DIODES
The ideal diode
D1 & D2 are ideal, find I and V
Solution:
Assume D1 & D2 are conducting (i.e., forward biased). Note that solution should verify
the assumption, otherwise a new assumption has to be made.
Since D1 is ON, VB = 0, and ∴V=0.
ID2 = (10-0)/10 = 1mA
I+ID2 = (0-(-10))/9.9
∴I+1 = 10/9.9 = 1.01mA
Since I >0, D1 is ON as assumed. D2 is ON by virtue that VB=0.
Terminal Characteristics of Real Junction Diodes
i = I s (e v / ηVT − 1)
Is = saturation current or scale current. Is doubles for every 10°C rise in temperature.
I S (T2 ) = I S (T1 )2 ∆T / 10
kT
VT =
q
I.e.,
k = Boltzmann’s constant = 1.38×10-23 joules/Kelvin
T = 273 + temperature in °C
q = 1.602×10-19 coulomb = electronic charge
At room temperature (20°C), VT ≈ 25mV
IS is on the order of 10-15A.
η = 1 for diodes fabricated as part of an I.C.
= 2 for discrete 2-terminal components.
For v >> η V T
i ≈ I S e v / ηVT
or
v = ηVT ln
i
IS
Vγ=
Vγ = threshold or transition voltage ≈ 0.5V for Si diodes.
Vγ Decreases by 2mV/°C, i.e., Vγ (T2) = Vγ (T1)-2(T2-T1)
Note for v<0, i ≈ -IS which is saturation current.
The Breakdown Region and Zener Diodes
VZK = Breakdown voltage = knee of the i-v curve.
In the breakdown region: a large increase in IZ results in only a small increase in VZ.
Voltage Regulators
Circuit Symbol for a Zener diode.
If RL is not connected (no load condition), VO = 6.8V.
I = (10-6.8)/0.5 = 6.4mA
If RL = 2kΩ be connected, IL = 6.8/2 = 3.4mA
∴IZ = 6.4-3.4 =3mA
If IZ < IZK, the Zener diode leave the breakdown region, and the diode becomes approx.
open circuit. VO = V
+
RL
R + RL
If R = RL V+ = 10V, VO = 5 < VZK the voltage required for the Zener diode to operate in
the breakdown region.
Zener Resistance
∆VZ
≈ ∆rZ
∆I Z
∆VO = ∆V
∆VO
∆V
rZ // RL
R + rZ // RL
voltage regulation