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+ V D i D Cathode Anode anode (+) cathode (-) + V D

Anode - engineering.purdue.edu

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Page 1: Anode - engineering.purdue.edu

+

VD

iD

Cathode

Anode

anode (+)

cathode (-)

+ VD −

Page 2: Anode - engineering.purdue.edu
Page 3: Anode - engineering.purdue.edu
Page 4: Anode - engineering.purdue.edu

n+-type Material

p-type Material

n-type Material

Page 5: Anode - engineering.purdue.edu
Page 6: Anode - engineering.purdue.edu

iC

iB

iB

iC

Current controlled Voltage controlled

Page 7: Anode - engineering.purdue.edu

𝛼dc =𝑖𝐶

𝑖𝐸𝛽dc =

𝑖𝐶

𝑖𝐵

iE

iC

iE

iC

iB

Page 8: Anode - engineering.purdue.edu

𝑖𝐸 = 𝑖𝐶 + 𝑖𝐵

𝑖𝐸

𝑖𝐶= 1 +

𝑖𝐵

𝑖𝐶

1

𝛼dc= 1 +

1

𝛽dc

𝛽dc =𝛼dc

1−𝛼dc ↔ 𝛼dc =

𝛽dc

𝛽dc+1

iC

+ VCE

− VBE −

iB

iE

Page 9: Anode - engineering.purdue.edu

𝑖𝐶 ≈ 0, 𝑅𝐶𝐸 ≈ ∞

𝑖𝐶 = 𝑖𝐵 𝛽

− 𝑉𝐶𝐸 → 0

𝑖𝐶 ≈ 𝑖𝐶(limit), 𝑅𝐶𝐸 ≈ 0

iC

+ VCE

− VBE −

iB

iE

VCE

𝒊𝑪

𝒊𝑩 𝜷

Page 10: Anode - engineering.purdue.edu

𝑖𝐶 𝑖𝐵

β

𝑖𝐶 = 𝑖𝐵 𝛽; 𝑉𝐵𝐸 = 𝑉𝐹; 𝑉𝐶𝐸 > 𝑉𝐹

𝑃 = 𝑖𝐶 ⋅ 𝑉𝐶𝐸

iC

+ VCE

− VBE −

iB

iE

VCE

𝒊𝑪

𝒊𝑩 𝜷

Page 11: Anode - engineering.purdue.edu

VCE

𝑖𝐶

VCE

𝑖𝐵 = 0; 𝑖𝐶 ≈ 0; 𝑉𝐵𝐸 < 𝑉𝐹; 𝑉𝐶𝐸 ≥ 0

iC

+ VCE

− VBE −

iB

iE

VCE

𝒊𝑪

𝒊𝑩 𝜷

𝑖𝐵 >𝑖𝐶

𝛽; 𝑉𝐵𝐸 = 𝑉𝐹;

𝑉𝐶𝐸 = 𝑉𝑆𝐴𝑇 ≈ 0.2 𝑉

Page 12: Anode - engineering.purdue.edu

𝑖𝐵 ≈ 0 𝑉𝐼𝑁

− 𝑖𝐶 ≈ 𝑖𝐸 ≈ 0 ⇒ VOUT ≈ VCC

𝑖𝐵 > 𝑖𝐵(𝑠𝑎𝑡)

𝑉𝐼𝑁

− 𝑉𝑂𝑈𝑇 = 𝑉𝐶𝐸(𝑠𝑎𝑡)≈ 0.2 𝑉

𝑖𝐵 =𝑉𝐼𝑁 − 𝑉𝐵𝐸(𝑆𝐴𝑇)

𝑅𝐵; 𝑖𝐶 ≈

𝑉𝐶𝐶 − 𝑉𝐶𝐸(𝑆𝐴𝑇)

𝑅𝐿

𝒊𝑪

VCE

iB (sat)

iB ≈ 0 1

LR

𝑖𝑐

𝑉𝑜𝑢𝑡𝑉𝑖𝑛

0

0 𝑅𝐶𝐸 = ∞

𝑅𝐶𝐸 = 0

Increasing 𝑅2

+

𝑉𝐼𝑁

+

𝑉𝑂𝑈𝑇 −

𝒊𝑪

𝒊𝑬

𝒊𝑩

𝒊𝑪

Page 13: Anode - engineering.purdue.edu

𝑉𝑖𝑛

𝛽dc > 20

𝑖𝐵 >𝑖𝐶(limit)

10

𝒊𝑪

VCE

iB (sat)

