electronics - rowan universityusers.rowan.edu/~krchnavek/rowan_university/... · zener diode •...
TRANSCRIPT
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ElectronicsRobert R. Krchnavek
Spring, 2020
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Review – Principles of Electric Circuit Analysis (PECA)
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Major Concepts in PECA• Basic Elements – R, L, C, M
• Sources – independent and dependent
• Analysis – Node voltage, mesh current, loop current, CAD.
• Complex impedance
• H(s) vs h(t)
• Op amps
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Minimum Required Skills• Recognize and solve a voltage divider. More detailed analysis, e.g. nodal
analysis, may be required but often is not.
• Handle impedances (R, L, and C) in series and parallel.
• Recognize and solve first-order systems (e.g., 1 pole, low-pass filter)
• transient response (t=0+).
• steady state response (t -> ∞)
• Solve op amp circuits: inverting, noninverting, differential, instrumentation, and others.
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Op Amps Real vs Ideal?
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Op Amps Ideal op amp
Power supplies: ignored Output current capability: Bias current: 0 Offset voltage: 0 Open loop gain: Bandwidth: Input impedance:
∞
∞∞∞
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Op AmpsA comparison between two old op amps. Op amps keep getting better.
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Bode Plots
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Bode Plots
• Invented by Hendrik Wade Bode in the 1930s.
• A frequency response curve for a system.
• Usually includes both magnitude and phase.
• Various aspects of Bode plots allow one to easily account for the frequency aspects/requirements of a system.
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Bode PlotsConsider the following RC circuit:
Vi(s) Vo(s)R
C
The steady-state (SS) response of this system is given by
H(s) =Vo(s)Vi(s)
=1
1 + sRC
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Bode Plotsτ = RC
ω0 =1
τs = 𝚥ω
Substituting the following:
H(s) =Vo(s)Vi(s)
=1
1 + sRC=
1
1 + sω0
=1
1 +𝚥ωω0
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Bode Plots
x-axis: logarithmic (base 10). y-axis: linear
Magnitude
Mag
nitu
de
0
0.25
0.5
0.75
1
1.25
Frequency (rad/s)1 10 100 1000 10000 100000 1000000
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Bode Plots• Constant ⟶ constant in magnitude and phase angle
• Zero ⟶
• Simple zero ⟶
• Pole ⟶
• Simple pole ⟶
5 Basic Responses:
1 + s ⇒ 1 + 𝚥ωω0
s ⇒ 𝚥ωω0
1
1 + s⇒
1
1 + 𝚥 ωω0
1
s⇒
1
𝚥 ωω0
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Bode PlotThe low-pass filter is an example of a pole:
1
1 + s⇒
1
1 + 𝚥 ωω0
https://www.electronics-tutorials.ws/filter/filter_2.html
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Bode PlotThe high-pass filter is an example of a zero:
https://www.electronics-tutorials.ws/filter/filter_2.html
1 + s ⇒ 1 + 𝚥ωω0
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Bode Plots• Constant
• Zero ⟶
• Simple zero ⟶
• Pole ⟶
• Simple pole ⟶
5 Basic Responses:
1 + s ⇒ 1 + 𝚥ωω0
s ⇒ 𝚥ωω0
1
1 + s⇒
1
1 + 𝚥 ωω0
1
s⇒
1
𝚥 ωω0
ω0
ω0
ω0
ω0
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Bode PlotsExample 1 – Develop a “model” circuit using ideal components that models the frequency response of a real op amp.
Consider the AD712 op amp. Obtain the data sheet and look up the bandwidth (BW) and open loop gain of the op amp.
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Bode Plots
Open loop gain in dB:
A(dB) = 20log10(150,000) ≈ 100
BW – 3 MHz. Open loop gain is 1. Unity gain or crossover frequency.
-20 dB/decade
Example 1 – Modeling the frequency response of an op amp
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Bode Plots
R1
Vi
Vo
R2
R
C
Assumptions: Gain is 150,000 dB. Roll off starts at 100 Hz.
≈ 100
Vo
Vi= 1 +
R2
R1= 150,000
R1 = 1kΩR2 = 150MΩ
Let then
f0 = 100Hzω0 = 2πf0ω0 = 628(rad/s)
Let R = 1kΩ
RC = τ =1
ω0
1000C =1
628then
C = 1.6μF
Example 1 – Modeling the frequency response of an op amp
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Bode PlotsExample 1 – Modeling the frequency response of an op amp
10 1000 106 f, Hz
ideal op amp
filter response
|A( !ω) |
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Bode PlotsExample 1 – Modeling the frequency response of an op amp
10 1000 106 f, Hz
ideal op amp
filter response
total response
|A( !" ) |
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Bode PlotsExample 1 – Modeling the frequency response of an op amp
10 1000 106 f, Hz
∠A(j!)
ideal op amp
filter response
total response
|A( !" ) |
-90
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Bode PlotsExample 2 – Bass Boost Amplifier
2. Design a bass amplifier that provides gain between 25Hz and 100Hz.
K
25 20,000100 f, Hz
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Bode Plots
K
25 20,000100 f, Hz
K HPF f1 = 25 Hz
LPF f2 = 100 Hz
Obtain the gain K with an ideal op amp circuit with gain.
