SEP600
Embedded System
Module 2: Transistor and Op-Amp
Course Outline
Module 1: Introduction and Review
Module 2: Transistor and Op-Amp
Module 3: Digital I/O and Analog Output
Module 4: Analog Input, Resolution, and Sampling
Module 5: Serial, SPI, I2C, CAN
Module 6: Timers, Interrupts, Task
2
Operational Amplifier (Op-Amp)
●
Amplification
●
Buffer/Voltage Follower
●
Addition, Subtraction,
Integration
●
Filtering
Source: Wikipedia: Op Amp
3
Op-Amp Characteristics
An operational amplifier can be described
as a dependent voltage source on the
differential input voltage between its input
terminals.
Ideal Op-Amp Actual Op-Amp
●
Rin = ∞ ●
Rin = 300kΩ to 1000GΩ
●
I+ = I - = 0 ●
I+ = I - ≠ 0
●
Amax = ∞ ●
Amax = 104 to 106
●
Rout = 0
●
Rout = 10Ω to 5kΩ
●
BW = ∞
●
BW = 0.5 to 15MHz
●
Input Offset = 0
●
Input Offset = 0.1 to 5mV
– Vin to get Vout = 0
4
Open Loop Op-Amp
Source: Electronics Tutorials
5
Open Loop Op-Amp: Comparator
Convert an analog signal into
ON or OFF depending on the
Non-Inverting: reference (threshold):
(-) as reference
V
Inverting:
(+) as reference
Source: Electronics Tutorials
6
Example
An 0-5V sinusoidal signal is connected to the positive input
of a comparator. Plot the output signal of the comparator if
the negative input is connected to 2V.
7
Closed Loop Op-Amp
Positive Feedback Negative Feedback
●
Hysteresis ●
Voltage Follower / Buffer
●
Inverting Amplifier
●
Non-Inverting Amplifier
●
Summing Amplifier
●
Difference Amplifier
●
Instrumentation Amplifier
8
Op-amp Comparator with Hysteresis
Feedback Fraction:
Non-Inverting:
(-) as reference R1
β=
R1+ R2
Inverting:
(+) as reference
9
Op-amp Comparator with Hysteresis
Feedback Fraction: Amount of Hysteresis:
V hysteresis =V UTP −V LTP
R1 V hysteresis =+ β V cc −(−β V cc )
β=
R1+ R2 V hysteresis =2 β V cc
UTP – Upper Trip Point, LTP – Lower Trip Point
10
Op-amp Comparator with Hysteresis
Example:
Find the upper and lower switch points and hysteresis width
of an op-amp comparator with hysteresis circuit with R1 =
10kΩ, R2 = 90kΩ, +Vcc = 10V, -Vcc = -10V.
11
Op-amp Comparator with Hysteresis
R1 10 k Ω
β= = =0.1
R 1 + R 2 10 k Ω+90 k Ω
R1
UTP= ×(+V cc )=β ×(+V cc )=0.1×(+10 V)=+1.0 V
R1 + R 2
R1
LTP= ×(−V cc )=β ×(−V cc )=0.1×(−10 V)=−1.0 V
R1+ R2
V HYS =UTP−LTP=1.0 V−(−1.0 V)=2.0 V
12
Voltage Follower Circuit
V o =V i
13
Inverting Amplifier Circuit
Rf
V o =− V i
Rs
14
Example with Inverting Amplifier
Assume all op-amps are ideal, find ia.
−47
V b= (1V )=−4.7 V
10
15
Example with Inverting Amplifier
−47
V b= (1V )=−4.7V
10
−220
V a= (150 mV )=−1V
33
V a −V b −1V −(−4.7 V )
ia = = =3.7 mA
R 1kΩ
16
Non-Inverting Amplifier Circuit
R s+ Rf
V o= Vi
Rs
17
Summing Amplifier Circuit
(
V o =−
Rf
Ra
Rf Rf
V a+ V b+ V c
Rb Rc )
18
Example with Summing Amplifier
Assume all op-amps are ideal, find Vout.
19
Example with Summing Amplifier
Assume all op-amps are ideal, find Vout.
