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課程資訊

課程名稱:電子學(一)
授課教師:副教授 林憶霞
指定教材:Microelctronic Circuits, Sedra&Smith

Microelctronic Circuits

Diodes


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  • 二極體(Diode):
    是一種電子元件,由半導體材料(如矽或鍺)製成,具有兩端極性(正極和負極)。其主要特性是單向導通性,即允許電流在一個方向流動,而在相反方向基本阻止電流流通。

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    當導通時,Diodes前後沒有電壓差,不導通前後有電壓差。
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Rectifier

  • 整流器:
    是一種電子裝置,用於將交流電(AC)轉換為直流電(DC)。這個過程稱為整流,是電力電子學中的基本技術之一。整流器廣泛應用於電源系統中,為電子設備提供穩定的直流電。

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    上圖即是運用二極體的特性確保電流符合直流電流。
Example 4.1

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  • Figure 4.4(a) shows a circuit for charging a 12-V battery. If
    vs
    is a sinusoid with 24-V peak amplitude,find the fraction of each cycle during which the diode conducts. Also, find the peak value of the diode current and the maximum reverse-bias voltage that appears across the diode.

(a).找出導通的週期比例
導通條件:

24cosθ=12cosθ=12
角度:
θ=60,2θ=120

導通周期比例:
比例=120360=13

導通的週期比例為

120360所以是
13

(b).找出最大的電流和反向偏壓差
Id=2412100=0.12A

0.12A和36V

Terminal Characteristics of Junction Diodes

  • 二極體接面的端點特性是描述二極體兩端電壓與電流關係的基本物理特性,這些特性可以通過二極體的I-V特性曲線進行可視化表示。
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    更多細節的版本
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    1.The forward-bias region, determined by
    v>0.

    2.The reverse-bias region, determined by
    v<0.

    3.The breakdown region, determined by
    v<VZK
The Forward-Bias Region
  • In the forward region the
    iv
    relationship is closely approximated by

    iIs(ev/VT1)(3.1)

    In this
    Is
    is a constant for given diode at a given temperature. It is usually called the saturation current,For small-signal diode, which are small-size diode intened for low-power applocations,
    Is
    is on the order of
    1015
    A.

    As s rule of thumb,
    Is
    doubles in value for every
    5C
    rise in temprature.

    v=VTlniIs(3.4)

    Evaluate the current
    I1
    corresponding to a diode voltage
    V1
    :

    I1=IseV1/VT

    Similarly, to
    I2
    :

    I2=IseV2/VT

    combined two equations
    I2I1=e(V2V1)/VT

    rewritten as
    V2V1=VTlnI2I1

    In base-10 logarithms, the diode voltage drop changes by
    2.3VT
    60mV

    通過對比兩個不同電壓對應的電流,推導出電流與電壓之間的對數關係,並且證明了對數型變化的規律。突出了在許多應用中,二極體的 i-v 關係是一個非常穩定且易於測量的特性。
The Reverse-Bias Region
  • 根據公式3.1我們知道
    i=Is
  • VT25mV
  • As s rule of thumb,
    Is
    doubles in value for every
    10C
    rise in temprature.A
The Breakdown Region
  • VZK
    is knee 在擊穿區域,反向電流迅速增加,但電壓的增長非常小。這是由於二極體的i-v曲線在此區域近似為一條垂直線,因此其電壓基本不再隨反向電流增長而顯著變化。
    雖然在擊穿區域運行通常不會對二極體造成損壞,但若反向電流過大,會產生過多的熱量,導致二極體損壞。因此,通常會通過外部電路限制二極體的反向電流,防止功率過大,從而避免損壞。

