Electric diode has two terminals and this two terminals make the diode call ‘di‘-‘ode‘, which is translated as it “two” “paths”.
The characteristics of a diode can be represented by a resistor and conductance. Basically diodes make the electric flow one direction discarding the current flow coming from other directions. An Ideal diode follows this characteristic very strictly.
If diode is in forward bias ideal diode simply let the current flow otherwise not. Forward biased ideal diode does not have any voltage difference across it.
In this example we have to determine V, and we will consider both of the diode as ideal diode.
We already know that ideal diode does not drop any voltage and as long as current is going forward diode should be open but the real problem of this problem is, at the same time Node A can’t be 3V and 1V, so either of the diode is cut off.
Now calculate the voltage differences:
If we consider 1V as active then, the voltage differences is: 1V – (-3V) = 5V
If we consider 1V as active then, the voltage differences is: 3V – (-3V) = 6V,
for 3V the voltage difference is huge so, for this case we will consider that diode (D3V) as conductive and other (D1V) as cut offed.
Now applying KVL: ( 3V – (-3V) ) =2k i
=> i = 6/2k
=> i = 3mA
Idol diode is what we wanted the diode to work like, but in reality idol diode does not exist, the practical characteristics of a diode looks almost like exponential.
I=Is(e VD / (n VT) – 1)
I=Ise VD / (n VT)
I is the diode current,
Is is the reverse bias saturation current (or scale current),
VD is the voltage across the diode,
VT is the thermal voltage, and
n is the ideality factor
VT is can be described with the equation, VT =kT/q
in room temperature VT =25mV.
From diode equation,I = Ise VD / (n VT)
I = 10 -12 e 0.5/ (1 x 25m)
I = 4.85 x 10 -4 A
In reality most of the case across the diode we get few more or less 0.7V.
In constant voltage model without taking help from our previous equations, we consider that Voltage across Si diode is always 0.7V, and for Ge it is 0.3V.
at Node A:
(5-0.7)/10k + (-5-0.7+0.7)/5k = i
=>i = 0.43mA – 1mA
=>i = -0.57mA,
i is negative so diode is reverse biased, and no current will go through this. i=0
So, V = i2 5k – 5
= 0.71 x 5 -5