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## Re: Simple transistor

 Subject: Re: Simple transistor "Greg Neill" Thu, 20 Apr 2006 10:31:59 -0400 sci.electronics.basics
 ```"John Fields" wrote in message news:l33f42pf3ng8t51dc7bjm11q5fo6i2kmcs@xxxxxxxxxx > On Thu, 20 Apr 2006 08:30:14 -0400, "Greg Neill" > wrote: > > >"Boki" wrote in message > >news:1145526532.183920.52120@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx > >> Hi All, > >> > >> There is a PN diode, P connects 3.3V, N connects to ground. > >> We knew that Vf is 0.7V. > >> > >> How to explain that the measurement Vf ( P to N ) is 3.3V now but not > >> 0.7V ... > > > >Diodes are not perfect... If you push enough current > >through them, the voltage drop rises. Check out the > >I vs V curve of the diode. Take a look at: > > > >http://www.st-andrews.ac.uk/~jcgl/Scots_Guide/info/comp/passive/diode/chars/chars.htm > > --- > How much current would you have to feed through the diode to get a > 3.3 volt drop across it in the forward direction? Taking a typical 1N914B diode as an example, and assuming a simple square-law model, the data sheet gives a maximum forward voltage of 1V at 100mA which makes the model: I = (100mA/1V^2)*Vf So you'd need to push about 960mA across the junction. This is not something that you'd want to do on a continuous basis! But you could get away with it for short duration pulses (less than a mSec). If "Boki" is measuring 3.1V DC across the diode, it's not a diode anymore. ```
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