Capacitors & conservation of charge

• posted

Hi all,

While messing with buck/boost circuits I had some thoughts which don't hold up in real tests....

A simple example (aside from losses) is that if transfer energy from say 10V 10uF into 1uF the voltage will increase to preserve the charge.

Now, in my circuit, I charge 25V 1,000uF capacitor and charge a 10uH inductor. When the switch turns off, all the energy should be in the

10uH inductance. So if I have a 100pF capacitor, then the voltage should be like 100,000volts according to my workings out.

Now I ran a computer simulation on this, at best I can only obtain

12KV on the 100pF. So most of the energy is lost in switching losses I assume.

In realworld tests, I end up with less voltage than I started out with, So I am trying to find out why ?

I know charging 22uH inductor at 100khz can be used as a buck/boost supply, I built a simple 12V to 30V inverter, can switch 10amps easily. Though I am not running at 100khz, only 100hz. Though the current pulse rises to something like 500amps over 500uS.

I am not sure I follow all this exactly, Or even if it will work ? AFAIK, The longer a inductor has current pumped across it the more charge it obtains over time. So at turn off, all the energy given to a coil is recovered. It works well, even with my simple buck/boost circuit.

So I am slightly confused as to why pushing 500A into a coil has no effect. I can only assume I have a huge loss somewhere, Or I do not follow the idea correctly ?

Cheers, Chris

• posted

You need to aquaint yourself with the buck-boost topology and theory of its operation. If you go to TI's web site and download their buck- boost app note no. SLVA059, studying this thoroughly, you will understand why.

In a nutshell, the 1000 uf input capacitor did not transfer ALL of its energy to the inductor. The output voltage is related to the input voltage and the duty cycle as follows:

Vout =3D Vin * (D/(1-D)) not accounting for losses.

The energy transferred to the inductor is half the inductance times the difference between the square of the final current and square of the original current. This is determined by the voltage across said inductor multiplied by the on time. During the power switch off time, the volt-seconds must equal that of the on time. The app note covers this. The value of the input capacitor does not determine the output voltage. The same goes for the output cap value. The values of both caps, however, are important regarding ripple. noise, and transient response.

Did I help? BR.

• posted

Inductors do not hold charge so, so you can't talk about "charging" and inductor. Instead, you induce a magnetic field, or "energies" an inductor.

Yes, in normal circumstances in electronics, charge can neither be destroyed nor created. The description of your posting is a bit too vague for me to answer remainder of your questions.

• posted

i haven't gotten the whole detail, but you seem to start with a bit of a misunderstanding which i will now endeavor to correct:

you charge a capacitor by loading a voltage into it, then open circuiting it; but, an inductor being essentially the opposite, you charge it by loading a current into it and shortcircuiting it; whereas a charged capacitor holds its energy in the electric field with zero current, a charged inductor holds its energy in the magnetic field, with zero voltage. If you short circuit a capacitor it loses its charge, if you open circuit an inductor it loses its current. In real life, the series resistance in the inductor windings decays the charged current a lot more quickly than the leakage resistance in the capacitor decays the charged voltage.

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So where is my misunderstanding? What have you "corrected"? I know the operation of inductors and capacitors. I looked over my previous post and found nothing in need of correction. What specific point have I made that you take issue with?

I've been developing buck-boost converters and SMPS for more than 2 decades, and have 2 published papers on SMPS in world wide peer reviewed magazines. I've been a regular attendee at the Unitrode/TI power seminars since the mid-1980's. I know my power supplies. I referred you to a TI app note so that you can acquaint yourself with the basics of buck-boost operation. Reading it is my advice to you.

Are you really looking for help. or someone to argue with? I offer you free expert advice, and then I get a lecture on L and C basics? Best regards.

Claude

• posted

This group is full of loonies.

I'm considering doing a bulk abuse report to keep them out of the sci or energy groups.

Graham

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