Mechatronics 2

Question 1

Transistor current I_E

Answer 1

I_E = I_B + I_C

Question 2

Transistor voltage V_BE

Answer 2

V_BE = V_B - V_E

Question 3

Transistor voltage V_CE

Answer 3

V_CE = V_C - V_E

Question 4

Transistor - Common emitter configuration

Answer 4

Acts as a quasi-ideal current source

Question 5

Transistor base/collector currents i_c

Answer 5

i_c = h_fe * i_b

Question 6

Transistor as a switch (common emitter)

Answer 6

Low base resistance, increase base voltage over 0.6V to activate Collector to Emitter current, which scales with i_b

Question 7

Transistor as an amplifier (common emitter)

Answer 7

Operation between cut-off and saturation. As base current increases gently, collector current increases almost linearly with it.

Question 8

Transistor - Common collector configuration

Answer 8

V_cc at collector kept constant.
V_in at base can be increased
V_e at emitter (load voltage) rises

Question 9

Emitter-follower equation: V_L = V_load = …

Answer 9

V_L = (V_in - V_be) / (1 + 1/h_fe * R_b/R_L)

Question 10

Operational Amplifiers Basic Equation

Answer 10

v_o = A (v₊ - v_-)

Question 11

Op-amp current through v+ or v-

Answer 11

No Current

Question 12

Op-amp with v+ > v-

Answer 12

Outputs Vs+
(Positive power supply limit)

Question 13

Op-amp with v+ < v-

Answer 13

Outputs Vs-
(Negative power supply limit)

Question 14

Op-amp with feedback

Answer 14

v+ = v-

Question 15

Buffer Amp AKA Voltage Follower

Answer 15

• Op-amp with v- connected directly to v_o
• Ideal voltage source
• V_in = v+

Question 16

Non-inverting voltage amplifier

Answer 16

• Op-amp with v- connected to a PD between v_o and GND.
• Can only increase the voltage.
• v_in = v+
• v_o = ( (R1 + R2) / R2 ) Vin

Question 17

Inverting voltage amplifier

Answer 17

• Op-amp with v- connected through a resistor R2 to v_o
• v_in is connected through a resistor R1 to v-
• v+ is grounded.
• A_v (voltage gain) = -R2/R1

Question 18

Sampling op-amp stage

Answer 18

• Buffer amp (v- tied to v_o)
• But v+ has a capacitor to GND
• and v_in connects to v+ through a switch (open when holding a value)

Question 19

Current-to-Voltage Amplifier

Answer 19

• Op-amp with v- connected to v_o through a resistor
• v+ is tied to GND, keeping v- also at 0V
• i_in at v- determines the voltage across the resistor, hence v_o
• v_o = - i_in * R

Question 20

Summing Amplifier

Answer 20

• Many input voltages are “turned into currents” over resistors, and then go through a current-to-voltage amplifier
• v+ so pulled to GND through a resistor R0 which is the parallel sum of the input resistors.

Question 21

Difference Amplifier

Answer 21

v_o = k( v_i2 - v_i1 )
- Resistor to v_o is kR1
- Resistor to GND is kR2

Question 22

Complex 1/j

Answer 22

1/j = -j

Question 23

Impedance Phase from Phasor

Answer 23

Φ = tan^-1(^Im()/_Re())

Question 24

Resistor Impedance

Answer 24

Z_R = R + 0j
Φ = 0°

Question 25

Inductor Impedance

Answer 25

Z_L = 0 + Lωj Φ = 90°

Question 26

Capacitor Impedance

Answer 26

Z_C = 0 - j/ωC Φ = -90°

Question 27

Complex Phase and Gain, H (Voltage or Current)

Answer 27

H = |H| ∠ Φ

Question 28

Gain, |H|

Answer 28

|H| = |A_out| / |A_in|

Question 29

Phase Shift, Φ

Answer 29

Φ = Φ_out - Φ_in

Question 30

Inverting gain...

Answer 30

Represented by +ive gain and 180° Phase Shift

Question 31

|H|_dB

Answer 31

|H|_dB = 20 log₁₀(|H|)

Question 32

Corner / Cut-off Frequency (ω_c)

Answer 32

ω_c = The frequency at which the output power is half of the input power. - AKA Half Power Point - When |H| = root(1/2) - Or when |H|_dB = -3 dB

Question 33

Gain Bode Plot

Answer 33

|H|_dB vs log(frequency) Can be for voltage or current

Question 34

Phase Bode Plot

Answer 34

Φ(degrees) vs log(frequency)

Question 35

Passive Low-Pass Filter Layout

Answer 35

Capacitor and resistor in series over v_i, output is voltage over the capacitor, like a potential divider.

Question 36

Passive Low-Pass Filter Cutoff

Answer 36

ω = 1/RC

Question 37

Passive Low-Pass Filter Behaviour

Answer 37

As ω → 0, |H| → 1, Φ → 0 As ω → ∞, |H| → 0, Φ → -90 So high frequencies are removed

Question 38

Passive High-Pass Filter Layout

Answer 38

Same as Low-Pass, but with capacitor and resistor swapped Like a potential divider where V_out is over the resistor.

Question 39

Passive High-Pass Filter Cutoff

Answer 39

ω_c = 1/RC

Question 40

Passive High-Pass Filter Behaviour

Answer 40

As ω → 0, |H| → 0, Φ → 90 As ω → ∞, |H| → 1, Φ → 0 So low frequencies are removed

Question 41

Passive Filter Cutoff (Relating to voltage and limiting frequency)

Answer 41

ω_c = 1/RC = ω / sqrt( (Vi/Vo)² - 1 )

Question 42

Schmitt Trigger V_i1

Answer 42

V_i1 = (V_cc/R₁ + **V_S-**/R₃) * (R₁||R₂||R₃)

Question 43

Schmitt Trigger V_i2

Answer 43

V_i2 = (V_cc/R₁ + **V_S+**/R₃) * (R₁||R₂||R₃)

Question 44

Voltage Integrator Gain/Phase

Answer 44

H = j/ωRC

Question 45

Voltage Differentiator Gain/Phase

Answer 45

H = -jωRC

Question 46

Active Filter Cutoff

Answer 46

ω_c = 1/RC₂

Question 47

Active Low-Pass Filter DC Gain

Answer 47

R2/R1

Question 48

Active High-Pass Filter DC Gain

Answer 48

0

Question 49

Active High-Pass Filter HF Gain

Answer 49

C1/C2

Question 50

First Order Process H(s)

Answer 50

H(s) = K_H / ( 1 + τs )

Question 51

Unity Feedback Proportional Control Transfer Function

Answer 51

G(s) = K_pH(s) / ( 1 + K_pH(s) )

Question 52

Proportional Control Steady State Value

Answer 52

lim[s→0] (s·Y(s)) = K_pK_H / (1 + K_pK_H) * c

Question 53

Proportional Control Steady State Error

Answer 53

c / (1 + K_pK_H)