E lectron ics
S p r i n g 2 0 2 0 R a n Ya n g
http://physics.wm.edu/~ran/
L e c t u r e i i i
What are we doing everyweek?
Design
• Physics, Math, Aesthetics
Simulation
Build in real
world
Trouble Shooting
What are we doing
every week?
Design
• Pre-lab Homework
• Lecture
Simulation
• Pre-lab Homework
• Lecture
Build in real world
• Lab
Trouble Shooting
• Lab
• Simulation
Capac i tor s
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Electrolytic Capacitors
• 1µF to 1 F
• 5V - 500V
• Most are polarized
Tantalum electrolytic capacitors
• All polarized
• 0.1µF to 1000µF
• 2V to 50V
Ceramic capacitors
• Non-polarized
• pF to 500nF to µF
• 1000V
Film Capacitors
• Non-polarized
• 50pF - 1000µF
• 10V - 1000V
Mica Capacitors
• 1pF to 10nF
• 1000V
• non-polarized
Capac i tor s
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Capac i tor s
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• cheap
We mostly use ceramic capacitors
• larger capacitance,
• larger in physical size
And film capacitors
• Great, you won’t burn anything!
Both are non-polarized
Capac i tor formu la
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• 𝑄 = 𝐶𝑉
A capacitor C:
•𝑑𝑄
𝑑𝑡= 𝐼 = 𝐶 ∙
𝑑𝑉
𝑑𝑡
The time rate of the charge: I
• pF, 10nF, 20µF… etc
Capacitance C, unit: Farad
Capac i t o r i n a c i r c u i t examp leDi s c harg i ng
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• 𝑉𝑖𝑛 − 𝑉𝑜𝑢𝑡 = 𝐼𝑅 𝐼 = 𝐶 ∙𝑑𝑉𝑜𝑢𝑡
𝑑𝑡
Current flows from source to R to C
• 𝑉𝑖𝑛 − 𝑉𝑜𝑢𝑡 = 𝑅𝐶 ∙𝑑𝑉𝑜𝑢𝑡
𝑑𝑡
At any given time t
• 𝜏 = 𝑅𝐶
Time constant 𝜏 (Ω ∙ 𝐹 = 𝑠)
Capac i t o r D i s c harg i ngmore math
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• 𝑉𝑖𝑛 − 𝑉𝑜𝑢𝑡 = 𝑅𝐶 ∙𝑑𝑉𝑜𝑢𝑡
𝑑𝑡𝜏 = 𝑅𝐶
Let’s solve this little differential equation
• 0 − 𝑉𝑜𝑢𝑡 = 𝜏 ∙𝑑𝑉𝑜𝑢𝑡
𝑑𝑡
•𝑑𝑉𝑜𝑢𝑡
𝑉𝑜𝑢𝑡= −
𝑑𝑡
𝜏
at 𝑡 = 0, 𝑉𝑖𝑛 = 0
• 𝑑𝑉𝑜𝑢𝑡
𝑉𝑜𝑢𝑡= −
𝑑𝑡
𝜏
• 𝑑𝑉𝑜𝑢𝑡
𝑉𝑜𝑢𝑡= ln 𝑉𝑜𝑢𝑡 + 𝐶1 −
𝑑𝑡
𝜏= −
𝑡
𝜏
Integral on both side
• ln 𝑉𝑜𝑢𝑡 = −𝑡
𝜏+ 𝐶1
• 𝑉𝑜𝑢𝑡 = 𝑒−𝑡
𝜏+𝐶1
• 𝑉𝑜𝑢𝑡 = 𝑒𝐶1 ∙ 𝑒−𝑡
𝜏 (𝐶 = 𝑒𝐶1 )
• 𝑉𝑜𝑢𝑡 = 𝐶𝑒−𝑡
𝜏
Integral on both side, continued.
