0
\$\begingroup\$

enter image description here

Q: Suppose this diode is silicon. So it can't have more then 0.7 voltage drop. If,after the diode has aquired 0.7 volts, more voltage comes from the voltage source,then what happens? Does the diode "passes" the voltage and takes more from source?

\$\endgroup\$

2 Answers 2

4
\$\begingroup\$

Looking at just one aspect of the diode (the 0.7V forward drop) tends to distort your understanding of the bigger picture.

enter image description here

The graph above shows a typical diode's characteristics.

In order for a forward current to flow there needs to be a certain voltage drop across the diode. However, once the diode begins to conduct this voltage drop will increase as a function of the current.

enter image description here

The size of this current will depend upon the supply voltage and the load.

The voltage drop is also dependant upon the absolute temperature.

The 0.6/0.7V is not a fixed (and only) value but a point on a continuous function. This value is used to simplify problems relating to the diode use under 'normal' conditions (e.g in bridge rectifiers, diode clamps etc.)

\$\endgroup\$
1
\$\begingroup\$

it can't have more then 0.7 voltage drop

Wrong. High-current diodes generally drop more voltage than signal-level diodes. I've measured drops of around 1V across a TO-220 packaged silicon diode.

Does the diode "passes" the voltage and takes more from source?

Any excess voltage not dropped across the diode is dropped across the load. This is Kirchhoff's voltage law.

\$\endgroup\$
3
  • \$\begingroup\$ So does the "excess voltage" skips the diode, jumps and fall across the load? (Assuming load comes after the diode). \$\endgroup\$
    – ema
    Commented Apr 19, 2015 at 5:53
  • 1
    \$\begingroup\$ @ema The voltage drop across the diode is because of the current through the diode. Higher current will lead to a slightly bigger drop. You may think of the diode "taking" whatever voltage it needs, and then leaving the rest for the rest of the circuit. \$\endgroup\$
    – efox29
    Commented Apr 19, 2015 at 6:44
  • \$\begingroup\$ @efox29 I understood your point. Actually clippers(diode limiters) made me confuse because in their case,diode do not let voltage drop to pass. In that case ,the load is limited to 0.7,while in other diode circuits, the load do takes the excess drop. \$\endgroup\$
    – ema
    Commented Apr 19, 2015 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.