Applied Electronic Circuit

Presentation on theme: "Applied Electronic Circuit"— Presentation transcript:

1 Applied Electronic Circuit
Report 3 최법경

2 Contents Complex number calculation Bode plot
Complex Exponential Linear system Phasor RC-circuit Defferentiator Integrator Bode plot Negative Resistance Converter supplying source into Op Amp circuit

3 Complex number calculation
Complex Exponential 2𝜋 𝑇 =2𝜋𝑓=𝑤(단위시간당 도는 정도) If w=const(정해졌다면) 𝜃=𝑤𝑡 (rcoswt, rsinwt)=rcoswt+jrsinwt → 𝑟𝑒 𝑗𝑤𝑡

4 Complex number calculation
𝐴 1 𝑒 𝑗 𝜃 1 = 𝑧 1 , 𝐴 2 𝑒 𝑗 𝜃 2 = 𝑧 2 𝑧 1 = 𝐴 1 𝑐𝑜𝑠 𝜃 1 +j 𝐴 1 𝑠𝑖𝑛 𝜃 1 𝑧 2 = 𝐴 2 𝑐𝑜𝑠 𝜃 2 +j 𝐴 2 𝑠𝑖𝑛 𝜃 2 𝑧 1 𝑧 2 = 𝐴 1 𝐴 2 𝑒 𝑗( 𝜃 1 + 𝜃 2 ) 𝑧 1 𝑧 2 = 𝐴 1 𝐴 2 𝑒 𝑗( 𝜃 1 − 𝜃 2 ) 𝑧 1 + 𝑧 2 𝑧 1 𝑧 1 − 𝑧 2 𝑧 2

5 Complex number calculation
x(t) H y(t) Linear system Acoswt H AHcoswt 𝐻 𝑎 1 𝑥 1 + 𝑎 2 𝑥 2 = 𝑎 1 𝑦 1 + 𝑎 2 𝑦 2 크기, 위상만 바뀐다. 𝐻=𝐻 𝑒 𝑗𝜃 𝑧=𝐴𝑐𝑜𝑠𝜃+j𝐴𝑠𝑖𝑛𝜃 =𝐴 𝑒 𝑗𝑤𝜃 =𝐴∠𝑤𝑡

6 Complex number calculation
Phasor 𝐴𝑐𝑜𝑠𝑤𝑡=𝑅𝑒 𝐴 𝑒 𝑗𝑤𝑡 𝐴𝑐𝑜𝑠 𝑤𝑡+𝜃 =𝑅𝑒 𝐴 𝑒 𝑗 𝑤𝑡+𝜃 =𝑅𝑒 𝐴 𝑒 𝑗𝑤𝑡 ∗𝐴 𝑒 𝑗𝜃 Time function Phasor Acoswt 𝐴 𝑒 𝑗0 =𝐴, 𝐴∠0 Acos(wt+𝜃) 𝐴 𝑒 𝑗𝜃 , 𝐴∠𝜃

7 Complex number calculation
Phasor에서 time function 구하기 𝐴 𝑒 𝑗𝜃 𝑝ℎ𝑎𝑠𝑜𝑟 ∗ 𝑒 𝑗𝑤𝑡 𝑅𝑒{𝐴𝑒 𝑗𝜃 ∗ 𝑒 𝑗𝑤𝑡 }=𝐴𝑐𝑜𝑠(𝑤𝑡+𝜃)크기와 위상을 표현 𝑑 𝑑𝑡 𝐴𝑐𝑜𝑠𝑤𝑡=−𝑤𝐴𝑠𝑖𝑛𝑤𝑡=𝑤𝐴𝑐𝑜𝑠 𝑤+ 𝜋 2 =𝑅𝑒{𝑤 𝐴𝑒 𝑗𝜋 2 𝑒 𝑗𝑤𝑡 } Time function Phasor 𝑑 𝑑𝑡 𝐴𝑐𝑜𝑠𝑤𝑡 𝑤 𝐴𝑒 𝑗𝜋 2 =𝑤𝐴∠ 𝜋 2 =𝑗𝑤𝐴 𝑑 𝑑𝑡 𝑗𝑤

