Why there is no current flowsin a reverse-biased arrangement?

[Yh Hoo]

Why there is no current flowsin a reverse-biased arrangement??

Once the polarity of the external voltage is the same as that of the junction voltage, electrons from the P-TYPE semiconductor will flow out and be attracted towards the positive terminal of the battery and then to the p-type semiconductor to fill the positively charged holes in the P-region. Likewise holes flow from p-type to n-type. However everything here will stop until an equilibrium between charge carriers and the repelling depletion is reached. So before that ARENT THE FLOW OF CHARGE CARRIERS NOT CONSIDERED AS ELECTRIC CURRENT ?
from the graph, there is a small current !

 

[uart]

Once the polarity of the external voltage is the same as that of the junction voltage, electrons from the P-TYPE semiconductor will flow out and be attracted towards the positive terminal of the battery and then to the p-type semiconductor to fill the positively charged holes in the P-region. Likewise holes flow from p-type to n-type. However everything here will stop until an equilibrium between charge carriers and the repelling depletion is reached. So before that ARENT THE FLOW OF CHARGE CARRIERS NOT CONSIDERED AS ELECTRIC CURRENT ?
from the graph, there is a small current !

Hi YhHoo, what you are describing is the redistribution of charge carriers under the conditions of changing (increasing in this case) reverse bias. This is a capacitive effect (the diode behaving like a capacitor). Yes, it most certainly does constitute a current, however this current is transient and will cease when the voltage stops changing. This current component is not seen on the static V/I characteristics you posted.

 

[uart]

The static reverse current shown on your diagram is predominantly due to thermally generated carriers in (or near) the depletion region. It's scale is exaggerated for the purpose of showing it easily on the diagram. Similar the scale of the reverse voltage is compressed.

 

[Yh Hoo]

Hi YhHoo, what you are describing is the redistribution of charge carriers under the conditions of changing (increasing in this case) reverse bias. This is a capacitive effect (the diode behaving like a capacitor). Yes, it most certainly does constitute a current, however this current is transient and will cease when the voltage stops changing. This current component is not seen on the static V/I characteristics you posted.

thanks for your reply uart ! ok u meant that the current will only be constituted in this circuit when there is a changing voltage applied to it, is it? and what is actually meant by my graph while the voltage applied is below the Vbr ?

 

[uart]

thanks for your reply uart ! ok u meant that the current will only be constituted in this circuit when there is a changing voltage applied to it, is it?

Yes that is correct (for the type of current you described in the original post).

and what is actually meant by my graph while the voltage applied is below the Vbr ?

(I'm assuming that by "below" you mean less negative than). That is the static reverse leakage current as I described above. It is only very small but is exaggerated on your graph.

 

[Yh Hoo]

The static reverse current shown on your diagram is predominantly due to thermally generated carriers in (or near) the depletion region. .

Thx for your reply. Sorry what is meant by the thermally generated carriers in the depletion region? as i know, in the depletion region there is no mobile charge carriers, do u mean that the thermal energy form the surrounding causes the electrons in the uncompensated negatively charged acceptor ions in P-region to break free thus constituting to this static reverse current??

 

[uart]

Thx for your reply. Sorry what is meant by the thermally generated carriers in the depletion region? as i know, in the depletion region there is no mobile charge carriers, do u mean that the thermal energy form the surrounding causes the electrons in the uncompensated negatively charged acceptor ions in P-region to break free thus constituting to this static reverse current??

Not "uncompensated negatively charged acceptor ions in P-region", but in fact just the normal Si atoms in the lattice. Thermal energy can cause them to release a free electron. At the same time this process leaves behind a hole in the Si atom that is free to move to adjacent atoms and so on. In effect an electron-hole pair has been thermally generated. This is the predominant cause of the reverse current, but as I said it's typically only very small.

 

[Yh Hoo]

I said it's typically only very small.

O. Then i understand. That means this very small current contribute to the very small rise in reverse current right ? And how about in a forward biased arrangement ? will the thermally generated charged carriers constitute a small current ? ?

 

[Yh Hoo]

Sorry for bothering you again... Maybe you are busy but i have been hassled by the principle of the Transistor. Can you explain for me? Thanks.

 

[es1]

O. Then i understand. That means this very small current contribute to the very small rise in reverse current right ? And how about in a forward biased arrangement ? will the thermally generated charged carriers constitute a small current ? ?

yes, it is always present. But it is on the order of uA. See the Irm spec in this data sheet for an example of real life values for a typical diode (notice it changes with temp. it can be much smaller with diodes designed to minimize it too). http://www.diodes.com/datasheets/ds28002.pdf

because it is uA, and forward current is so much larger than uA, people just ignore this current altogether in the forward region to keep the math simpler but thermally generated charge carriers are always present.

 

[Yh Hoo]

O i see. But can you please have a look at the graph i posted, before the magnitude of the applied positive voltage (with its polarity opposing the junction voltage) is greater than the built-in junction voltage, will there be any current flows ?? The applied voltage i meant here is steady voltage.