have to limit this voltage to charge 2.7 V super capacitor.charging time is important for me \$\endgroup\$
    – isa.b
    Mar 29, 2018 at 9:09
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    \$\begingroup\$ @isa.b You are using constant current, so the time to charge to 2V7 does not depend on your maximum output voltage. You can only dump a maximum of 20mA into it, irrespective of voltage. \$\endgroup\$
    – Dirk Bruere
    Mar 29, 2018 at 9:27
  • \$\begingroup\$ BTW, using 20mA to charge a 100F capacitor to 1 Volt will take more than 1 hour, so to fully charge it from a 20mA source to 2V7 will take almost 3 hours \$\endgroup\$
    – Dirk Bruere
    Mar 29, 2018 at 9:33
  • \$\begingroup\$ Tanks for your guide.but 3hour is very long time.isn't any way to chrge it fast?if the current be 4ma what I can do?! :O \$\endgroup\$
    – isa.b
    Mar 29, 2018 at 9:56
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    \$\begingroup\$ If you only have 20mA or less to charge the capacitor there is nothing you can do to make it faster. You need an external power supply such as a solar cell or small wind turbine \$\endgroup\$
    – Dirk Bruere
    Mar 29, 2018 at 10:15

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