6.301 Solid-State Circuits
Recitation 25: More on Charge Control and Space-Charge Lasers
Prof. Joel L. Dawson
First, let’s do some cleanup from the last recitation. We said that we would treat the problem of
emitter switching as an example of how to use the charge control equations.
iC
! I E u(t)
Note that from our experience with Lab 2, we might expect this common base stage to be relatively
fast. What does charge control have to say about that?
Use charge-control equations for emitter current. Notice that I follow my own sign convention
here…I choose to define current flow out of the emitter as positive.
" 1
1 % dqF
iE = qF $ +
+
= I E u(t)
# ! F ! BF '& dt
If we solve this, we find that
(
qF = ! I E 1 " e"t /!
)
Where ! is defined by
1 1
1
=
+
! ! F ! BF
Writing this out:
!=
! F! BF
"! F 2
"
=
=
! F = # F! F
! F + ! BF ! F + "! F " + 1
6.301 Solid-State Circuits
Recitation 25: More on Charge Control and Space-Charge Lasers
Prof. Joel L. Dawson
So the time constant involved is smaller than ! F ! Recall that the time constant associated with base
current drive was ! BF . Now we have all we need to calculate iC :
iC =
(
qF
= " F I E 1 # e#t /" F ! F
!F
)
And for iB , we apply KCL (again remembering the sign convention that I have chosen):
iB + iC = iE ! iB = iE ! iC
(
= I E ! " F I E 1 ! e!t /" F # F
)
Initially, then, the base current = iE !
CLASS EXERCISE
Consider the following circuit:
IC
S1
qF
!F
q
dq
iB = F + F
! BF
dt
iC =
# 1
1 & dqF
iE = "q f % +
"
$ ! F ! BF (' dt
IB !
At time t=0, we open switch S1. Derive and sketch IC as a function of time.
(Workspace)
Page 2
6.301 Solid-State Circuits
Recitation 25: More on Charge Control and Space-Charge Lasers
Prof. Joel L. Dawson
Take a moment to remind yourself of what is physically happening. We have assumed that
recombination in the base is negligible. Thus, the collector current continues just as long as it takes to
clear qF out of the base via reverse injection.
Space Charge Layers
This, unfortunately, is where things get messy. The upside is that working through the math here will
give you a very detailed understanding of what goes on in a bipolar transistor.
Recall that the BC and the BE pn junctions have capacitances associated with their depletion
regions. We are able to define a nonlinear, voltage-dependent capacitance associated with these
depletion regions, and relate charge and voltage as
dQ = C(V )dV
To figure out how much charge it takes to get to a voltage V0, we must integrate
!
Q0
0
dQ =
!
V0
Q0 =
!
V0
0
0
C(V )dV
C(V )dV
As a check, we note that with a capacitance that is independent of voltage, we have
Q0 =
Just as we expect.
!
V0
0
V0
CdV = C ! dV = CV0
0
Now for this BJT, we have
qVC
qVE
Page 3
6.301 Solid-State Circuits
Recitation 25: More on Charge Control and Space-Charge Lasers
Prof. Joel L. Dawson
We have to store charge qVC on Cµ , and qVE on C je (part of C! ). This charge has to be supplied to
the base, and that is why these capacitances are relevant to us here. For Cµ , we have
Cµ =
Cµ 0
(1 ! VBC / " C )
mC
= Cµ 0" C mC (" C ! VBC )
! mC
If we’re curious about the charge on Cµ , we must perform the integration
!
QVC
0
dQVC =
!
VBC
0
Cµ (V )dV
VBC
QVC
) # Cµ 0" C mC &
1! mc ,
= +! %
"
!
V
(
)
.
C
BC
(
+* $ 1 ! mC '
.- 0
m /
= K C )+" C mC / ! (" C ! VBC ) C ,.
*
-
Where K C =
Cµ 0! mC
1 " mC
and mC ! = 1 " mC
That’s a lot of work, and for what? Well, for one, we can now write down the charge on these
nonlinear capacitors as a function of VBE and VBC . A typical mC for a base-collector junction is 1/2,
and for a good base-emitter junction is 1/3. We have
1/2
QVC = K C #! C1/2 " (! C " VBC ) %
$
&
QVE = K E #! E 2 / 3 " (! E " VBE )
$
Page 4
2/3
%
&
6.301 Solid-State Circuits
Recitation 25: More on Charge Control and Space-Charge Lasers
Prof. Joel L. Dawson
And the additional charge control terms
iB =
dQVC dQVE
+
dt
dt
iC = !
dQVC
dt
iE = !
dQVE
dt
Ultimately, these monstrous expressions for QVC and QVE will be useful because, once we compute
the total charge necessary to get to a particular VBC or VBE , we will sweep the nonlinearity
completely under the rug.
This calls for an example.
Consider the following circuit with a base current drive.
iB = 200 µ A
RL = 1k!
VCL = 10.2V
" F = 100
"R = 5
# F = 1ns
# R = 5ns
# S = 29ns
RL
iB !
We ask, how long does it take to get out of cutoff? In order to get out of cutoff, VBE must equal
600mV. In order to get VBE to 600mV, we must charge QVE and QVC appropriately. The total
amount of charge needed is
!QVE = QVE (0.6V ) " QVB (0V ) = 3.8 pC
!QVC = QVC ("9.6V ) " QVC ("10.2V ) = 0.6 pC
Page 5
6.301 Solid-State Circuits
Recitation 25: More on Charge Control and Space-Charge Lasers
Prof. Joel L. Dawson
Total charge that must be supplied by iB to get out of cutoff: 4.4pF.
And here is where we sweep aside the nonlinearity. We say that
!t =
=
!Q !QVE + !QVC
=
iB
iB
4.4 pC
= 22ns
0.2mA
Good for a ballpark figure.
Page 6
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6.301 Solid-State Circuits
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