Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
https://overunity.com/5553/privacy-policy/
If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding

User Menu

Custom Search

Author Topic: Question about minimizing losses?6  (Read 2408 times)

captainpecan

  • Hero Member
  • *****
  • Posts: 552
Question about minimizing losses?6
« on: September 18, 2008, 05:57:28 PM »
Surely this has been talked about before, but I cant seem to find any info anywhere about it. Could someone feed me some of their 2 cents and explain it for me?

Here's my question, lets say you are simply passing a magnet over a coil, and it generates 1.4 volts at 25ma. Technically from my understanding that a normal silicon diode has a loss of .7 volts when you pass voltage through it. Knowing this, if you passed that 1.4 volts through a bridge rectifier, which would simply be placing a diode on both cycles, wouldn't that theoretically give you 0 volts out? Basically just dissipating 100% to heat (or close to it). Now this is my understanding of it anyway.

So therefore, if you took the same 1.4 volt 25ma charge, and first  passed it through a step up transformer that turned it into, let's say 100volts. I know the amperage would be virtually non existant at this point, but now would you end up with 98.6 volts after it passes through the bridge? This would virtually allow you to be able to pass a very small charge through a component that before you never would have even gotten it through due to resistance and losses.

Are my thoughts correct? Could someone help me understand if I am wrong?

Second, after this scenario, now we would have a high voltage but virtually useless current due to almost no amperage. If you could step it back down with another transformer, you would again retain usefull amps and end up with like a 1.3 volts at 24ma or close to it anyway. Thereby cutting most of the losses out of the circuit, similar to how a power company does it by ramping up the voltage to send it long ways through the power lines. Hopefully you get my question. I'm not trying to be critical with the numbers, just the concept here.
Now, on a side note, it would be DC coming out of the bridge, and a transformer in my understanding can only be AC, so you theoretically could not even use a transformer to step down the voltage, but will one work with pulsed DC current? And if not, how can you get the amperage back after you passed through the bridge?

Any help understanding these concepts are greatly appreciated!

mscoffman

  • Hero Member
  • *****
  • Posts: 1377
Re: Question about minimizing losses?6
« Reply #1 on: September 18, 2008, 06:53:35 PM »

Yes, That sounds like a good idea. The diodes step forward voltage is just a voltage - it is not power but putting diodes
in series does increase the step size to as large as you want. It's kind off like putting a small battery backwards
into the circuit but the battery never discharges. Power is not lost in the step. For example putting two silicon diodes
reversed in parallel is often seen in the antenna section of a shortwave receiver. Any voltage over ~.7 volts will be shorted
to ground. The Voltage the receiver is looking for is in the microvolt range.

It does effect the voltage transfer efficiency of a device. Use Schotky or Germanium diodes which have a lower step voltage
if the silicon step voltage affects your design. It has to do with the chemical first ionization level of the material used, kind
of like the 90Vdc step of an NE2 neon bulb.

For example if I were going to parallel 12Vdc batteries in a car to protect one from accidental discharging I would tend to
want to find Germanium power diodes which have only a .2Vdc step.

For even lower resistance - but super component count inefficiency consider "synchronous rectification" using low "on" resistance FETs.

:S:MarkSCoffman