8 Complex number calculation
Acoswt H 𝑒 𝑗𝜃 Ahcos(wt+𝜃) 𝐴 𝑒 𝑗0 ∗𝐻 𝑒 𝑗𝜃 =𝐴𝐻 𝑒 𝑗𝜃 Re{입력의 phasor 전달함수 𝑒 𝑗𝑤𝑡 } =>출력

9 Complex number calculation
RC-circuit 𝑣 𝑡 =𝑉𝑐𝑜𝑠𝑤𝑡⇒ 𝑉 𝑝 =𝑉∠0 𝑖 𝑡 = 𝑉 𝑅 𝑐𝑜𝑠𝜃⇒ 𝑣 𝑟 =0 𝑉 𝑃 𝐼 𝑃 =𝑍=𝑅∠0=𝑅 𝑖 𝑡 =𝐶 𝑑𝑣 𝑡 𝑑𝑡 , v t =Vcoswt, 𝑉 𝑝 =v∠0 𝐼=𝐶𝑗𝑤 𝑉 𝑝 =𝑗𝑤𝐶𝑉 =𝑤𝑐𝑗 ∠0

10 Complex number calculation
Differentiator ( 미분기 ) 𝑖 1 =𝐶 𝑑𝑣 𝑖 𝑑𝑡 , 𝑖 2 = 𝑣 0 𝑅 𝑖 1 + 𝑖 2 =0 𝑣 0 t =−RC d 𝑣 0 𝑑𝑡 𝑣 𝑖 𝑖 2 𝑖 1 𝑣 𝑜

11 Complex number calculation
1 𝑗𝑤𝑐 Differentiator phasor 𝑣 𝑜 𝑣 𝑖 =− 𝑅 1 𝑗𝑤𝑐 =−𝑗𝑤𝑅𝐶 𝑖 1 = 𝑣 𝑖 1 𝑗𝑤𝑐 , 𝑖 2 = 𝑣 𝑜 𝑅 , 𝑖 1 + 𝑖 2 =0 𝑣 𝑜 =−𝑗𝑤𝑅𝐶 𝑣 𝑖 𝑣 𝑖 𝑡 =𝐴𝑐𝑜𝑠𝑤𝑡, 𝑉 𝑝 =𝐴∠0 𝑣 𝑜 =−𝑗𝑤𝑅𝐶𝐴=𝑤𝑅𝐶𝐴∠− π 2 (−wRCA∠ π 2 ) 𝑣 𝑜 𝑡 =−𝑤𝑅𝐶𝐴𝑐𝑜𝑠 𝑤𝑡+ π 2 =wRCAsin(wt) 𝑣 𝑖 𝑖 2 𝑖 1 𝑣 𝑜

12 Complex number calculation
𝐻= 𝑣 0 𝑣 𝑖 =−𝑗𝑤𝑅𝐶=𝑤𝑅𝐶∠− π 2 |H| ∠H wRC w wRC w

13 Complex number calculation
실제 쓰이는 미분기 회로 𝑣 𝑜 𝑣 𝑖 =− 𝑅 2 𝑅 𝑗𝑤𝑐 =− 𝑗𝑤 𝑅 2 𝐶 1+𝑗𝑤 𝑅 1 𝐶 =− 𝑤 𝑅 2 𝐶∠ π 𝑤 2 𝑅 1 2 𝐶 2 ∠ tan −1 𝑤 𝑅 1 𝐶 = 𝑤 𝑅 2 𝐶 1+ 𝑤 2 𝑅 1 2 𝐶 2 ∠− π 2 − tan −1 𝑤 𝑅 1 𝐶 𝑅 1 𝑅 2 0.707 1 𝑅𝐶 미분기 반전증폭기 High pass filter − π 4 − π 2

14 Complex number calculation
Integrator 𝐼 1 = 𝑉 1 𝑅 1 , 𝐼 2 = 𝑉 𝑗𝑤𝐶 =𝑗𝑤𝐶 𝑉 0 , 𝐼 3 = 𝑉 0 𝑅 0 𝐼 1 + 𝐼 2 + 𝐼 3 =0, 𝑉 1 𝑅 1 +𝑗𝑤𝐶 𝑉 0 + 𝑉 0 𝑅 0 =0 𝑉 0 =− 1 𝑅 𝑅 0 +𝑗𝑤𝐶 =− 𝑅 2 𝑅 𝑗𝑤𝐶 반전 증폭기이기 때문에 𝜋만큼 shift 𝐼 1 𝐼 3 𝐼 2 0.707 1 𝑅𝐶 − π 4 − π 2

