# Transient _switching_ response - of ECSE

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```					       Chapter 12-2. Transient (switching) response

(Section 12-2 in text, pages 449 – 458)

BJT finds extensive use as an electronic switch in complex logic
circuits.

Provides faster switching speeds compared to a pn junction diode.

The time-delay between “on” and “off” states can be attributed to
the build-up and removal of excess minority carrier charge from
the base region.

1
Transient response

Consider a pnp transistor

E         B                  C

IE        p+         n                  p            IC

pB(0)                  IB
QB       pB(WB)

pB0

xB

2
Transient response: Base current

In the quasi-neutral region of the base, the base lead supplies
electrons for:

Recombination with excess holes (= QB/B)
Increasing or decreasing of excess hole charge in base (dQB / dt)
– Under steady state, this part is zero (excess hole
concentration is equal to excess electron concentration in
neutral region)
Injection of electrons to emitter, IEN

3
Transient response (continued)

If we neglect the electrons emitted to the emitter (i.e.,  = 1) then,

QB dQB           where IB and QB are time dependent and  = 1.
IB     
B   dt
dc QB
When the transistor is in forward active mode,      IC 
B
QB dQB dc QB
IE = IB + IC =        
B   dt   B

Holes
Hole current      Steady-state loss    Rate of change
collected by
injected from   = of holes in base by + of hole conc. +
collector
emitter          recombination          in base.

Apply these concepts to turning on a BJT.
4
Idealized switching circuits

S

Let us say, base current is suddenly changed from zero to a value
IB, by turning on the switch. In the above circuit, the base current
changes from zero to VS/RS, if we assume that VS >> VBE, where
VBE is the forward voltage drop of emitter-base junction (approx.
0.7 V).
5
Analysis of switching transients

Using the charge control model we just discussed, we can
write,      Q    dQB
IB  B            where IB is constant for t > 0.
  B  dt
dQB       Q
Or           IB  B
dt       B
       t   
The general solution of this equation is:    QB t   I B  B 1  e  B



            
assuming the boundary condition that                                       
QB(t) = 0 when t = 0 (i.e. starting from “off” state).

As QB(t) increases from zero to IBB, the collector current will
also increase since the collector current is given by:
dc QB
IC          dc I B
B                                                    6
Analysis of switching transients (continued)

But the collector current cannot increase continuously since IC is
limited by the value of RL and VCC. Once, IC reaches VCC / RL, then
IC cannot increase even though QB continues to increase. At this
value of IC, VCE is close to zero, and C-B junction gets forward
biased, and the transistor is said to be in saturation. IC is not equal
to dc IB anymore in saturation.

dc QB                      t 
IC                   1  exp 
 dc I B                   for 0 < t < tr
B                       B 


I C  VCC / RL                             for t > tr

During 0 < t < tr , the BJT is in active mode. At t = tr , the collector
current reaches the maximum value of VCC/RL and does not
increase further. The time tr is called turn-on time of the BJT.           7
Analysis of switching transients (continued)

          tr 
The turn-on time tr can be obtained from:   dc I B 1  exp 
               IC
          B 


        1     
Rearranging and solving for tr:                 1  I  I  
tr  B ln               
       C dc B 

Note that one can reduce the turn-on time by increasing the base
current (faster storage charge build up). However, if you make the
base current too large, you will be storing too much charge in the
base during “on” so that it will affect the turn-off time.
8
Methods to speed-up turn-off transients

Introduce R-G centers in base.
Use Schottky diode clamp to prevent BJT going into “deep”
saturation.

9
Examples of transient response
IB

VS
IB 
RS
t
QB
I BB                                                       
I BB 1  et B   
t
IC

Vcc / RL
dc
QB
B

 dc I B 1  e t  B   
t
tr
10

```
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