Assume that the diode has a constant voltage drop in forward mode of
, and that all
transistor baseemitter junctions have a forward voltage of


. Calculate the value of
and the collector current of Q3 and Q4 (which is the primary bias current for this circuit).
Class Notes
–
Additional BJT Problems
Subscribe to view the full document.
237
Solution
On the surface, this looks like a very complex circuit with many components. However, when we
break down the circuit, we can see that it is nothing more than small buildingblock circuits which
we have already studied. To begin, since we are solving the circuit at DC, we know that the
capacitors will be opencircuit. Thus,
is not connected, and the capacitor between the bases of Q1
and Q2 has no effect:
Let us try to use iteration to produce the answer to this question
–
it would be quite difficult to
create and solve a direct set of equations with so many transistors. To begin this process, we need an
initial guess for
,
, and
, our goal values. Working backwards, we know that
for Q3/4 will
depend on base current, which in turn depends on the collector currents of Q1/Q2, which in turn
depend on the base currents of Q1/Q2. We do not have an easy, direct method for finding these base
currents yet. So, as in previous examples, we can assume that the base current will be very small,
effectively zero for the first iteration:
This is typically a safe assumption if we expect the transistor to operate in active mode, and/or if the
transistor has a high value for
. Since this is identified as an amplifier circuit, and
, we have
both conditions, meaning this is a relatively safe starting assumption. We can now calculate the
values of
and
directly by first finding the current,
, in the seriesconnected string of
R2
D1
R3
R4:
(
)
(
)
( )(
)
238
Note: Be careful with the number of significant digits which you carry for
; it actually has a large
effect in this problem. If you carried only three, then
, which would not be sufficient for
the BJT to be in a conducting mode. We will try to avoid introducing this issue on exam problems,
but this is not always possible
–
when in doubt, carry a few extra digits, and apply basic sanity
checks to your solutions (i.e., identifying in the above that 0.699 would put the BJT in cutoff mode).
( )( ) (
)( ) (
)
We now know the emitter voltages of Q1 and Q2, assuming that both are in either conducting mode
with


:
Note that we are now working with PNP transistors. They use all the same equations as the NPN
transistors, except the polarities of currents and voltages are inverted:
PNP Transistor
Notice that current flows into the emitter, and out of the collector and base. The fundamental
equation of the BJT still holds:
. Given the above directionality of currents, the equations
of the PNP transistor in active mode also hold:
. To check the conditions, we use the same
conditions for the NPN transistor with reverse polarity for all voltages. In other words, what was
before becomes
, and so on. This will become clearer when we check the conditions for this
problem.
Subscribe to view the full document.
 Spring '17