4.3 Modeling the Diode Forward Characteristic

  • 在二極體的端點特性之後,我們要開始去考慮導通時,有關二極體的更多特性。並且了解與之相關的電路模型。
4.3.1 The Exponential Model
  • 這個模型可以最精確地描述二極體在順偏區的運作,但是這個模型的特質為嚴重的非線性性質(severely nonlinear nature)所以非常難以使用。
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    • 假設
      VDD0.5V
      ,此時,
      ID
      將會遠大於
      IS
      ,並且彼此為指數關係。

      ID=ISeVD/VT(Eq.4.6)
    • 也可以運用歐姆定律化簡為。
      ID=VDDVDR(Eq.4.7)
4.3.2 Graphical Analysis Using the Exponential Model
  • 圖像的分析模型,如同其名稱,直接將Eq.4.6和Eq.4.7 運用i-v的座標圖來表示:
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    根據Eq.4.7 我們可以看出直線為負載線(load line),其與二極體特性曲線的交點為作用點(operating point)
    Q
    ,作用點對應的i,v即為
    ID
    VD
4.3.3 Iterative Analysis Using the Exponential Model
  • 可以運用Eq.4.6和Eq.4.7去做迭代分析,此處課本運用例題舉例說明:
  • Example 4.4
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    找出
    IDVD
    在Fig. 4.10的電路中,
    VDD=5VandR=1kΩ
    .
    假設
    1mA
    的電流下二極體電壓為0.7V。

ID=VDDVDR4.3mA
  在前面我們曾經提過 Eq.4.5 他可以協助我們精確的找出
VD

V2V1=2.3VTlogI2I1

  
2.3VT60mV
,所以我們得到:

V2=V1+0.06logI2I1

  
V1=0.7VI1=1mAI2=4.3mA
:

V2=0.7+0.06log4.3mA1mA=0.738V

  得到第一次迭代
VD=0.738V,ID=4.3mA
,帶入並計算第二次迭代:

ID=VDDVDR=4.262mA

V2=0.7+0.06log4.262mA4.3mA=0.738V

  得到第二次迭代
VD=0.738V,ID=4.262mA

意思就是第一次,先運用一些模型去猜測一個解,再逐步往下求更精確地解。一個不斷近似的過程。

4.3.5 The Constant-Voltage-Drop Model
  • 恆壓降模型,是最簡單並且運用最廣泛的模型,基於觀察和經驗得知
    VD
    介於0.6V到0.8V,所以直接近似
    VD
    為0.7V。

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    此處課本示範如果為到Ex4.4,如何求解?
    VD=0.7V

    ID=50.71k=4.3mA

    相比於指數模型,我們不需要多次求精確解,只要善用近似的特性求解即可。
The Ideal-Diode Model
  • 如果
    VDDVD
    ,將
    VD=0
    這時候就恰好符合理想二極體的情境,完全導通並且沒有電壓降,所以得名理想二極體模型。

    VD=0V

    ID=51k=5mA
The Small-Signal Model

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  • 小信號模型,類似Fig4.13 (b)時候,可以看到一個DC電源串聯一個微小的AC信號,此模型即是處理這個情況。由於此時交流信號變化非常小(
    5mV
    ),所以可以將其造成的非線性線性化,來進行電路分析。

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    vD(t)=VD+vd(t)(Eq.4.8)

    iD(t)=ISevD/VT(Eq.4.9)

    帶入Eq.4.8到Eq.4.9得到:
    iD(t)=ISe(VD+vd/VT

    展開為
    iD(t)=ISeVD/VTevd/VT(Eq.4.10)

    運用Eq.4.6可得到:
    iD(t)=IDevd/VT(Eq.4.12)

    如果現在v_d信號的振幅足夠小,就會滿足
    vdVT1
    。依據泰勒展開式可以得出以下公式Eq.4.14。

    iD(t)ID(1+vdVT)(Eq.4.14)

    這是小信號模型的近似值,對於振幅小於5mV有效。
    iD(t)=ID+IDVTvd(Eq.4.15)

    iD=ID+id(Eq.4.16)

    根據歐姆定律可以得到:
    vd=idrd,rd=VTID(Eq.4.18)
  • Example 4.5
    圖片

    圖片

    R=10kΩ,DC=10V,60Hzsinusoidof1Vp

    ID=100.710k=0.93mA

    rd=250.93=26.9Ω

    vd=10.026910+0.0269=2.68mV

    該值非常小,符合小信號模型。

Example

  • 3.4 Find the values of I and V in the Circuits shown in Fig. E4.4.

