CHAPTER 2 The Architecture and Decide the Specifications
G
GA = PAVN / PAVS
ed from the network / Power available from the
(2.2)
GA物理意義:前級考慮最低雜訊的匹配,即 ;後級共軛匹配,
即Γ
低雜訊放大器的設計,通常將輸出端設定為共軛匹配即 ,以達
到後 當的
態 本論文
穩定性(stability)描述了放大器電路是否能避免振盪的程度,為放大器 設計
圖2.9 Stability of two-port network 示意圖
T物理意義:前級與後級皆不共軛匹配,即Γs ≠Γin*、ΓL≠Γout*。
= Power deliver source
in* s =Γ Γ Γ *
= out ,適合用於LNA的設計。
L
out* L=Γ Γ
端最低的反射功率及最大的輸出增益。輸入端則選擇適 ,使得 Γ *
=
Γ ,以達到輸入端較低的反射係數且產生較低雜訊的最佳狀 ;因此 即採用Available power gain,G
Γs
in s
A的功率增益定義來設計低雜訊放大 器。
步驟中一項重要的考慮因素。而穩定性可以由電晶體的S(S-parameter) 參數、各匹配網路以及終端阻抗型式…等等原因來決定並得之一個滿足無 條件穩定(unconditionally stable)的電路,可以在電路輸入及輸出兩端,接 上任何被動元件,都能保持穩定工作而不致引發振盪,如圖 2.7 所示;而 在已知頻率下,依反射係數的定義,達到無條件穩定的條件為:
Z
LV
S+
- Transistor
Г
OUTCHAPTER 2 The Architecture and Decide the Specifications
Γs <1 (2.3)
ΓL <1 (2.4)
) 0 ] (Re[
<1 Z 1 22
12
11 21 >
Γ
− + Γ
=
Γ IN
L L
IN S
S
S S (2.5)
以及,
) 0 ]>
(Re[ UT
1
12 <
= −
Γ O
S
OUT S S Z
S .6)
上式中的所有係數均都已對特 阻 進行正規化的動作。討論至
,我們可得到以下的結論:
當
(2
11 21
Γ ΓS 1 S
22 +
性 抗Z0
此
1
S11 < 、S22 <1且 k>1、∆ <1 (2.7)
其中,
(2.8)
21 12
2 2
22 2
11
S S 2
S S
k 1− − + ∆
=
21 12 22
11S S S
S −
=
∆ (2.9)
符合上述條件,則稱該網路必為無條件穩定。
在射頻電路當中,有一些非理想的特性,包括存在於任何環境中的雜
訊,而這會對電 度的干 響到所要 信號;且再來是
路本身的非線性應,使之無法避免信號間的相互干擾,而或因為元件本 身的限制,影響電路的性能。
2.3.2.1 諧波失真(Harmonic Distortion) 如
2.3.3 非線性效應
路造成相當程 擾,並影 的
電
在非線性系統中,若輸入之信號為fc在經過放大器後,除了主要的信 外,還會產生多次項的諧波(註.1),其示意如圖2.8所示:
號
CHAPTER 2 The Architecture and Decide the Specifications
大器輸入與輸出之頻譜示意圖 圖2.10 非線性放
2.3.2.2 增益壓縮點(Gain Compression)
增益壓縮點(1dB Compression Point)(註.2)(註.4) (註.5),因為非線性效 應的關係,放大器的增益在小訊號時為線性,但功率加大時增益就會逐漸
小。為了表示放大器的工作範圍,我們通常以輸出增益比線性增益小1dB 的 出功 ,稱為1dB增益壓縮點 其示意如圖2-9所示。另外,一般放大 的1–dB compression point OIP3 10dB,定義如下式:
減
時 輸 率 ,
器 皆低於 大約為
1 ) dB ( G ) dB (
G1dB = 0 − (2.10) (
P1dB dBm)=G1dB(dB)+IP1dB(dBm) (2.11)
當S11 <1、S22 <1 (2.12)
圖2.11 1dB增益壓縮點示意圖
G0: 線性放大的增益量 P1dB: 1dB 增益壓縮點
IP1dB: 在1dB 增益壓縮點時的輸入訊號功率
VIN VOUT
VIN
VOUT
fc fc 2fc 3fc
f f
…
1dB Compression point
CHAPTER 2 The Architecture and Decide the Specifications 2.3.2.3 交互調變(Intermodulation)
在多個頻 時由於電路
件的非線性效應,會造成不同頻率的信號產生交互調變(Intermodulation)
(註.3) (註.4) (註.5) 落在我們所要的
頻道之內, 率(BER),
其示意如圖2.10 所示 個鄰近干擾訊號,經過非
線性放大器時,其 在所要的頻道內,惡化
了訊號品質。故三階截斷點 IIP3及OIP3分別為
輸入及輸出的三階截斷點 ,一般而
言,交會點越高,或者IIP3及OIP3越大,代表線性
道的系統或是無線通訊的環境之下,訊號在接收 元
的作用,若交互調變的非線性失真信號
會造成訊號的相互干擾,增加訊號解調後的位元錯誤
,而無法由濾波器濾除的兩 三階交互調變失真(IM3)信號會落
(IP3),如圖2.11所示,其中
(input or output third-order intercept point) 度越好。
圖2.12 三階非線性現象示意圖
圖2.13 三階截斷點(IP3)示意圖
PIN(dBm) POUT
(dBm) PIP3 (dBm)
IMD (dBc)
X 2X
IM3 (dBm)
1
3 PIP3 = POUT+IMD/2 Fundamental
IM3 X
CHAPTER 2 The Architecture and Decide the Specifications 註.1 當輸出訊號壓縮 ω1與3ω1之振幅大小為何?證明之。
