lecture iii - physics

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

http://physics.wm.edu/~ran/Pages/fc.html

http://physics.wm.edu/~ran/

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

Ran Yang || [email protected]

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


Capac i tor s


• 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


• 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


• 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



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


•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



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


• 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


• 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


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



𝑉𝑖𝑛

Decibel voltage gain: 𝐴𝑣(𝑑𝐵) = 20 log𝐴𝑣

I n te rpre t Bode p lo t



𝑉𝑖𝑛


• 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



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


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


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 ?


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 ?


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

lecture iii - physics

Documents