    圖片
    a.
    圖片

    b.
    圖片

      先猜測導通,計算後發現不符合。改用不導通計算,因為不導通不會有電流,自然也不會消耗電壓。
    c.
    圖片

    d.
    圖片

    e.
    圖片

      3V導通,但是2V, 1V不會導通。
    f.
    圖片

      電流分配:每顆二極體會根據其導通電壓和電源電壓的差異來分配電流。具有最低正向電壓的二極體1V會有最大的電壓差,因此它會承擔最多的電流。

  • Ex4.10 For the circuit in Fig. 4.10, find I D and VD , Use (a) iteration and (b) the constant-voltage-drop model

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  • Ex 4.11

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BJTs


  • 雙極體接面電晶體(Bipolar Junction Transistors),有三個電極emitter(E),base(B),collector ( C)。每個電極之間都有一個pn接面,emitter–base junction (EBJ)和collector–base junction (CBJ)依照他們的bias會有相對應的模式,active mode(導通), cutoff mode(截止),saturation mode(飽和)。
    image

    image

    image

6.1.2 Operation of the npn Transistor in the Active Mode

  • 導通的情況是我們最關注的,也因此他應該要是最先介紹的,觀察Fig6.3 我們可以一些推導來開展BJT的分析。
    image

    以電子流(Current Flow)角度, 當BJT受到一個電場下,且EBJ為順偏置CBJ為逆偏置時為導通,Emitter的電子會跨過EBJ其中少數的電子會和Base的少數電洞結合並且回到Emitter,大多數的電子會隨著CBJ逆偏穿越到Collector,並且回到Source形成一個封閉的迴路。
  • 相關影片資源:
    Animated BJT – How a Bipolar Junction Transistor works | Intermediate Electronics
  • 在這裡先來解釋集極電流的來源會和BE接面相關的問題,因為其電流的動力就是BE接面他直接的影響了電流的大小。
  • 在此處我們要開始以公式的導入來協助接續的分析:
    The Collector Current(
    iC
    )這是集極的電流情況,結合前幾章節的內容可以推導出:

    iC=ISeVBE/VT(Eq6.1)

    The Base Current(
    iB
    ):

    iB=iCβ(Eq6.2)

    iB=(ISβ)eVBE/VT(Eq6.3)

    The Emitter Current(
    iC
    ):

    iE=iC+iB(Eq6.4)

    整理可得:
    iE=β+1βISeVBE/VT(Eq6.6)

    iC
    的角度:

    iC=αiE(Eq6.7)

    The constant
    α
    :

    α=ββ+1

    The constant
    β
    :

    β=α1α

6.1.4 Operation in the Saturation Mode

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  • Saturation Mode發生在EBJ和CBJ皆為順偏置時,此時,BJT失去放大信號的功用,只是單純的開關因為
    VCE0V

    image
  • 經由實驗和微觀物理層面的分析我們知道
    VBC0.4V
    :

    iC=ISeVBE/VTISCeVBC/VT(Eq6.14)

    藉由過去的結論我們知道可以運用類似的方式表達
    iB
    :

    iB=(IS/β)eVBE/VT+ISCeVBC/VT(Eq6.15)

    此時
    iCiB
    的比值會遠小於Active Mode。

    βforced=iCiB(Eq6.16)

    比較
    C,E
    的電壓差異:

    {VCEsat=VBEVBC0.10.3V

    經本上在saturation mode
    VCE=0.2V
    and
    VBE=0.7V
    .