Example 2 – Bass Boost Amplifier
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Op Amps
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Op Amps
• What are the differences between an ideal op amp and a real op amp?
• What causes these differences?
• How can you minimize their impact on circuit design?
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Op Ampsv+
v+
vo
v-
v-
vo=A(v+- v-)
Ideal • • • • • • • • • •
Real • • • • • • • • • •
AOL = ∞AOL ≠ f( 𝚥ω)Rin = ∞Rout = 0
ignore power supplies
v+ − v− = vd = 0
common mode gain = 0
BW = 0slew rate = 0
drift = 0
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Op Amps
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Diodes
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Diodes• A non-linear, two-terminal, passive device.
• Many types:
• Signal or Rectifier diodes, e.g, Si “pn” junction diodes. Also Ge, SiC.
• Schottky diodes, e.g., metal-semiconductor diodes.
• Light-emitting diode (LED), multiple colors available. High-brightness, blue LEDs (1993) led to the revolution in lighting we see today. See history in Wikipedia.
• Laser diodes. Used in optical communication systems.
• Zener
• PIN
• Gunn
• Tunnel
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Diodes
ID = IDSS(eVDVt − 1)
Vt =kTq k = 1.38(10−23) J/K
T in Kelvin
q = 1.6(10−19) C
IDSS or IS is 10−6 → 10−14 A
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Diodes
• Clamps
• Rectifier
• Peak Detector
• Doublers/Triplers
Applications
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Diode – Applications
Vi, (v+-v-)1-1
V+
V-
Vo
v+
v+
vo
v-
v-
vo=A(v+- v-)
+V
-V
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Diode – Applications
Vi, (v+-v-)1-1
V+
V-
Vo
v+
v+
vo
v-
v-
vo=A(v+- v-)
+V
-V
Never put a voltage on the input greater than .± V+
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Diode – ApplicationsRectifiers
vs
vL
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Diode – ApplicationsRectifiers
vs
vL
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Diode – ApplicationsRectifiers
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Diode – ApplicationsRectifiers
D1
C
D2
T1
CTvL < 0
What is vL?
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Diode – ApplicationsRectifiers
D1
C
D2
T1
CTvL < 0
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Diode – ApplicationsRipple and the size of the capacitor
vLC1
T1
D1
D2D4
D3
Load
iL
LOAD
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Diode – ApplicationsRipple and the size of the capacitor
Equivalent circuit when the capacitor is discharging through the load:
C iL
iC
vLvC
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Diode – ApplicationsCapacitor ESR
vLC1
T1
D1
D2D4
D3
Load
iL
LOAD
iL = 2A vripple = 1V
60 Hz
C= 16.7 mF
Find the following:
Assume transformer secondary is 12VAC and ideal diodes.
Ton Vs = 17VT = 16.7 msT/2 = 8.33 msT/4 = 4.17 msTon = 1ms
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Diode – ApplicationsCapacitor ESR
vLC1
T1
D1
D2D4
D3
Load
iL
LOAD
iL = 2A vripple = 1V
60 Hz
C= 16.7 mF
Find the following:
Assume transformer secondary is 12VAC and ideal diodes.
Ton Vs = 17VT = 16.7 msT/2 = 8.33 msT/4 = 4.17 msTon = 1ms
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Diode – Applications• Low loss requires a low ESR capacitor.
• Low ESR capacitors are usually expensive.
• The lifetime of a capacitor is usually inversely proportional to its temperature – hot capacitors die early.
• Another capacitor specification that must be met: WVDE - working volts, DC.
• Vpk on the capacitor should be 1.5 x Vpeak of your circuit. In our 12VAC supply, Vpeak is 17 so Vpk should be 25.
Capacitor ESR
NOTE: Similar details must be addressed for the transformer. Ipeak, cannot saturate the transformer, etc.
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Diode – Applications• Vripple cannot be made low enough (mV) by only using a capacitor.
• A “regulator” is used to further lower Vripple. (Note: The capacitor is still required.)
• There are many categories of regulators: linear vs nonlinear, shunt vs series. LM317 is a popular linear, series regulator.
• We will consider a simple, linear, shunt regulator using a Zener diode.
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Zener Diode• All diodes exhibit the reverse
“breakdown” shown in the curve.
• Most will fail in this region.
• Zener diodes are designed to work in this regime.
• A minimum current is required to achieve breakdown. See spec sheet. Typical is 1mA.
• Must consider the power dissipation in this region.
iDiZ
vD
vZ
VDVF
VZ
ID
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Zener Regulatorunregulated
power supply
load
vLvunreg RLD1
Rlim
voltage regulator
Design the Zener regulator:
•Assume vunreg = 16V •vL = 5.0V •iL,max = 2A
•Find a Zener diode. •Calculate PD,max •Calculate Rlim
![Page 52: Electronics - Rowan Universityusers.rowan.edu/~krchnavek/Rowan_University/... · Zener Diode • All diodes exhibit the reverse “breakdown” shown in the curve. • Most will fail](https://reader031.vdocuments.site/reader031/viewer/2022011817/5e7f9fe2e7f9b3568600ada3/html5/thumbnails/52.jpg)
Precision Rectifier
RL
vi
vo
Find vo/vi.
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Diode – ApplicationsWhat about the input to the supply?