−50
V a= V in =−5V in
10
5
V b= (15V )=5 V
10+5
V o =− (
10
10
10
(−5 V in )+ (5V )
5 )
V o =−(−5V in +10)=5 V in −10 V
20
Differential Amplifier Circuit
R d (R a + Rb ) Rb
V o= V b− V a
R a (R c + R d ) Ra
21
Differential Amplifier Circuit
R d (R a + Rb ) Rb
V o= V b− V a
R a (R c + R d ) Ra
Ra Rc
If: =
Rb Rd
Rb
Then: V o = (V b −V a )
Ra
22
Differential Amplifier Circuit
R d (R a + Rb ) Rb
V o= V b− V a
Differential Mode Input:
R a (R c + R d ) Ra
V dm =V b −V a
1
Then: V a =V cm− V dm
2
Common Mode Input (ie. Noise):
(V a +V b ) 1
V b =V cm + V dm
V cm = 2
2
23
Differential Amplifier Circuit
R d (R a + Rb ) Rb 1 1
V o= V b− V a V a =V cm− V dm V b =V cm + V dm
R a (R c + R d ) Ra 2 2
V o=
[
(R a R d −R b R c )
R a (R c + R d ) ] [
V cm +
2 R a (R c + R d ) ]
R d (R a + R b )+ R b ( R c + R d )
V dm
V o = A cm V cm + A dm V dm
common mode gain differential mode gain
24
Common-Mode Rejection Ratio (CMRR)
CMRR is used to evaluate how near ideal a differential
amplifier is.
CMRR=
Ideal differential amplifier:
| |
A dm
A cm
A cm=0 large A dm
R a R d −R b R c =0
25
Example with Differential Amplifier
Assume all op-amps are ideal, find the CMRR.
26
Example with Differential Amplifier
Assume all op-amps are ideal, find the CMRR.
(10)(100)−(96)(10) 40
A cm= = =0.0363636
(10)(10+100) 1100
(100)(10+96)+(96)(10+100) 21160
A dm= = =9.6181818
2(10)(10+100) 2200
A dm 9.6181818
CMRR=| |= =264.5
A cm 0.0363636
27
Differential Amplifier Application
Wheatstone Bridge Differential Amp.
Source: Nature
Applications: Strain Gauge, Force Sensor, Electromyography
(EMG), Capacitive Sensing, etc.
28
Example with Differential Amplifier
Design a differential amplifier circuit that can convert a 0-0.5V differential
signal to a 0-3.3V microcontroller input. The input resistance should be
10kΩ per leg.
29
Example with Differential Amplifier
Design a differential amplifier circuit that can convert a 0-0.5V differential
signal to a 0-3.3V microcontroller input. The input resistance should be
10kΩ per leg.
3.3 Rb Rd
A= =6.6= =
0.5 Ra Rc
R a=10 k Ω R c + R d =10 k Ω
R b=(6.6) R a =66 k Ω 7.6 R c =10 k Ω
R c =1.315 k Ω
R d =8.684 k Ω
30
Bipolar Junction Transistor (BJT)
Region of Operations:
●
Active – as amplifier, IC = β * IB
●
Saturation – Fully-ON, IC = ISat
●
Cut-off – Fully-OFF, IC = 0
Circuit Configuration:
Source: Electronics Tutorials
31
Bipolar Junction Transistor (BJT) as Switch
To operate as a switch, we want the transistor to only be in
the Saturation (ON) or Cut-off (OFF) regions.
●
Base-Emitter voltage VBE > 0.7V
●
Max Collector current flows (IC = VCC/RL)
●
VCE = 0 (ideal saturation)
●
Operating as “closed switch”
●
Base-Emitter voltage VBE < 0.7V
●
No Collector current (IC = 0)
●
Operating as “open switch”
Source: Electronics Tutorials
32
Digital Logic Transistor Switch
NPN Transistor Switch PNP Transistor Switch
●
Allow a HIGH digital signal (ie. From ●
Allow a LOW digital signal (ie. From
microcontroller) to control turning microcontroller) to control turning
ON a load ON a load
Source: Electronics Tutorials
33
Bipolar Junction Transistor (BJT) as Switch
Example:
Find the base current, IB, and maximum resistor RB that can control a 100mA
load current, IC. The control voltage is Vin = 5V and the transistor gain, β = 200.
RBE is a high resistance discharge / initialization resistor and can be ignored.
Source: Electronics Tutorials
34
Bipolar Junction Transistor (BJT) as Switch
Example:
Find the base current, IB, and maximum resistor RB that can control a 100mA
load current, IC. The control voltage is Vin = 5V and the transistor gain, β = 200.
RBE is a high resistance discharge / initialization resistor and can be ignored.
I C 100 mA
I B= = =0.5 mA
β 200
V IN −V BE 5.0 V−0.7 V
RB= = =8.6 k Ω
IB 0.5 mA
Source: Electronics Tutorials
35
Darlington Transistor Switch
Allow a lower gain
transistor to control a
higher gain transistor
to control a large load
current.
βTotal =β1×β2
In our example earlier, if a
darlington configuration with β1 =
200 and β2 = 50 is used, βTotal =
1000 so only IB = 10µA is required.
Source: Electronics Tutorials
36
Course Outline
Module 1: Introduction and Review
Module 2: Transistor and Op-Amp
Module 3: Digital I/O and Analog Output
Module 4: Analog Input, Resolution, and Sampling
Module 5: Serial, SPI, I2C, CAN
Module 6: Timers, Interrupts, Task
37