6.2 Current–Voltage Characteristics

image
image

6.2.4 An Alternative Form of the Common-Emitter Characteristics
  • 共射極特性的不同形式表達,前面定義我們都是以
    iC
    為分析的主體 但是現實上 分析時使用
    iB
    更為容易也符合事實。

    圖片

VCE deep in saturation region
0.2V

Example

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MOSFETs(MOS Field-Effect Transistors)


  • 金屬氧化物半導體場效電晶體
    圖片
  • MOSFET分為NMOS和PMOS,觀察Fig 5.1這是一個NMOS結構,基板為P型材料,並搭配三個主要端點:S(source)、G(Gate)、D(Drain)其中S和D是放置在一塊N型材料上,而G則位於一層絕緣材料(氧化物,通常為SiO₂)上。另外在基板以下還有接上一個B(Body)端通常做為電壓的基準點。

Operation with Zero Gate Voltage

  根據PN接面的運作原理,當G輸入電壓為0時,S和D兩端都不會導通,這類似於兩個背對背連接的二極體。此時,NMOS的通道並未被反向偏壓形成導通區域,因此兩端並不會導通。

Creating a Channel for Current Flow

圖片

  •   觀察Fig. 5.2,在Gate上接上一個正電壓
    VGS
    時,Gate與基板之間會產生電場作用,類似於電容器的行為。這會使得基板表面產生電子,並將基板中的電洞推向更深的區域,從而在表面形成一個源漏通道(具體來說這可以被視為一個反轉層)。可以藉由控制S和D的電壓來控制電流。
  • VGS
    大到足以導通時,這個電壓稱為臨界電壓(threshold voltage)
    vt
    ,這個數值大約在於0.3V~1.0V之間。而當
    vGS>vt
    時,會有一個過驅電壓(overdrive Voltage)
    vOV

    vOV=vGSvt(Eq.5.1)
  • 關於通道中的電子電荷量,我們使用庫侖定律:
    gDS=|Q|=Cox(WL)vov(Eq.5.2)

    其中C指的是氧化層的電容。
    Cox=ϵoxtox(Eq.5.3)
  • ϵox
    是氧化層的介電常數,約是
    3.45×1011F/m2
  • tox
    是氧化層的厚度

Applying a small
vDS

圖片

  • 在Drain接上一個小的正電壓
    vDS
    ,使的S和D的通道導通。

    iD=gnμnE,|E|=vDS/L,n(vGSvt)

    帶入整理為一條算式:
    iD=[(μnCox)(WL)(vGSvt)]vDS(Eq.5.7)

    以下補充跨導參數,用來描述 MOSFET 的極控制效能,即在一定的閘極電壓下,MOSFET 可以產生多少的電流。
    kn=μnCox

    kn=μnCox(W/L)

    然後因為
    g=IV
    所以可以推出以下:
    rDS=1/gDS

    圖片

Operation as
vDS
Is Increased

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  • 總結上述分析MOS的精隨 就是閘極可以控制通道電流和電壓的上限。
  • MOS的工作區域分為,Triode和Saturation region。由於Triode和前面的關係相比並非線性所以另外推導我們可以得到另一條通道電流的公式:
    iD=kn(WL)(vOVvDS12vDS2)(Eq.5.15)

5.1.6 Operation for vDS ≥ V OV: Channel Pinch-Off and Current Saturation

圖片

  • 飽和區和前一小節的差異在於,
    vOVvDS
    ,就過去的討論我們知道,這樣是不會導通的,所以此時SD通道是Pinch-off的,但是我們同時也知道即使彼此不導通,但仍然會有極少數的墊子跨越。
  • 將電壓數值帶入Eq.5.15可以得到:
    iD=kn12(vGSvt)2(WL)(Eq.5.20)

5.2 Current-Voltage Characteristics

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5.3 MOSFET Circuits at DC

  • 在這章節開始