• 𝑉𝑜𝑢𝑡 = 𝑉0𝑒−𝑡
𝜏
Constant C can be determined by initial condition at 𝑡 = 0 𝑉𝑜𝑢𝑡 0 = 𝐶 = 𝑉0
Capac i tor D i s c harg ing Cur ve
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•𝑉𝑜𝑢𝑡 = 𝑉0𝑒−𝑡
𝜏
• 𝑡 = 𝑅𝐶 = 𝜏, 𝑉𝑜𝑢𝑡 = 0.37𝑉0
Constant C can be determined by initial condition at 𝑡 = 0 𝑉𝑜𝑢𝑡 0 = 𝐶 = 𝑉0
Appl i ca t ion : RC c i r cu i t
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• 𝑉𝑖𝑛 − 𝑉𝑜𝑢𝑡 = 𝑅𝐶 ∙𝑑𝑉𝑜𝑢𝑡
𝑑𝑡
Luckily, it’s the same circuit we just discussed
• input signal:
• 𝑉𝑖𝑛 𝑡 = cos𝜔𝑡 + 𝑗 sin𝜔𝑡 = 𝑒𝑗𝜔𝑡
• output signal:
• 𝑉𝑜𝑢𝑡 𝑡 = 𝑉0𝑒𝑗𝜔𝑡. (V0 is a complex
number)
Time-dependent input voltage (we discussed last week)
• 𝑒𝑗𝜔𝑡 − 𝑉0𝑒𝑗𝜔𝑡 = 𝑅𝐶 ∙ 𝑗𝜔𝑉0𝑒
𝑗𝜔𝑡
• 1 − 𝑉0 = 𝑅𝐶 ∙ 𝑗𝜔𝑉0
• 𝑉0 =1
1+𝑅𝐶𝜔𝑗
Plug in
• 𝑉0 =1
1+𝑅𝐶𝜔𝑗
Greatly simplified math by working with complex numbers
• 𝑥 + 𝑗𝑦 = 𝑥2 + 𝑦2𝑒𝑗 tan−1 Τ𝑦 𝑥
Recall:
• 𝑉0 =1
1+ 𝑅𝐶𝜔 2𝑒𝑗 tan−1 𝑅𝐶𝜔
=1
1+ 𝑅𝐶𝜔 2𝑒−𝑗 tan
−1 𝑅𝐶𝜔
• Amplitude: 1
1+ 𝑅𝐶𝜔 2
• Phase: tan−1 𝑅𝐶𝜔
We can write our solution
Appl i ca t ion : RC c i r cu i t
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• Amplitude: 1
1+ 𝑅𝐶𝜔 2
• Phase: tan−1 𝑅𝐶𝜔 unit: 𝜔: 𝑟𝑎𝑑/𝑠
solution
• 𝜔 → 0 low frequency
• 𝐴 → 1 𝜙 → 0° 𝑉𝑜𝑢𝑡 ≈ 𝑉𝑖𝑛• No amplitude or phase difference
• 𝜔 → ∞ high frequency
• 𝐴 =1
𝑅𝐶𝜔→ a very small value
• 𝜙 → −90°
Frequency analysis
• "Open" at low frequency
• "Short" at high frequency
For a capacitor
Appl i ca t ion : low pass f i l te r
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• 𝜔 → 0 low frequency
• 𝐴 → 1 𝜙 → 0° 𝑉𝑜𝑢𝑡 ≈ 𝑉𝑖𝑛• No amplitude or phase difference
• 𝜔 → ∞ high frequency
• 𝐴 =1
𝑅𝐶𝜔→ a very small value
• 𝜙 → −90°
Frequency analysis
l ow pass f i l te r : Bode p lo t
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Our plot uses logarithmic scale on both x and y axis
Bode plot: The y axis uses a linear scale and the x axis uses a logarithmic scale
First define voltage gain : 𝐴𝑣 =𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
• we define Decibel power gain as : 𝐴𝑝 𝑑𝐵 = 10 log 𝐴𝑝 = 10 log𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
• And we discussed last week: 𝑃 =𝑉2
𝑅
• Plug in: 10 log𝑃𝑜𝑢𝑡
𝑃𝑖𝑛= 10 log
𝑉𝑜𝑢𝑡2
𝑉𝑖𝑛2 =20 log 𝐴𝑣
• And just in case : 𝑥 = 10𝑦 𝑦 = log10 𝑥 = log 𝑥
Decibel voltage gain: 𝐴𝑣(𝑑𝐵) = 20 log 𝐴𝑣
