# A simple question about diodes.

#### farmosh203

Joined Nov 26, 2006
20
Hello I have a quick question about diodes that I seem to be getting a bit confused on.

If I have a diode like this:

Vi ------[>|------- Vo

I know that when Vi is greater than .7 volts, then it becomes a short circuit. When Vi is less than .7 volts it becomes an open circuit.

Now where I am confused is this case:

Vi -----|<]---------Vo

I am confused on when it becomes a short circuit. I would imagine that since the polarity of the battery in the diode is reversed that anything less than -.7 volts would make it short circuit. Anything greater than -.7 volts is open circuit.

Or is it the other way around, where anything less than .7 volts makes it a short circuit and anything greater than .7 volts makes it open circuit?

If someone could answer this question, it would be greatly appreciated, thanks.

#### hgmjr

Joined Jan 28, 2005
9,029
Check out the tutorial on diodes here at All About Circuits...

hgmjr

#### Dave

Joined Nov 17, 2003
6,970
Hello I have a quick question about diodes that I seem to be getting a bit confused on.

If I have a diode like this:

Vi ------[>|------- Vo

I know that when Vi is greater than .7 volts, then it becomes a short circuit. When Vi is less than .7 volts it becomes an open circuit.

Now where I am confused is this case:

Vi -----|<]---------Vo

I am confused on when it becomes a short circuit. I would imagine that since the polarity of the battery in the diode is reversed that anything less than -.7 volts would make it short circuit. Anything greater than -.7 volts is open circuit.

Or is it the other way around, where anything less than .7 volts makes it a short circuit and anything greater than .7 volts makes it open circuit?

If someone could answer this question, it would be greatly appreciated, thanks.
The answer stems from a semiconductor physics explanation. The diode is constructed from two pieces of opposite doped semiconductor material sandwitched together - in the case of your example this material would be silicon. These two pieces are the P-type (positively doped, i.e. more holes) and N-type (negatively doped, i.e. more electrons), hence PN junction/diode. When these two materials are sandwitched together a process of diffusion occurs that creates a depletion region between the P and N sides, i.e. a region where the net charge is zero.

If you apply a positive voltage to the P-side of the diode electrons are drawn across the depletion region towards the positive voltage, conversely holes will be drawn across the depletion region in the opposite direction. This whole effect results in a narrowing of the depletion region. At 0.7V (for silicon) the above process has happened such that the depletion region has disappeared and this is why the diode now conducts (or as you put is a short circuit).

On the otherhand, if you apply a negative voltage to the P-side of the diode electrons are repelled from the depletion region, conversely holes from the depletion region are repelled in the opposite direction. This whole effect results in a broadening of the depletion region. Hence further increasing the voltage will result in a further widening of the depletion region and in this case no current will be able to flow. In reality there is a small saturation current that is attributed to thermal effects, and at a certain reverse applied voltage avalanche breakdown will occur which results in a significant current flow across the diode, however this is a another issue altogether.

Dave

#### neeraj

Joined Feb 18, 2007
1
in the first case when vi is above .7 volt then diode will act as a short ckt,
first make it clear that you are talking about 'si' p-n junction diode when p
is connected to positive diode becomes forward bias so conduction will be there ,ideally diode act as a short ckt but practically a large amount of current flow through it (you can refer to v-i charecters.. of diode in your book).
if vi is less then .7 volt (which is the knee voltage) diode output current is
very small not zero.so diode will not act as a open ckt (diode will be open ckt only when positive supply is connected to 'n' termnal (it is reverse bias condition).
so make it clear that .7 volt is not boundary line for open and short ckt.
now comming to your question in second diagram ..........
always make sure that if supply to p terminal is greater then supply to n then diode will conduct (or in your words short ckt) or vice versa.
if you are applying -.7 volt as input to n termnal and it is greater then -.7 volt then diode conducts otherwise no conducion (only reverse saturation current exists).

#### farmosh203

Joined Nov 26, 2006
20
Thanks for clearing that up, makes sense, thanks.

#### jimjohn

Joined Jan 28, 2008
8
Hi,
In simple words
"To start conduction anode of the diode should a voltage greater than 0.7V whatever at the cathode"