|
|
"John Fields" <jfields@xxxxxxxxxxxxxxxxxxxxx> wrote in message
news:l33f42pf3ng8t51dc7bjm11q5fo6i2kmcs@xxxxxxxxxx
> On Thu, 20 Apr 2006 08:30:14 -0400, "Greg Neill"
> <gneillREM@xxxxxxxxxxxxxxxxxx> wrote:
>
> >"Boki" <bokiteam@xxxxxxxxxxxxxx> 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.
|
|