SYSC 3203
Quiz 2
Sept 24, 2014
Question 1: Electrical isolation
Consider two circuits, using 1) a BJT transistor, and 2) an optoisolator.
In both, the transistor terminals (Emitter and Collector) are exposed to the powerline voltage (120V).
Describe what might happen in each. Show possible electrical pathways from the powerline to a patient.
How does the optoisolator provide better electrical safety than the BJT transistor?
Solution:
In the case of the BJT, there is a direct electrical path (through the transistor’s base electrode)
from the exposed terminals to the patient. There
is a danger of shock to the patient whenever the
terminals are exposed to a suitably high voltage.
In the optoisolator, the patient is electrically isolated from the exposed terminals since a portion
of the signal path is purely optical. There is no
danger of shock to the patient provided the voltage on the exposed terminals does not exceed the
breakdown voltage of the device.
SYSC 3203
Basic circuit theory, Page 2 of 5
Sept 24, 2014
Question 2: Bipolar transistor
An npn bipolar junction transistor (BJT) is connected as shown in the following circuit with
RC = 6.8kΩ and RE = 3.3kΩ. Assume the device is operating in the forward active mode such
that its collector current and base current may
be related using iC = βiB . If vBE = 0.7 V and
β = 150, calculate
(a) the voltage vE at the junction of the transistor’s emitter and the resistor RE
(b) the emitter current iE through RE
(c) the collector current iC through RC
(d) the voltage vo
when VCC = 9 V and vi = 3 V . Is the assumption
about the operating mode justified? Briefly (< 50
words) explain why.
Solution: See for example Sedra & Smith Ex. 4.2/4.3 for a nice workthrough of this configuration,
and Ex. 4.4 for the same configuration operating in cutoff mode.
(a) vE = vi − vBE = 3 − 0.7 = 2.3 V
(b) iE = vE /RE = 2.3/3.3k = 0.697 mA
(c) iE = iB + iC = iB (1 + β), and iC = βiB . Hence, eliminating iB , we find that iC =
150
151 × 0.697 mA = 0.692mA
β
β+1 iE
=
(d) vo = VCC − iC RC = 9 − 0.690mA × 6.8kΩ = 4.29 V
The assumption about the operating region is justified, because the base-collector is reverse biased
(vi − v0 = 3 − 4.29 = −1.29 V ) while the base-emitter is forward biased (vi − vE = 0.7 V ); this is the
classic operating condition corresponding to the forward active mode of operation - see for example
Sedra & Smith § 4.2.
SYSC 3203
Basic circuit theory, Page 3 of 5
Sept 24, 2014
Question 3: Transimpedance amplifier
An ideal operational amplifier (or “op-amp”) has infinite input impedance and infinite open-loop gain.
When operating in a negative feedback configuration, what are the rules for
(a) the current into its inverting (-) and non-inverting (+) inputs; and
(b) the potential difference across its inverting (-) and non-inverting (+) inputs.
A certain biomedical sensor is modeled as a current source Is = 200 µA, and its signal is to be amplified
using the transimpedance amplifier shown at right with R = 1 kΩ. Using your answers above or
otherwise, express the output voltage vo as a function of input current. Briefly (< 50 words) explain the
meaning of the word transimpedance in this context.
Solution:
(a) the currents are zero i.e. i+ = 0 and i− = 0 (assumption of infinite input impedance)
(b) the potential difference is zero i.e. v+ = v− (assumption of perfect negative feedback)
Since i− = 0, the feedback current if = (vo − v− )/R must be equal to the sensor current is . Then
since v+ = v− and v+ = 0, we have simply that is = if = vo /R or
vo = Ris
It is an impedance because it converts voltage to current; it is a transimpedance because it relates
voltage out to current in.
SYSC 3203
Basic circuit theory, Page 4 of 5
Sept 24, 2014
Question 4: Summing amplifier
An ideal operational amplifier is configured as shown in the circuit below. Show that the output voltage
vo is proportional to the sum of the input voltages v1 + v2 + v3 . How would you modify the circuit (i.e.
choose resistor values) so that vo = 2v1 + 3v2 + v3 ?
Solution:
Note: this is essentially the same as Sedra & Smith Ex. 2.10 except we have added a third input.
Starting with the feedback network: as usual, write (vo − v− )/R2 = (v− − 0)/R1 . Re-arranging we
get
vo = v− (1 + R2 /R1 )
Aside: You will see this factor (1 + R2 /R1 ) all the time for op-amps in the non-inverting configuration; in the simple non-inverting amplifier you will have v+ = vi and hence G = vo /vi =
(1 + R2 /R1 ), however in the case of an adder we need to do a bit more work, as follows.
Moving to the + terminal, we can write the total current (v1 −v+ )/R+(v2 −v+ )/R+(v3 −v+ )/R = 0
(since no current flows into +). Re-arranging,
v1
v2
v3
V+
+
+
=3
R
R
R
R
v1 + v2 + v3 = 3V+
From the feedback analysis, we have v− = vo / (1 + R2 /R1 ), and since v+ = v− ,
v1 + v2 + v3 =
3vo
1 + R2 /R1
and hence we can achieve the required vo = v1 + v2 + v3 by setting 1 + R2 /R1 = 3, or R2 = 2R1 .
Final part (harder!): let’s introduce new factors α, β and γ such that the three input resistors have
values R/α, R/β and R/γ respectively. The current equation becomes
SYSC 3203
Basic circuit theory, Page 5 of 5
Sept 24, 2014
α(v1 − v+ ) β(v2 − v+ ) γ(v3 − v+ )
+
+
=0
R
R
R
αv1 + βv2 + γv3 = (α + β + γ)v+
Exactly as before, we substitute for v+ (using the v+ = v− ideal op-amp rule) from the feedback
analysis, giving
αv1 + βv2 + γv3 =
vo =
α+β+γ
vo
(1 + R2 /R1 )
1 + R2 /R1
(αv1 + βv2 + γv3 )
α+β+γ
To achieve the required vo = 2v1 + 3v2 + v3 we can choose α = 3, β = 2 and γ = 1, and must then
satisfy 1 + R2 /R1 = 6, or R2 = 5R1 . NB the resistor values are R/2, R/3, R not 2R, 3R, R
i.e. inputs with a smaller resistor have greater weight in the sum.
To reduce it to the two-input case of Sedra & Smith Ex. 2.10, set γ = 0 (corresponding to infinite
resistance on the v3 input) and then for vo = 6v1 + 4v2 choose resistors 12/6 = 2kΩ, 12/4 = 3kΩ
and 1 + R2 /R1 = 6 + 4 = 10, or R1 = 1kΩ, R2 = 9kΩ).
Solution: Alternate solution using superposition
Temporarily connect v2 and v3 to GND. Then the current goes - it flows between the voltage
we simply have a “regular” non-inverting op-amp sources as required, in order to keep i+ = 0.
whose input v+ is taken from a potential divider
consisting of resistances R and R||R i.e.
vo = (1 + R2 /R1 ) v1
= (1 + R2 /R1 ) v1
= (1 + R2 /R1 )
R||R
R + R||R
R/2
3R/2
v1
3
Repeat for the other two combinations i.e.
grounding v1 ,v3 then v1 ,v2 respectively, and add
the results. This method makes it clearer where