PermalinkSubmitted by guiferra on Sat, 06/18/2016 - 03:55
Excellent, understandable and efficient explanations, my compliments.
However, something it is not completely clear with the explanation of number of carriers available for conduction in N-type semiconductors. In fact, if it is true that the fifth electron of dopant phosphorous has no silicon atom with an available covalence bound in its surroundings to be bound with, it is still true that that electron should still be bounded to the phosphorous atom. So why it is available for conduction, is the bound to its atom so weak when it does not take part of a covalent bound? Which energy is necessary to take it out from its atom compared with the energy necessary to break a covalence bound?
I will appreciate any comment to this.
PermalinkSubmitted by skyline1397 on Tue, 09/12/2017 - 21:59
If I have an ideal diode and I apply -9 V in the anode (P-type junction) and -10 V at the cathode (P-type junction) would the diode become forward biased??
I get that the current-voltage relationship of an ideal diode shows that as long as the voltage across the ideal diode is non-negative i.e. V≥0, the ideal diode looks like a short circuit but the graphs refer to V. Is this V equal to V(anode)-V(cathode)?
Where V potential of anode with respect to cathode
Comments
Doped semiconductors
Excellent, understandable and efficient explanations, my compliments.
However, something it is not completely clear with the explanation of number of carriers available for conduction in N-type semiconductors. In fact, if it is true that the fifth electron of dopant phosphorous has no silicon atom with an available covalence bound in its surroundings to be bound with, it is still true that that electron should still be bounded to the phosphorous atom. So why it is available for conduction, is the bound to its atom so weak when it does not take part of a covalent bound? Which energy is necessary to take it out from its atom compared with the energy necessary to break a covalence bound?
I will appreciate any comment to this.
Diode potential of anode with respect to cathode
If I have an ideal diode and I apply -9 V in the anode (P-type junction) and -10 V at the cathode (P-type junction) would the diode become forward biased??
I get that the current-voltage relationship of an ideal diode shows that as long as the voltage across the ideal diode is non-negative i.e. V≥0, the ideal diode looks like a short circuit but the graphs refer to V. Is this V equal to V(anode)-V(cathode)?
Where V potential of anode with respect to cathode