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About c3r14l.k1l4

  • Birthday 11/16/1983

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  1. Nop. Sorry... You explained me it's functioning... and I already knew that... I just want to know how to calculate the source voltage regarding the one applied to it's gate. Check this: In the BJT circuit the emmitter is 5-0.7V so aprox 4.3V. In the FET one the source is 5-X=1.431 so X=3.569V? Where does this value come from? Is it the pinch voltage? I think so... just to make shure...
  2. Good luck man! I'm just waiting for the results of one disciplin (DataBases 2), the others are done. (I'm from informatics engineering).
  3. Because those devices are current limited (it's like they have resistors in series inside them). They wont give you more current for the purpose of not exceeding the maximum rating power (humm, burn :D).
  4. Hi, you'll need a supply with the voltage the schematic was designed to. About it's current, the more the better (the circuit will only use the required amperes). The problem is that if your transformer has less power, the circuit will cause the transformer to dissipate more power (incresing it's temperature) and therefore... kaput :P Don't forget to have a rectifier bridge with support for more than 1.5x the maximum current you'll need.
  5. Greetings, I'm new here and this is my first post. All my electronics knowledge has been built from experience and books and I intend to help whenever I can. But as I'm not God I've got some doubts. One of them wich is bugging me is how to get a FETs source voltage value. For instance, we know that in a BJT the emmitter is always the base - junction drop voltage (~0.7V), but in the books I can't find a similar relationship for the FETs. Only the drain current in function of the gate voltage (the parabola function). Can anyone explain this to me? And... sorry for my english, as it's not my native language... Cheers!
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