Transistor Example – Fall 2023
V1
12 Volts
R1
R2
R3
47k Ohms
1k Ohm
R
R4
10 k Ohms
Q1
NPN
R5
R6
470k Ohms
220 Ohms
Vce2 = 6 Volts
Q2
NPN
Mission: Assuming that each transistor is a silicon device and has 𝛽 = 150, find a
value for R3 that sets the collector-emitter voltage on Q2 to 6 Volts. (Half the supply
voltage, as shown).
Analysis: Let’s assume that Q1 and Q2 are in active mode until we see evidence
otherwise.
Let’s also start by working backwards. If 𝑉!"# = 6 𝑉 for Q2, then the current through
R3 must be Vs/(2R) = (6)/R and this is also the collector current since R3 is in series
with the collector of Q2. The base current entering Q2 must be:
𝐼$# =
1
𝑉%
𝐼!# =
𝛽
2𝛽𝑅
The voltage on the base of Q2 should be 0.7 Volts relative to the grounded bottom
node since it’s a silicon device. The current through R4 is the base current of Q2 so:
𝐼$# =
𝑉!& − 0.7
𝑅'
On the LHS of the circuit, let’s do KCL at the base of Q1:
𝑉( − 𝑉$&
𝑉$&
= 𝐼$& +
𝑅&
𝑅)
If Q1 is also silicon, as we’re told, we can write:
𝑉$& = 0.7 + 𝑉"&
Where VE1 is the voltage at the emitter of Q1, relative to ground.
The emitter current leaving Q1 is by definition 𝐼"& = (1 + 𝛽)𝐼$& so the voltage drop
on R6 is
𝑉"& = 𝑅+ 𝐼"& = (𝑅+ )(1 + 𝛽)𝐼$&
Solve the three equations to get the base current of 134 microAmps.
So, the voltage across the 220 Ohm is 4.42 Volts and the current entering the
collector Q1 is 20.1 mA.
KVL tells us that the collector-emitter voltage on Q1 plus the voltage on R2 plus the
voltage on R6 must add up to 12 Volts, so Q1 and R2 share 12.00-4.42 = 7.58 Volts.
How to figure this out though?
Well, if we do KCL at the collector of Q1:
From before:
𝑉( − 𝑉!&
= 𝐼!# + 𝐼$#
𝑅#
𝐼$# =
𝑉!& − 0.7
𝑅'
Since we know the collector current to be 20.1 mA, we have two equations and can
solve and get a negative value of Vc1.
STOP: this is a problem. Do you see why?
REASON: We’re predicting a collector voltage that is less than the emitter voltage,
which in turn means that the base-collector junction is forward biased. This
transistor is NOT in the active mode as we initially assumed.
HOW TO CORRECT: The problem turns out to be that there is too much base
current entering Q1 which tries to cause more collector current to flow than the
resistors allow. Notice that the max amount of current that can flow into the
collector is on the order of 12V/(1220) = 9.83 mA. Our setup wanted 20.1 mA to flow
– Q1 was in hard saturation.
Let’s back off the base current by increasing the value of R1 to 330k Ohms. (One of
the jobs of resistors is to limit current flow!) The base current will drop to 27.9
microAmps and the collector current will drop to 4.19 mA – less than half the
maximum.
OK. Repeat the calculation and we find that the emitter voltage on Q1 is 0.927 Volts
and we get a collector voltage of 7.17 Volts. Much better. J
So, the base current into Q2 is
𝐼$# =
7.17 − 0.7
= 647 𝜇𝐴
10𝑘
From earlier, we had:
𝐼$# =
Plug in:
1
𝑉%
𝐼!# =
𝛽
2𝛽𝑅
647 𝜇𝐴 =
12
2(150)𝑅
Solve:
𝑅 = 61.8 Ω