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Rectification, Super-Caps, Output Smoothing Circuit

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ACharnley

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Hi,

I have a current limited power supply driving an active-rectifier which will operate a PWM to charge a super-capacitor. In the attached circuit the PWM has been replaced with a diode but the principle is the same (say 100% duty cycle).

Because the super-caps overwhelm the power supply the voltage increases steadily up to 5V at which point the PWM will end. At around 4.5V the power may enable to a USB device and since there's no inductance on the output, that's where the fun starts.

According to LTSpice the ripple is minimal but when I built it in real-life at around 4.75V the USB device I had attached (a Garmin) had visual issues and crashed. I believe (I'm awaiting a new oscilloscope to verify) that the ripple when the capacitor reaches saturation point is greater than simulated. I haven't built it with a PWM yet only using a diode, but that's my theory.

As the AC frequency is low this requires a very large inductor (10H as shown) to smooth the ripple. I'm not an analog engineer and do wonder if there's another way?

Regards,

Andrew

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Not sure what you mean by saturation point of the capacitor, but...

Do you not have a capacitor on the output side of the inductor? A large inductor next to the load without a sufficiently large capacitor is going to cause the load to experience huge voltage drops whenever there is a transient current increase causing brownouts and such. Conversely, whenever current demand suddenly drops you're going to get a voltage increase that the cap needs to be there to absorb or in worse cases suppressed or clamped because it's basically inductive flyback. Basically the transient response of your power supply is too slow and there is insufficient decoupling.

You would not see this in the simulation since you are using a static load rather than a dynamic one. Replace the load with ...I don't know...two resistors in parallel with each other where one resistor is in series with a switch that is constantly switching on and off and watch the voltage dip and spike rather than maintaining that nice ripple you originally saw.

But if you want to filter out ripple, have you tried using a multi-order LC filter? Where you just chain a bunch of LC filters together? You can get away with smaller inductors and capacitors if you do that. You just have more of them and the dropoff should be higher than if you throw the sum of all those Ls and Cs into a single order LC filter.
 
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OK! This is above my working knowledge but I'm keen to learn it. You're right about the inductive flyback... the Garmin is toast. :)

The multiple LC approach sounds great, I certainly can't squeeze 10H onto the PCB. I'll work an update to the circuit and see what I can come up with. Thanks!
 
OK! This is above my working knowledge but I'm keen to learn it. You're right about the inductive flyback... the Garmin is toast. :)

The multiple LC approach sounds great, I certainly can't squeeze 10H onto the PCB. I'll work an update to the circuit and see what I can come up with. Thanks!

Yeah with 1 or 2H you get zapped just by running a 9V battery through it and disconnecting it. Happened at work with a giant coil we were testing. And you have 10H!
 
I was only running something small as a tester, probably 22uH but it fried it anyway. I'm a bit lost here, I get very little ripple without the inductor but the Garmin crashed so something's up. Adding the LC seems to be adding ripple. Perhaps it needs the oscilloscope to see what is really going on.

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I was only running something small as a tester, probably 22uH but it fried it anyway. I'm a bit lost here, I get very little ripple without the inductor but the Garmin crashed so something's up. Adding the LC seems to be adding ripple. Perhaps it needs the oscilloscope to see what is really going on.

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Well, part of the reason is your L is much lower than what you had before so that's to be expected. Also, now you appear to have a dynamic load so that's to be expected too. Put a equal value resistor in parallel with your dynamic load so it's a bit more reflective of something real. It's pretty harsh for a load to just suddenly draw no current momentarily.

It could also be resonance of any LC filter you put into the circuit which is one "gotcha" of using LC filters.

But do you even need the inductor there to begin with? It may not be necessary depending on what you're after.
 
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I assumed it crashed the Garmin (this was before I blew it using the inductor) due to ripple/over-voltage, but LTSpice isn't showing much at all. Even with the LC the ripple limits are minimal.

Effectively my plan was to over-load the power supply with a suitably sized super-cap knowing that any transients wouldn't get through, so long as I cut power before it reached it's saturation point.
 
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What is your voltage source? 16 volts 0.5A under a short. + A huge inductance.

A dynamo hub. I'm using 16V and that frequency for the simulation, the AC runs from about 10-100Hz and the voltage from 0-36V.

Under a short it's nigh on 0V - a short collapses the field.
 
Playing with some values, but I doubt the ripple/noise here would have too great an effect.
 

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A Charnley,

I think your whole plan has a flaw. (another one)
Compare Super Cap to a Battery.
1) Pretend we have a 5V battery. Likely it will have a voltage of 4.75V at 10% charge and 5.25V at 90%. You can run the USB directly off the battery. Just like your project states.
2) A capacitor stores energy by CxVxV. Capacity times Voltage squared. 4.75^2 =22 while 5.24^2 =27 With a capacitor, you will charge until you have 27 unites if power, then discharge down to 22, then back up to 27. So the ability to store (and use) is from 22 to 27.
x) Battery can store from 10 to 90 while the cap is storing from 22 to 27. Most of the cap is not used.
dynamo hub
I think the dynamo needs to be run where it works best.
AND
The capacitor needs to have a very large voltage swing, to store more power.

Over at Linear.com they have some examples where they charge up a cap to (any voltage) then use a PWM to regulate to get 5.0V out.
Example: Capacitor = 10V. Output is 5V for USB. Then use a "buck/boost" PWM to take any voltage from 3V to 10V and make 5V. Now the capacitor voltage can range from 3V to 10V. (V squared = 9VV to 100VV) Now you can store and use power from 9 to 100 units. The ratio of 9:100 is much larger than 22:27. (and the output is 5.0V at full charge and at minimal charge)

My point is use a battery, or battery + PWM OR use a capacitor (with a large voltage swing) + PWM.

I reread and I was not clear: The way you have things use a battery not a cap. Use a cap with a very large swing from Vmin to Vmax, then regulate to 5.0V.
 
Hi Ron,

This is by no means the whole circuit, there's a boost working from 0.95v, however the USB side of it will only work from about 4V upwards. This is part of the spec, below this the power is used elsewhere. The last part of my long winded project is ensuring > 5V doesn't reach the USB. I haven't quite managed it.
 
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