15 Bode plot 기울기=20dB/dec W가 10배 늘어나면 20씩 떨어진다. 3dB 20𝑙𝑜 𝑔 10 |𝐻|
log 10 𝑤

16 Bode plot 20 log 𝑅 2 𝑅 𝑤 2 𝑅 2 2 𝐶 2 𝑖 𝑤= 1 𝑅𝐶 인 𝑤를 𝑤 1 이라하고 𝑤<0.1 𝑤 𝐻 =20 log 𝑅 2 𝑅 1 𝑖𝑖 𝑤<10 𝑤 1 𝐻 =20 log 𝑅 2 𝑅 𝑊 𝑅 2 𝐶 =20 log 𝑅 1 𝐶 1 𝑤 =20 log 𝑅 𝐶 1 −20 log 10 𝑤 𝑖𝑖𝑖 𝑤= 𝑤 1 일때 𝐻 𝑑𝑏 =20 log 𝑅 2 𝑅 1 −3 상수

17 Negative Resistance Converter(NCR)
𝑖 𝑖 = 𝑣 𝑖 − 𝑣 𝑜 𝑅 3 , v i = R 1 R 1 + R 2 v o ⇒ 𝑅 𝑖 = 𝑣 𝑖 𝑖 𝑖 =− 𝑅 1 𝑅 2 𝑅 3 𝑣 𝑖 >0, 𝑖 𝑖 <0 𝑣 𝑖 𝐼 𝑖 𝑣 𝑖 𝑅 𝑖 = 𝑣 𝑖 𝑖 𝑖 =− 𝑅 𝑟𝑞 𝑖 𝑖 𝑅 𝑒𝑞 1V -1V

18 supplying source into Op Amp circuit
+ + 여기에 수식을 입력하십시오. Bypass capacitor Decoupling Capacitor = + Capacitor는 고주파에서 short 상태 이므로 노이즈를 제거 할 수 있다.

19 supplying source into Op Amp circuit
𝑖 𝑐𝑐 = 𝑖 𝑒𝑒 + 𝑖 𝑜 Op Amp가 부하에게 전류 공급. 𝑖 𝑒𝑒 = 𝑖 𝑐𝑐 + 𝑖 𝑜 Op Amp가 부하로 부터 전류를 Sink. 𝑖 𝑜 =0일때 𝑖 𝑐𝑐 = 𝑖 𝑒𝑒 = 𝐼 𝑄 𝑑𝑎𝑡𝑎 𝑠ℎ𝑒𝑒𝑡 참조 저전력 Op Amp는 𝐼 𝑄 가 작은 Op Amp 𝑖 𝑐𝑐 𝑖 𝑜 𝑖 𝑒𝑒

20 supplying source into Op Amp circuit
𝐸𝑥) 𝑖 𝑐𝑐 = 𝑖 𝑒𝑒 + 𝑖 𝑜 = 𝐼 𝑄 + 𝑖 𝑜 𝑖 𝑒𝑒 = 𝐼 𝑄 + 𝑖 𝑜 𝑖 𝑐𝑐 𝑖 𝑜 𝑣 𝑜 >0 𝑣 𝑖 >0 𝑖 𝑒𝑒 𝑣 𝑖 <0 𝑣 𝑜 >0

21 supplying source into Op Amp circuit
𝑖 𝑐𝑐 = 𝑖 𝑜 + 𝐼 𝑄 𝑖 𝑒𝑒 = 𝑖 𝑜 + 𝑖 𝑄 𝑣 𝑖 >0 𝑣 𝑜 <0 𝑣 𝑖 <0 𝑣 𝑜 <0

22 supplying source into Op Amp circuit
𝑣 𝑐𝑐 𝐼 𝑄 𝑣 𝑒𝑒