解:
We assum Acosω1t
And we can express the Vout = K0+K1Vin+K2Vin2+K3Vin3+… by use
Taylor series,now xpression about Vout,
out = ( K0+0.5 K2A2) + ( K1A+0.75 K3 A3)cosω1 1dB時,其ω1、2
VIN VOUT
e that Vin =
we instead of Vin in that e we can find that
V t + 0.5 K2A2cos2ω1t
+ 0.25 K3A3cos3ω1t + … now we can draw a diagram abou
So, t PIN(dBm) versus POUT(dBm)
the slope of fundamental∝|A|,the slope of 2nd∝|A|2 So wh rade 1dB (Pin),we can know the
Ps. Actuality,We now that our gain will compression because nonlin can know that K3 will be negative .
Ps1. 2nd is denote the Second Harmonic 3rd is denote the Third Harmonic
in it,we can find
and the slope of 3rd∝|A|3
en our input signal deg
magnitude of 2nd will degrade 2dB and the magnitude of 3rd will degrade 3dB .
can k earity,so we
VIN
Fundamental
2nd 3rd VOUT
PIN(dBm)
1 1
1
2 3
1
Saturated power PSAT
POUT(dBm)
CHAPTER 2 The Architecture and Decide the Specifications
註.2 當輸入一弦波訊號 Vin = A 明
之。
解:
We assume that Vin = Acosω t
And we can express the Vout = K0+K1Vin+K2Vin2+K3Vin3+… by use Taylor series,now we instead of Vin in that expression about Vout,So
Vout = K0 + K1Vin + K2Vin2 + K3Vin3 + …
θ)/2
5 K2A2) + (K1A+0.75 K3 A3)cosω1
cosω1t 時,其輸出訊號之大小為何?證
1
= K0 + K1(Acosω1t) + K2(Acosω1t)2 + K3(Acosω1t)3 + … Because,cos2θ = (1+cos2
cos3θ = (3cosθ+cos3θ)/4
So,we can find a result that Vout can express ,
Vout = (K0+0. t + 0.5 K2A2cos2ω1t
+ 0.25K3A3cos3ω1t + …
CHAPTER 2 The Architecture and Decide the Specifications 註.3 當輸入一弦波訊號 Vin = Acosω1t + Acosω2t 時,其輸出訊號之大小
為何?證明之。
:
We assume that Vin = Acosω1t + Acosω2t
And we can express the Vout = K0+K1Vin+K2Vin2+K3Vin3+… by use Taylor series,now we instead of Vin in that expression about Vout,So
Vout = K0 + K1Vin + K2Vin2 + K3Vin3 + …
= K0 + K1(Acosω1t + Acosω2t) + K2(Acosω1t + Acosω2t)2 + K3(Acosω1t + Acosω2t)3 + …
Because,(cosα+cosβ)2 = cos 2α + 2cosαcosβ + cos2β
= cos2α + cos(α+β) + cos(α-β) + cos2β
= 1+ (cos2α + cos2β)/2 + cos(α+β) + cos(α-β) (cosα+cosβ)3 = (cosα + cosβ)(cos 2α + 2cosαcosβ + cos2β) = cos3α + 3cos2αcosβ + 3cosαcos2β + cos3β in it,
3cosα(cosαcosβ) = 3[cos(2α+β) + 2cos(β) + cos(2α-β)]/4 3cosβ(cosαcosβ) = 3[cos(α+2β) + 2cos(α) + cos(-α+2β)]/4 So,
+ 3[cos(α+2β) + 2cos(α) + cos(-α+2β)]/4 + cos3β
/2 K2A2[cos2ω1t + cos2ω2t] 2nd 3rd (2ω2-ω1)t] IM3 解
(cosα+cosβ)3 = cos3α + 3[cos(2α+β) + 2cos(β) + cos(2α-β)]/4
So,
Vout = (K0 + K2A2) DC + (K1A + 3/4K3A3)(cosω1t + cosω2t) 1st + 1
+ 1/4 K3A3[cos3ω1t + cos3ω2t]
+ 1/2 K2A2[cos(ω2+ω1)t + cos(ω2-ω1)t] IM2 + 3/4 K3A3[cos(2ω1-ω2)t + cos
+ …
CHAPTER 2 The Architecture and Decide the Specifications 註.4 當輸入訊號為單一頻率時,其 1dB 增益壓縮點(P1dB)與三階截斷點 解:
