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ENGG 1203 Tutorial_05
Example 1 : Last Q5
1
Solution 1 : use KCL
I1
I2
From B to M
I4
2
Solution 1 :
I1
Solution 1 :
I2
I4
4
Solution 1 :
5
Example 2 :

You have connected the lamp, with Vcc = 12V.
The datasheet of the lamp states that it only
turns on when VL > 8V.

The lamp has an internal
resistance of 1k ohm.

What is the range of R
that would allow the
circuit to function
correctly with all input
combinations.
6
Solution 2 :
The range of R
 When the lamp is turned on, x = “1”, VL > 8


When the lamp is off,
x = 0, and VL < 8.
It will function
regardless of the value
of R.
Function as an AND gate
7
Example 3(a) :
A force sensitive resistor (FSR) is a resistor
with its resistance changed according to the
force applied to it.
 For simplicity sake, your partner has wired up
the FSR using a simple potential divider circuit

8
Solution 3(a) :
Calculate the following quantities
 when Rref = 10k ohm :

◦ Voltage across the FSR ;
◦ Voltage at Vout ;
◦ Current owing through the FSR.
9
Example 3(b) :

The output Vout is used to detect the presence
of a ball. Due to its light weight, the ball
produces only 0.5N when it is located on top
of the sensor. The rest of the system requires
that VIL = 2V and VIH = 10V
◦ VIL : Max. voltage that the system regards as logical
LOW
◦ VIH : Min. voltage that the system regards as logical
HIGH

Determine the range of value that Rref may take
for correct functioning of the circuit.
◦ It should output a logical HIGH when a ball is
presence and a logical LOW otherwise.
10
Solution 3(b) :
VDD: 12V
(Presence of the ball / 0.5N);
Rfsr = RL = 10k ohm
VIH: 10V; Minimum HIGH;
10V – 2V:
(Unknown; Not in use)
VIL: 2V; Maximum LOW;
Rfsr = RL = 1M ohm
(Absence of the ball / 0N)
VGND: 0V
11
Solution 3(b) :
When the ball is on the sensor, Rfsr = RH = 10k
ohm. The output voltage at that time should be
higher than VIH.
 When the ball is not on the sensor, then Rfsr =
RL = 1M ohm. The output voltage should be
lower than VIL.

12
Solution 3(c) :

Your partner suggests that it may be possible to use
2 FSRs connected to perform a logical OR
operation: When the ball rolls over either one of
the 2 FSRs, the output Vout is HIGH, and is LOW
otherwise.

What is the output voltage Vout ?
◦
◦
◦
◦
one of the FSRs is under pressure of 0.5N,
both FSRs are under a pressure of 0.5N each,
none of the FSRs is under pressure.
assume Rref is 100k ohm
13
Solution 3(c) :

If Rref = 100k,
◦
◦
◦
◦
◦

(i) one of the FSRs is under pressure of 0.5N,
(ii) both FSRs are under a pressure of 0.5N each,
(iii) none of the FSRs is under pressure.
Reqv is the equivalent resistance of the
parallel combination of the two FSRs.
VIL is not MORE than 2V and VIH is not
less than10V,  the circuit still
functioning correctly as a 2input OR
14
Solution 3(d) :

If there are 3 FSRs connected in parallel,
assume Rref remains at 100k
  THEN, will it behave as a 3-input OR?
 NOT,
Vout > 2 V
15
Solution 3 :

Yes, it works correctly as a 2-input OR gate
because the output is HIGH when there is a ball
on top of at least one of the input.

Even if we connect 3 FSRs in parallel, the circuit
cannot correctly function as a 3-input OR gate.
◦ In the case when there is no ball falling on the circuit, the
equivalent resistance of the parallel combination of the 3
FSRs drops too low that Vout > 2V .
◦ As a result, the output FAILED to represent a logical LOW
in this case.
◦  A new R’ref is needed
16
Operational Amplifier
uA
mA
Output stage
Input Differential
Pair
Ack : http://www.ti.com/lit/ds/symlink/ua741.pdf
https://www.triadsemi.com/2014/06/02/asic-integration-power-savings/
17
Ideal Op-Amp
Ideal Op-Amp
Ack : http://physics.tutorvista.com/waves/direct-sensing.html
• Infinite Gain K, Vo = K (V+ - V-)
• V+ = V• Input Current in Op-Amp is ZERO
• (Input Resistance is Ideally infinite)  
• Vo,max = Vcc,
• Vo,min = -Vcc
18
Recap : Vi to +
19
Recap : Vi to -
20
Recap : Buffer

Op-Amp Buffer
◦ Non-Inverting Op-Amp
◦ R1 = Infinite, R2 = 0

R1 = Infinite,
R2 = 0
GAIN = ??
21
Example 4 :

The circuit is controlled by the charge output
from a digital system.
◦ The digital signal represent a logic “1” using 2V, and
represents a “0” using 0V.