iB ≈ 0 1

LR

𝑖𝑐

𝑉𝑜𝑢𝑡𝑉𝑖𝑛

0

0 𝑅𝐶𝐸 = ∞

𝑅𝐶𝐸 = 0

Increasing 𝑅2

𝒊𝑪

+

𝑉𝐼𝑁

+

𝑉𝑂𝑈𝑇 −

𝒊𝑪

𝒊𝑬

𝒊𝑩

Page 14: Anode - engineering.purdue.edu

𝑡𝐷

𝑡𝑅

𝑖𝐶

𝑡𝑆

𝑖𝐶

𝑡𝐹

𝑖𝐶

t

VIN

t

iC

tD

tR

tS tF

+

𝑉𝐼𝑁

+

𝑉𝑂𝑈𝑇 −

𝒊𝑪

𝒊𝑬

𝒊𝑩

Page 15: Anode - engineering.purdue.edu

𝑡𝑂𝑁 = 𝑡𝐷 + 𝑡𝑅

𝑡𝑂𝐹𝐹 = 𝑡𝑆 + 𝑡𝐹

• 𝑡𝑑

• 𝑡𝑟

• 𝑡𝑠

• 𝑡𝑓

t

VIN

t

iC

tD

tR

tS tF

+

𝑉𝐼𝑁

+

𝑉𝑂𝑈𝑇 −

𝒊𝑪

𝒊𝑬

𝒊𝑩

Page 16: Anode - engineering.purdue.edu

VCC RC

RB

+

VIN

𝑅𝐵

𝑉𝐼𝑁

𝑉𝐼𝑁

ATmega32

Page 17: Anode - engineering.purdue.edu

VCC RC

RB

+

VIN

𝑅𝐵 >𝑉𝐼𝑁 − 𝑉𝐹

𝐼𝐷𝑂,𝑚𝑎𝑥=

4.3 V

40 mA= 107.5 Ω ≈ 110 Ω

To avoid overloading digital output, might actually double or quadruple RB

ATmega32

Page 18: Anode - engineering.purdue.edu

VCC RC

RB

+

VIN

𝑅𝐶

𝑉𝐶𝐶 = 5𝑉

ATmega32

Page 19: Anode - engineering.purdue.edu

VCC RC

RB

+

VIN

𝑖𝐶(max) =𝑃max

𝑉𝐿𝐸𝐷=

80 mW

2 V= 40 mA

𝑅𝐶 >𝑉𝐶𝐶 − 𝑉𝐿𝐸𝐷 − 𝑉𝐶𝐸(𝑠𝑎𝑡)

𝑖𝐶(max)=

5 − 2 − 0.2 V

40 mA= 70 Ω

To maintain hard saturation, 𝑖𝐵 > 4 mA ⟹ 𝑅𝐵 < 1075 Ω

ATmega32

Page 20: Anode - engineering.purdue.edu

VCC RC

RB

+

VIN

𝑉𝐶𝐶

𝛽

ATmega32

Page 21: Anode - engineering.purdue.edu

VCC RC

RE

𝒊𝑪

𝒊𝑬

𝒊𝑩

+

VIN

ATmega32

Page 22: Anode - engineering.purdue.edu

𝛽

Page 23: Anode - engineering.purdue.edu

Image: http://hades.mech.northwestern.edu/images/d/d6/Phototransistor.jpg

Page 24: Anode - engineering.purdue.edu

+

VOUT

Image: https://www.sparkfun.com/products/9299

Page 25: Anode - engineering.purdue.edu

VOUT

VCC RC

VIN

RIN

Image: www.sparkfun.com/products/314

Page 26: Anode - engineering.purdue.edu
Page 27: Anode - engineering.purdue.edu

𝑅1

𝑅2

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iD

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iD

D

G

S

Page 35: Anode - engineering.purdue.edu

𝑉𝑇

𝑉𝐺𝑆(on)

iD

+ VDS

− VGS

+

D

G

S

VDD

RD

Page 36: Anode - engineering.purdue.edu

• 𝑖𝐷

• 𝑖𝐷

𝑉𝐺𝑆 < 𝑉𝑇 ⟹ 𝑖𝐷 ≈ 0; 𝑉𝐷𝑆 ≈ 𝑉𝐷𝐷iD

+ VDS

− VGS

+

RD

VDS

iD (mA)

VGS – VT

Cutoff (VGS < VT)

BVDS

𝑉𝐺𝑆 > 𝑉𝑇 & 𝑉𝐷𝑆 < 𝑉𝐺𝑆 − 𝑉𝑇 ≪ 𝑉𝐷𝐷 ⟹ 𝑖𝐷 ≈ 𝑉𝐷𝐷/𝑅𝐷 ;

𝑉𝐺𝑆 > 𝑉𝑇 & 𝑉𝐷𝑆 > 𝑉𝐺𝑆 − 𝑉𝑇 ⟹ 𝑖𝐷 ∝ 𝑉𝐺𝑆 − 𝑉𝑇2

𝑃 = 𝑖𝐷 ⋅ 𝑉𝐷𝑆

Page 37: Anode - engineering.purdue.edu

𝑉𝐼𝑁 < 𝑉𝑇

• 𝑖𝐷 ≈ 𝑖𝑆≈ 0 ⟹ 𝑉𝑂𝑈𝑇 ≈ 𝑉𝐷𝐷

𝑉𝐼𝑁 > 𝑉𝑇

• 𝑉𝑂𝑈𝑇 = 𝑉𝐷𝐷 − 𝑉𝐷𝑆 = 𝑉𝐷𝐷 − 𝑖𝐷 𝑉𝐺1 ⋅ 𝑅𝐷

VG = VT

iD

VDS

VG1

iD +

VOUT

+ VIN

+ VDS(sat)

+ VIN

+

VOUT

−1

𝑅𝐷

Page 38: Anode - engineering.purdue.edu

VOUT VIN

QP

QN

VDD

VGS +

VGS

+

CMOS Inverter

VDD

QP

QN

VOUT

VDD

QP

QN

VOUT

VIN low VIN high

Page 39: Anode - engineering.purdue.edu

𝑖𝐶𝑖𝐵

𝑖𝐷𝑉𝐺

Page 40: Anode - engineering.purdue.edu
Page 41: Anode - engineering.purdue.edu