[One tone]:
We assume that Vin = Acosω1t
Vout = K1A --- Ideal (linear)
Vout = K1A + 3/4K3A3(K3 always is negative) --- Actuality (nonlinear) So,
(K1A - 3/4K3A3)|dB – (K1A)|dB = -1dB (take 20 logarithmic) (K1A - 3/4K3A3) / (K1A) = 10-1/20 = 0.89125
An
= 3/4K3A3
A2 = K1/0.75K
So,
10log[AIP32 / ] = 10log[(K1/0.75K3)/( 0.10875K1 / 0.75K3)] = On output ratio increased by 1dB
So,
POIP3 – P1dB =
(IP3out)為何?證明之。
PIP3
A2 = 0.10875K1 / 0.75K3 = AP1dB2 d,
K1A
3 = AIP32 AP1dB2
9.6377dB
10.6377dB 1
3
Fundamental P1dB
10.63dB
IM3(dBm)
PIN(dBm)
?dBc
CHAPTER 2 The Architecture and Decide the Specifications 註.5 當輸入訊號為兩根頻率時,其 1dB 增益壓縮點(P1dB)與三階截斷點
osω2t ( A1≠A2 )
we can express the Vout = K0 + K1Vin + K2Vin2 + K3Vin3 + … by So,
3Vin3 + …
= K0 + K1(A1cosω1t + A2cosω2t) + K2(A1cosω1t + A2cosω2t)2 + osω2t)3 + …
= K0 + K1(A1cosω1t + A2cosω2t) + K2[A12/2 + (A12cos2ω1t)/2 + A1A2 cos(ω1+ω2)t + A1A2 cos(ω1-ω2)t + A22/2 +
(A22cos2ω2t)/2] + K3[(3A13/4 + 3A1A22/2)cosω1t + (3A23/4 + 3A12A2/2)cosω2t + 3[A12A2cos(2ω1+ω2)t]/4 +
3[A1A22cos(ω1+2ω2)t]/4 + 3[A12A2 cos(2ω1-ω2)t]/4 + 3[A1A22cos(ω1-2ω2)t]/4 + A13cos(3ω1)t/4 + A23cos(3ω2)t/4]
+ …
= (K0 + K2 A12/2 + K2 A22/2) DC + 3K3A12A2/2)(cosω2t) 1st + [(K2A12)cos2ω1t]/2 2nd
22)cos2ω2t]/2 2nd 3rd
+ 3K3A12A2[cos(2ω1+ω2)t + cos(2ω1-ω2)t]/4 IM3 ω2)t]/4 IM3
Vout = K
Vout = K tive)
So,
(K1A1 -(take 20
(K1A1 - 13/4 - 3K3A1A22/2) / K1A1) = 10-1/20 = 0.89125 (IP3out)為何?證明之。
解:
[Two tones]:
When Vin = A1cosω1t + A2c Then
using Taylor series,now we instead of Vin in that expression about Vout,
Vout = K0 + K1Vin + K2Vin2 + K K3(A1cosω1t + A2c
+ (K1A1 + 3K3A13/4 + 3K3A1A22/2)(cosω1t) 1st + (K1A2 + 3K3A13/4
+ [(K2A
+ (K3A13)cos(3ω1)t/4
+ (K3A23)cos(3ω2)t/4 3rd + {K2A1A2[cos(ω1+ω2)t + cos(ω1-ω2)t]}/2 IM2 + 3K3A1A22[cos(ω1+2ω2)t + cos(ω1-2
+ …
1A1 --- Ideal (linear)
1A1 + 3K3A13/4 + 3K3A1A22/2 (K3 always is nega ---- Actuality(nonlinear)
3K3A13/4 - 3K3A1A22/2)|dB – (K1A1)|dB = -1dB logarithmic)
3K3A
CHAPTER 2 The Architecture and Decide the Specifications
A12 = 0.10875*4K1 /9 K3 = AP1dB2 And,
K1A1 =
A22 = K
Now we So,
10log[A On outp So,
POIP3 – .40692dB
15.4069 4.76922 dB
e P1dB will d
3/4K3A1A22 (The slope of the third order product is the same) 1/0.75K3 = AIP32
assume A1 = A2 = A,
IP32/ AP1dB2] = 10log[(K1/0.75K3)/( 0.10875*4K1 /9 K3)]
= 14.40692dB ut ratio increased by 1dB
P1dB = 15
Comparison the results of one tone and two tones:
2dB - 10.6377dB =
Now we can find if the fundamental signal are two tones input,th egrade 4.76922 dB
CHAPTER 3 Design Considerations of Radio Frequency Circuit