Diode Ciruit electronics

[paulmdrdo]

it says that assume that the diode fails open. what is the voltage across the diode and the voltage across the limiting resistor?

i don't know the meaning of "fails open". 'm new in electronics. thanks!

 

[pradhan.rachit]

By fails open, i think it means that the state when failure of diode will cause it to act as an open switch.

If the diode fails open, then the voltage across the diode will be the supply voltage unless no current flows through the circuit...

 

[paulmdrdo]

the answer from my book is V_d=5v voltage across the diode. V_lim = 0v voltage across limiting resistor.

i don't know ho it was calculated can you tell me why is that.

 

[pradhan.rachit]

I don't know why your book says Vd=5V.
If the diode acts as an open switch, then the whole voltage drop should be shifted at the diode.
So according to your circuit diagram, Vd should be 10V
Voltage across limiting resistor should be 0V since there is no current flowing through the circuit.

 

[paulmdrdo]

there's a dynamic resistance of 10 ohm.

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why is the voltage drop shifted at the diode, 0v at resistor?

 

[betwixt]

If the diode 'fails open' it normally means it is open circuit and therefore no current flows through it. It's dynamic resistance becomes unimportant, effectively it would be infinitely high. If no current flows, no voltage is dropped across the resistor and the voltage across the diode ends would be 10V.

Brian.

 

[paulmdrdo]

please bear with me, the diode is in series with the source and resistor. if there's no current flowing through the resistor, there's no current flow in the diode as well hence no voltage drop across the diode. why is the voltage drop occurs in the diode? I'm still confused. please help.

 

[betwixt]

The book is wrong! It's calculation is incorrect.

If you start with 10V and the limiting resistor drops 0V (V_lim = 0v) there is still 10v remaining.

Think of the "failed open" diode as not being there at all since it is open circuit, you can clearly see that no current can possibly flow through the circuit so no voltage is dropped.

Brian.

 

[paulmdrdo]

if that's the case the voltage drop across the diode is 0v. (I'm assuming that it is not there because it is"failed open")). Am I correct?

 

[betwixt]

No!

The diode is not dropping any voltage at all because no current is flowing though it, the voltage across the diode is still 10V.

If is was a good diode (not failed) it would pass current and the voltage would be dropped across the resistor but in this case the resistor drops (V=I*R) = 0*1,000 = 0V so all the 10V is still there.

Brian.

 

[paulmdrdo]

thanks a lot! now i get it!

 

[Analog Ground]

It is not necessary to think in terms of current. Think of the equivalent resistance of the diode when "failed open" as being very, very high such as 10 gigaohms. Then, think of the resistor and the diode forming a resistor voltage divider. Because the diode resistance is so high, all the voltage is across the diode. Vdiode = Rdiode/(Rlimit + Rdiode) * Vbias