Switching Time of a Diode
• Switching
Time of a diode is the time it takes to switch the diode between two
states (ON and OFF states).
•
Assume R S is large enough that all current flows through diode in forward bias
conditions.
Switching Time of a Diode
•
For Forward Bias:
•
•
•
o
•
•
(
)
Current source I D is an ideal diode current source: ID = IS e VD / φT − 1
C j represents the space-charge ( junction capacitance ).
C d represents the excess minority carrier charge ( diffusion capacitance ).
(Note that both of these are small-signal capacitances; to be applicable largesignal analysis, average capacitance values must be used).
For Reverse bias:
o
Eliminate the current source ID and the diffusion capacitance C d.
Transient response requires finding a solution to:
1
Switching Time of a Diode
• Transient
response:
o The exponential and the non-linear dependence of C d and
C j on V D make this difficult to do by hand.
o Consider
the simulated response:
Switching Time of a Diode
• Turn-off Transient.
o The
•
Region 1 :
•
Region 2 :
turn-off transient, has two operation intervals.
§ Diode
is on.
§ I 2 removes excess minority charge .
§ Voltage drop is small allowing C j to be ignored
since space-charge remains ~constant.
§ I D
~= 0 (diode off).
§ Space-charge changes while building a reversebias over the diode.
§ Therefore, C j dominates performance.
2
Derivation of Turn-off Transient.
o
Region 1 : Removal of excess minority charge .
§ Charge-control expression:
The term
dQ D (t)
dt
represents the current due to the removal of injected
carriers from the base region and the term
QD (t)
τT
Diffusion current where τ T is the Base transit time.
represents the normal
o Solving
this differential equation assuming that:
QD = I1 τT at time t = 0 or the initial value of QD and Iin = I2
(V −V )
V
Where I1 ≈ 1 D is the current under forward bias and I2 = 2
RS
RS
[
QD (t) = τT I2 + (I1 − I2 )e −t / τT
]
o
Yields:
o
The turn-off time is derived by solving for the time t = t 1 (Q D evaluates to 0):
⎡ (I − I )⎤
t1 = τT ln ⎢ 1 2 ⎥
⎣ − I2 ⎦
o Region 2 : Changing the Space Charge.
§ Diode
is off, circuit evolves toward steady state.
VD (t = ∝) = V2 = I2 RS
§ During
this time, a reverse voltage is built over the diode.
§ Therefore, space charge has to be provided.
3
§ The
change in excess minority charge can be ignored as well as the reverse bias
diode current I d.
§ This leaves us with a simple RC circuit (red capacitor model shown earlier).
I2 =
VD (t )
+ Cj
RS
of interest.
o
dVD
dt
Cj is the average junction capacitance over the voltage range
Assuming the value of V D at time t = t 1 is 0, the solution is the wellknown exponential:
− (t − t1 ) ⎤
⎡
R C
VD (t) = I2 RS ⎢1 − e S j ⎥
⎢
⎥
⎢⎣
⎥⎦
The 90% point is reached after 2.2 time-constants of RS Cj .
o
t2 − t1 = 2.2 RS Cj
•
Derivation of Turn-on Transient.
o Similar
considerations hold for the turn-on transient.
o Space Charge :
§
The transient waveform for the diode voltage (assume t = 0):
1 ⎤
⎡
−
R
C
VD (t ) = RS ⎢I1 − (I2 − I1 )e S j ⎥
⎢
⎥
⎢⎣
⎥⎦
⎡ I − I ⎤
t3 = RS Cj ln ⎢ 1 2 ⎥
⎣ I1 ⎦
o Excess
o
Assu min g VD = 0
charge :
− (t − t3 ) ⎤
⎡
QD (t ) = I1 τ T ⎢1 − e τT ⎥
⎢
⎥
⎣
⎦
It takes 2.2 time constants for Q D to reach 90% of its final value.
t4 − t3 = 2.2 τT
4