l ow pass f i l te r : Bode p lo t
Ran Yang || [email protected]
First define voltage gain : 𝐴𝑣 =𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
Decibel voltage gain: 𝐴𝑣(𝑑𝐵) = 20 log𝐴𝑣
I n te rpre t Bode p lo t
Ran Yang || [email protected]
First define voltage gain : 𝐴𝑣 =𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
Decibel voltage gain: 𝐴𝑣(𝑑𝐵) = 20 log𝐴𝑣
• 𝐴𝑣(𝑑𝐵) = 20 log𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
• 𝑉𝑖𝑛 = 1𝑉 𝑉𝑜𝑢𝑡 = 10𝐴𝑣(𝑑𝐵)
20 = 10−3
20 = 0.707𝑉
𝐴𝑣(𝑑𝐵) = −3𝑑𝐵
When 𝐴𝑣(𝑑𝐵) = −3𝑑𝐵, 𝑉𝑜𝑢𝑡
𝑉𝑖𝑛= 0.707
Well, what’s so special about 3dB?
I n te rpre t Bode p lo t : 3dB
Ran Yang || [email protected]
Decibel voltage gain: 𝐴𝑣(𝑑𝐵) = 20 log𝐴𝑣
Decibel power gain : 𝐴𝑝(𝑑𝐵) = 10 log𝐴𝑝
• 𝐴𝑝(𝑑𝐵) = 10 log𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
• 𝑃𝑖𝑛 = 1𝑊 𝑃𝑜𝑢𝑡 = 10𝐴𝑝(𝑑𝐵)
10 = 10−3
10 = 0.5𝑊
If power decreases 3dB: 𝐴𝑝(𝑑𝐵) = −3𝑑𝐵
When 𝐴𝑝(𝑑𝐵) = −3𝑑𝐵, 𝑃𝑜𝑢𝑡
𝑃𝑖𝑛= 0.5 =
1
2
That’s why we will use 3dB as a reference all the time.
When the power gain increases/decreases by a factor of 2, the decibel power gain increases/decreases by 3dB.
I n t e rp re t Bode p lo t : Cu to f f F requency 𝑓𝑐
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At cutoff frequency, the power is halved
• 𝑉𝑜𝑢𝑡 =1
1+ 𝑅𝐶𝜔 2(𝑉𝑖𝑛 = 1𝑉)
• At -3dB, half output power and 𝑉𝑜𝑢𝑡 = 0.707𝑉
• Plug in, solve 𝜔 =1
𝑅𝐶
• 𝜔 = 2𝜋𝑓
Back to the output voltage Amplitude formula:
• 𝑓𝑐 =1
2𝜋𝑅𝐶
• This example: 𝑓𝑐 = 219𝐻𝑧
f here is our cutoff frequency
High pass f i l te r
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Same process but switch the position of Resistor and capacitor
•What's output voltage amplitude formula for a high pass filter?
•Plot voltage gain vs frequency (Hz)
•Bode plot
Practice:
What have we l ea r ned t oday ?
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Different types of capacitors
Time dependent RC circuit
Passive filters
W h a t t o s t u dy fo r n ex t we e k ?
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Read Textbook Chapter 2 -Semiconductors
1
Read Textbook Chapter 3 – Diode theory
2
Try Chapter 4 and 5
• Ac to Dc
• Zener diode
• LED
3
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