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I am planning to design a magnetic energy harvesting circuit with a current transformer (CR8349-1500) to charge a super capacitor.

Load current rating is 0 to 35 amperes. From the current transformer datasheet, it is mentioned that the turns ratio (Te) is 1520, which means 0.6mA secondary current is produced for 1A primary load current.

I made a bridge rectifier circuit for AC/DC conversion and used a 3.3V Zener diode for charging the super capacitor. I have attached the diagram below:

enter image description here

My requirements are:

  1. Primary winding (load wire ) should be one, because of the CT hole size
  2. Capacitor needs to be full charged at 2.4V
  3. Capacitor should discharge at 1.8V with constant 10 mA.

The capacitor I am using is 1 farad that needs to power the microcontroller.

The problem is the super capacitor takes sometime to charge. I am expectit takes more than a minute to charge, but I need to power the microcontroller within seconds rather than waiting for several minutes to charge the super capacitor.

Is there any circuit or IC that can boost the charging process of this super capacitor?

Datasheet for the current transformer.

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    \$\begingroup\$ It's not clear to me what you're asking. You really have two options here... First is to use a smaller capacitor that can charge more quickly. The second is to get more power into it so that it charges more quickly. I suppose you could also design some sort of 2-stage network so that you can power the micro from the circuit immediately and then charge up the capacitor over time. \$\endgroup\$
    – jwh20
    Mar 29, 2022 at 15:00
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    \$\begingroup\$ it looks like you are designing an AC power supply ... why are you calling it magnetic Energy harvesting circuit? \$\endgroup\$
    – jsotola
    Mar 29, 2022 at 15:12
  • \$\begingroup\$ @jsotola probably because it's harvesting energy from the magnetic field around another circuit \$\endgroup\$ Mar 30, 2022 at 15:03

1 Answer 1

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Rough calculations


Your capacitor needs to acquire 2.88 joules of energy to charge 1 farad to 2.4 volts. That's basic energy equations for a capacitor. You want to do this in maybe 5 seconds rather than 1 minute? OK, in 5 seconds the average power throughput needs to be 2.88 / 5 watts = 576 mW.

The CT input power is 35 amps at some small voltage on the CT primary. That small voltage needs to be 576 mW divided by 35 amps = 16.5 mV RMS.

Is 16.5 mV feasible across the input of the wire that passes through the core of your CT?

It's a single turn and, it might have a localized inductance (due to the CT core) of 1 μH so, what voltage is developed across 1 μH when 35 amps flows. Let's assume an AC frequency of 50 Hz (noting that 60 Hz will give a slightly better result).

1 μH has a reactance of 0.314 mΩ at 50 Hz.

Multiply this by 35 amps and you get a terminal voltage of 11 mV. Not enough to charge the capacitor to 2.4 volts in 5 seconds.

But, it's worse than that because as the capacitor charges, the secondary voltage grows to a level that is larger than what it is intended or rated for. This means that the core will be likely start to saturate well before the capacitor reaches the value you want.

This causes heat and it also reduces the effectiveness of liberating a decent voltage from the secondary and things start to slow down AND, maybe a minute might be all you can expect from this idea.

Of course, if you had a much bigger CT you could get a quicker result.

You do the math.

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  • \$\begingroup\$ would you recommend me some big CT's?. It will be helpful \$\endgroup\$
    – Sri
    Mar 30, 2022 at 12:44
  • \$\begingroup\$ @Sri sorry I can't recommend any CTs but, if you find one that might work, just copy a link into the comments here and I'll take a look. You should also take the 2 minute tour to understand why folk give there time freely to help others. \$\endgroup\$
    – Andy aka
    Mar 30, 2022 at 13:23

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