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@6/20 2012
try to figure out how the "old" low tech IF amplifier works and what it provide. here is the practical parts i ever tried in debugging IF-B0 .
Vcc=11V~12V
Component: BJT 2sc9018 (ss9018)
Component: BJT 2sc9018 (ss9018)
The Hybrid-pi model is used to routhly design the LC tuned load IF amplifer.
for 2sc9018(or ss9018, they are almost same characters ):
* β = 120 (*which we are going to used)
* Rπ = 26(mA)/Ie*β
* Va >=74V (*Early voltage roughly Calculate from FIg.2 @Ie<5mA,60V, from Spice Model Vaf=74V )
*Ro = (Va + Vce)/Ie ~= 28K (@Ie=3mA, 168k@0.5mA)
Fig.2
Fig.3
Popular IFT on Chian Taobao
Popular IFT on Chian Taobao
TTF-1-1 36+127:10, Q>=80
TTF-1-2 RED: 36+127:8 Q>=80
TTF-1-3 yellow 139+25:11
TTF-2-1 : 45+117:7 , Q=80-120
TTF-2-2: 45+117: 10, Q=80-120
TTF-2-9: 48+114:25, Q=80-120
My stock IFT on hand
My stock IFT on hand
primary:
primary: 134.21 uH
primary tap-top: 89.447uH
primary tap-bottom:: 4.39uH
seconday: 0.1020H
Pri:Sec = sqrt(134.21/0.1020) ~= 36:1
Tap:primary= sqrt(134.21/4.39) ~= 5.5:1
yellow IF.T.:
Primary:402.7660 uH
Seconday:2.4462 uH
p:s = sqrt(402.766/2.4462) = 12.8:1
red coil:(NOT transformer)
2.3405uH default
max: 2.9784 2.9788 2.9777
min: 1.9965 1.9943
system error: 0.0230
big pink IFT:
primary: 4.3320uH
tap-top: 1.246uH
tap-bottom: 1.1989uH
secondary: 0.064uH
Primary:Secondary=sqrt(4.332/0.064) ~= 8:1
tap: 1.9:1
IF-B0 gain analysis
IF-B0 gain analysis
Rt=2.pi.f.L.Q ( the impedance of the tuned LC circuit)
Rp: total impedance in the primary of transformer
IF frequency: 465kHz
IFT characters
T1[yellow IFT]: primary: 350uH, 300pF, primary 6.3ohm Q ~=90 [calculated 150,practically, around 80-100, not by testing]
p:s = 12.8:1
Rt=Xl*Q = 2 * pi * f * L*Q =2*3.1415926*465*10^3*350*10^-6*90 ~=90k
T2(tap green IFT): primary 760pF, 134uH, 3ohm,Q=90 (practically, use Q=80-100 as estimation)
P:S= 36:1
tap = 5.5:1 (primary: tap)
Rt=Xl*Q = 2 * pi * f * L*Q= 2*3.1415926*465*10^3*136*10^-6*140 ~=35k
Stage 1 gain
Stage 1 gain
T1 secondary load: 126*(26/3mA)*12.8*12.8 ~= 178.9k
Q1 output resistance: 27k
total T1 load: Rp= 28K | 178.9 | 165 = 21K
gain of stage1: 20*log((2mA/26)*18.77k/12.8)= 42dB (insert loss around 3dB)
Qload = Rp/(2*pi*f*L) = 18
Bw=f.res /Q = 25kHz
Stage 2 gain
Stage 2 gain
Rt=35k, P:S= 36:1, tap = 5.5:1 (primary: tap)
T2 secondary load: 300R*36*36 = 388.8K (300R ,the estimation of detector input impedance )
Q 1 output resistance: 24K
total T1 load: { 28k|| 388.8k/(5.5*5.5)=12.85k || 35k/(5.5*5.5)= 1.934k } ~= 1k
Gain of stage1: 20*log(3mA/26*1k*5.5/36) = 24dB
total gain: 44dB + 24dB-5dB ~= 70dB
stage 2 change to IFT TF2-9 (48+114: 25)
stage 2 change to IFT TF2-9 (48+114: 25)
600uH, 200pF, 465khz, Q=100
p:s = 6.48:1
p:tap = 3.37:1
Rt=Xl*Q = 2 * pi * f * L*Q =2*3.1415926*465*10^3*600*10^-6*100 ~=175k
T2 secondary load: 300R*6.48*6.48 = 12.6K
Q 1 output resistance: 24K
