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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:

Current source I D is an ideal diode current source: ( )1eII TD /VSD = C j represents the space-charge ( junction capacitance ). C d represents the excess minority carrier charge ( diffusion capacitance ). o (Note that both of these are small-signal capacitances; to be applicable large-

signal 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:

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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 turn-off transient, has two operation intervals.

Region 1 : Diode is on. I 2 removes excess minority charge . Voltage drop is small allowing C j to be ignored

since space-charge remains ~constant.

Region 2 : I D ~= 0 (diode off). Space-charge changes while building a reverse-

bias over the diode. Therefore, C j dominates performance.

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Derivation of Turn-off Transient.

o Region 1 : Removal of excess minority charge . Charge-control expression:

The term dt

)t(dQD represents the current due to the removal of injected

carriers from the base region and the term T

D )t(Q

represents the normal

Diffusion current where T is the Base transit time.

o Solving this differential equation assuming that: T1D IQ = at time t = 0 or the initial value of QD and Iin = I2

Where ( )S

D11 R

VVI

is the current under forward bias and

S

22 R

VI =

o Yields: ( )[ ]T/t212TD eIII)t(Q +=

o The turn-off time is derived by solving for the time t = t 1 (Q D evaluates to 0):

( )

=

2

21T1 I

IIlnt

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.

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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).

( )

dtdV

CRtV

I DjS

D2 += Cj is the average junction capacitance over the voltage range

of interest.

o Assuming the value of V D at time t = t 1 is 0, the solution is the well-known exponential:

( )

=

jS

1CRtt

S2D e1RI)t(V

o The 90% point is reached after 2.2 time-constants of RS Cj .

jS12 CR2.2tt =

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):

( ) ( )

0VgminAssuIII

lnCRt

eIIIRtV

D1

21jS3

CR1

121SDjS

=

=

=

o Excess charge :

( )( )

=

T

3tt

T1D e1ItQ

o It takes 2.2 time constants for Q D to reach 90% of its final value.

T34 2.2tt =