The output of the circuit is the signal Vcharge.
◦ To charge the battery, Vcharge = 20V.
◦ When it is not charging, Vcharge = 0V.
22
Solution 4 :
Complete the following op-amp circuit by
i) determining the values of the resistors
R1, R2, R3, R4;
ii) determining the connection to point A and B.  
• For point A and B, indicate if it should be
connected to (i) charge if (ii) GND or (iii) 20V.
•
23
Solution 4 :
1
:
9
• Consider the case when charge is HIGH and
Vcharge is HIGH
• The configuration is possible
• when R3 = 0 (i.e. short circuit)
• R4 = infinite (i.e. open circuit),
• B = charge    2V
• Thus, A = GND  R1:R2 = 1:9
24
Solution 4 :
• When B = Low (0), Vcharge = 0
25
Example 5 :
• When V1 = 1V and V2 = 2V,
determine the current Ix
and the voltage VA.
•  Determine a general
expression for VA in terms
of V1 and V2.
26
Solution 5 :



When V1= 1V and V2= 2V, Ix = 1A
When V1= 1V and V2 = 2V, VA= 4V
A general expression for VA :
VA = V2 + Ix X 2
= V2 + ( V2 – V1 )/1 X 2
= V2 + 2( V2 – V1)
= -2V1 + 3 V2
​
27
Example 6 :
Use a single op-amp and
resistors to make a circuit
that is equivalent to the
following circuit.
Vn/Vi = (1+R2/R1) …….
Vo/Vi
= (Vo/Vn)(Vn/Vi)
= (1+R4/R3)(1+R2/R1)
=1+ (R1R4+R2R3+R2R4)/R1R3
28
Example 7
Use the ideal op-amp model (V+ = V-) to determine
an expression for the output current Io in terms of
the input voltage Vi and resistors R1 and R2.
Vx = (Vi+Vx) R2 /(R1+R2)
Vx = ViR2/R1
Io = Vx/R2
=(ViR2/R1)/R2
= Vi/R1
Example 8(a)
• The shaft angle of the output pot tracks that of
the input pot
• If the person turn the left potentiometer (the
input pot), then the motor will turn the right
potentiometer (the output pot)
30
Solution 8(a)
• Pot resistances depends on shaft angle
• Lower part of the pot is αR
• Upper part is (1 − α)R, where R = 1000Ω.
• α is from 0 (least clockwise) to 1 (most clockwise)
• If αi >αo, then the voltage to the motor (VM+ − VM−)
is positive, and the motor turns clockwise (so as
to increase αo)
31
Solution 8(a)
• Determine an expression for VM+ in terms of αi, R,
and VS.
• The output of the voltage divider is
• The op-amp provides a gain of 1, so VM+ = V+.
V+
32
Solution 8(b)


The following circuit produces a voltage Vo that depends
on the position of the input pot. Determine an expression
for the voltage Vo in terms of αi, R, R1, R2, and VS.
The positive input to the op-amp is connected to a voltage
divider with equal resistors so

The input pot is on the output of the op-amp, so

In an ideal op-amp,V+ = V− so
33
Solution 8(c)


The following circuit produces a voltage Vo that depends on
the positions of both pots. Determine an expression for Vo in
terms of αi, αo, R, and VS.
The positive input to the op-amp is connected to pot 1 so
that

The output pot is on the
output of the op-amp, so

In an ideal op-amp,V+ = V− so
34
Solution 8(d)


Assume that we are provided with a circuit whose
output is αi/αo volts. We want to design a controller of
the following form so that the motor shaft angle
(proportional to αo) will track the input pot angle
(proportional to αi).
Assume that R1 = R3 = R4 = 1000Ω and VC = 0.
◦ Is it possible to choose R2 so that αo tracks αi ?
◦ If yes, enter an acceptable value for R2.
Solution 8(d)
Assume that R1 = R3 = R4 = 1000Ω and VC = 0
 If R3 = R4 then the right motor input is 5V. If αi =
αo then the gain of the left op-amp circuit must be
5 so that the motor voltage is 0.
 The gain is (1 + R2/R1), so R2 must be 4000Ω.

36
Lab (5)

Can Vp go to 6V ? & How ?
Vp max =
VccRpot/(Ro+R pot)
37
Lab (5)
Ack : http://www.electronics-tutorials.ws/diode/diode_4.html
Vdrop
38