total T1 load: { 24k|| 12.6/(3.37*3.37)=1.1k || 175k/(3.37*3.37)= 15k } = 1k
Gain of stage1: 20*log(3mA/26*1k*3.37/6.48) = 35dB
total gain: 44dB + 35dB-5dB ~= 75dB
Change to TF-1-3 yellow 139+25:11
Change to TF-1-3 yellow 139+25:11
600uH, 200pF, 465khz, Q=80
p:s =14.9:1
tap = 5.56
colector:Sec=2.27
Rt=Xl*Q = 2 * pi * f * L*Q =2*3.1415926*465*10^3*600*10^-6*100 ~=140k
T2 secondary load: 300R*6.48*6.48 = 66K
Q 1 output resistance: 24K
total T1 load: { 24k|| 12.6/(5.56*5.56)=2.1k || 175k/(3.37*3.37)= 15k } = 1.7k
Gain of stage1: 20*log(3mA/26*1.7k*5.56/14.9) = 37dB
total gain: 44dB + 37dB-5dB ~= 76dB
IF-B0 : Stable Margin estimation
IF-B0 : Stable Margin estimation
IF Stage stable constrain
IF Stage stable constrain
A stable amplifier, gm, Rb, Rc limited by :
2*pi*f*c*gm*RbRc < 2
[refer to <<Principles of Transistor Circuits>> Ninth Edition@2000 Appendix C]
To provide a stable UN-neutralization amplifier, need margin factor 4, So
2*pi*f*c*gm*RbRc = 2/4 = 0.5
f: IF frequncy in HZ c: in Fala
Rb: the impedance which ideal amplifier see from it's input
Rc: the impedance which ideal amplifier see from it's output
Suppose Stages BJT's base impedance just the stage's input impedance(@update 2012/12/11, this is wrong, in normal case the BJT base see the impedance domain by the pre-stage's loaded tank's Rt) Then,
@455khz, c(Cob)=1.7pF(9018), gm=Ie/26, input resistance: Rb=beta*(26/Ie) Ie: mA
Rc*Rb <= 0.5/(2*pi*f*c*gm)= 0.5/(2*pi*f*c*(Ie/26)))
= 2.67*10^6/Ie
9018 IF Stage stable constrain
9018 IF Stage stable constrain
9018 IF sage stable conditions (Rc,Rb: in Kohm, Ie in mA , IF=455kHz, c=1.7pF ) [@updated 2012/12/11]
Rc*Rb*Ie<=2.67 (stable factor n=4)
Rc*Rb*Ie<=10.68 ( start oscillation)
stable margin of stage 2 :the IFT TF2-9 (48+114: 25) @2012/12/11
stable margin of stage 2 :the IFT TF2-9 (48+114: 25) @2012/12/11
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600uH, 200pF, 465khz, Q=100
p:s = 6.48:1
p:tap = 3.37:1
Rt=Xl*Q = 2 * pi * f * L*Q =2*3.1415926*465*10^3*600*10^-6*100 ~=175k
Rc=detector||(Rt/3.37/3.37)|| ro=28k ~=1kohm
(detector ~=300*6.48*6.48/3.37/3.37)
Rb= first stage Rt || BJT-Rin~= 600R || (100*26/0.5) = 600R
test result: Ie from 1mA to 10mA get gain from 25dB to 45dB, be stable.
@3mA, Ie=3mA,Rc=1k, Rb=0.6K , Gain=20*log(Rc*Ie/26/1.7) = 35dB
stable constrained: Rc*Rb*Ie=0.6*1*3 = 1.8 < 2.67
@10mA Ie=10mA, Rc=1k, Rb=0.6k, Gain=20*log(Rc*10/26/1.7) = 46dB
stable constrained: Rc*Rb*Ie =1*0.6*10 = 6 >= 2.7, stable factor dropped to 1.7
Why oscillation in my first trying use IFT
Why oscillation in my first trying use IFT
@2012/12/11
p:s = 12.8:1,Q ~=90 , Rt=90k
3mA BJT-Ro = 44k
Rb= first stage Rt || BJT-Rin= (90k/12.8/12.8) || (100*26/0.5) = 0.6k
Rc= Rt|| Detector Impedance|| BJT-Ro = 90k||(300*12.8*12.8) || 44k ~= 18K
(Ie~=3mA)
if Ie use 2.7mA as first designed
Rb*Rc*Ie =18×0.6x2.7 = 29, far away from stable!!!!(STABLE SHOULD < 10/n)
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