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Everything posted by walid

  1. Hi Alun If this is a VCO used in a FM Tx transmitt at 100MHz, the tank circuit contains these C and L, tell me an example of their values. thank you.
  2. Hi audioguru once u told me in one of your replies that u geuss VE to be 2.25V and ... From that deduce your design decisions which are: IC=10mA VE=2.25V Vcc=5V Q2 = 2N3904 has hfe=230 at the operating current (IC) .... If I use these decesions with the same 2n3904, then I measure IC and IB and they give a different hfe, say 160 and not 230. You know that at this case IB and IC became different, What to do ?
  3. Hi Audioguru THE VCO STAGE TANK CIRCUIT (LC) L=100nH, C=5-35pF, so the freq range from 87MHz at c=35P to 225MHz at 5p. To make the coil of 100nH, u may wound 5.5 turns of 1mm enameled copper wire on 5mm in dia former, the turns must be close so that the length of the coil = 5-6 mm. When I looked to the photo of your actual TX, I noticed that u wound 10 turns on about 2.5mm in dia former and of length = 10mm (1) What the difference btn using the two coils. Notice that my coil is half in number of turns and in length and double in diamete. (2) If-as u told me- MIC provid Q1 by 5mV signal, and we know that Av about 10, so the sig at the base of Q2 is about 50mV, this 50mv voltage applied to base of Q2, and we know that VB=VE+0.72=2.25+0.72=2.97V, so, Vb changes btn 2.97V+0.05 and 2.97V-0.05. From all that I see that u waste the current. You need that the Q2 not saturated nor cutoff, so VB need to be only 0.72+0.05=0.77V or 0.8V. What u say?
  4. Why you may don't hear about FM Tx osc. your answer is bad, but thank u
  5. Hi I put C4 = 50nF instesd of 100nF in the Pspice commands that follows, to see how this affect on the pre-emphasis: Audioguru FM Tx 1st stage (Audio preamp), V- divider configuration Ver.02 VCC 1 0 5V Vmic 8 0 AC 1 *Vmic 8 0 SIN( 0 5M 1K 0 0 0 ) Rmic 8 7 3.3K R2 1 2 150K R3 2 0 39K Q1 3 2 4 Q_2N3904 R4 1 3 10K R5 4 0 470 RL 6 0 21.6k C1 7 2 330N C2 2 4 100P C3 3 6 330N C4 4 0 50N .MODEL Q_2N3904 NPN (Is=6.734f Xti=3 Eg=1.11 Vaf=74.03 Bf=416.4 Ne=1.259 + Ise=6.734 Ikf=66.78m Xtb=1.5 Br=.7371 Nc=2 + Isc=0 Ikr=0 Rc=1 Cjc=3.638p Mjc=.3085 Vjc=.75 Fc=.5 + Cje=4.493p Mje=.2593 Vje=.75 Tr=239.5n Tf=301.2p + Itf=.4 Vtf=4 Xtf=2 Rb=10) .TRAN 1ns 2Ms .OP .AC DEC 20 1Hz 20kHz .probe .end the freq. response curve shown below (Fig.freq responce C2=50n). My questions: (1) I don't so any resonable difference btn this curve and the simillar one above, comment. (2) In our calculations above we conclude that IC =0.3 mA and IB =1.3 uA, and we correct IB to be 2.6uA to maintane a ratio of 10:1, then, I calc Hie to be = DC hfe*re = 230*83.3 = about 19Kohm and audioguru correct me to Hie = AC hfe*re and tell me that AC hfe from datasheets = 100, so, Hie=100*83.3 = 8.3K ohm. Also we conclude that the intended v.gain = 10 and Zout =RC =10k. the following is the result of the program above: NAME Q1 MODEL Q_2N3904 IB 9.91E-06 <=== ours =1.3 or 2.6uA IC 2.05E-04 <=== ours = 0.3 mA VBE 6.24E-01 ===> its very good as exactly audioguru said VBC -2.22E+00 VCE 2.85E+00 ===> about VCC/2 = 2.5V BETADC 2.07E+01 <=== 20.7 it is so far from 230 GM 7.91E-03 $$$$ I don't know what this RPI 3.24E+03 <=== it is Hie = 3.24K and our was 8.3K RX 1.00E+01 unknown RO 3.71E+05 the trasistor o/p Z and when this in //RC ++> Zout=RC CBE 8.69E-12 CBC 2.38E-12 CJS 0.00E+00 BETAAC 2.57E+01 this Ac hfe far from 100 CBX 0.00E+00 FT 1.14E+08 the voltage gain appears in the second Figure (Fig.Av 1st stage). From this figure we can say that Av = 43.8/5 = about 9 very close to 10, I accept this. Why these differences? pease comment to know where to depepnd on Pspice and where not. thank u.
  6. Hi audioguru (1) In my calculations above hie= 10.9Kohm at Ic=0.3mA, and Fig 4 above reveal this it is about 10Kohm, so I can say that my calcuklations are true. what would u say? (2) I have the same datasheets from FAIRCHILD. If u look at Fig.2 above u may notice that VBE at IC=0.3mA = 0.63V and not 0.62V if u want to be more accurate. (3) Why i have not the AC hfe curves in my datasheets as u, mine is also from TAIFCHILD as yours, anyway are u mean that hfe in hie=hfe*re is the AC hfe and not DC hfe? (4) When i face a problem like you in not having the resistors that make the desired ratio 10:1, is this the best solution to disturb the design and make the ratio near 20:1? (5) What did u mean by The Mod-4 bypass cap, i think you mean the the Ce cap in your fourth version (C4 in the first design). Are u made 4 versions of this Tx, why? (6) If I want to boost a high audio frequencies of 10KHz using a a bypass cap for a transistor's emitter resistor, how can I calculate its value? my answer: Its Xc must be <= 0.1RE at the desired frequency, correct me. (7) I see from all what u tell me that millions of things must be taken into the mind when trying to design a preamp circuit, and man can't do this. so I think one must use a computer tool to do this huge work for him! (8) below u can see a graph (Fig.freq responce ver02) represents the frequency response of this preamp, i use a pspice to do it, what u comment about it. (9) In my calculations above u comment only about VBE, and Hie only. Can I assume that the remaining calc. are true? (10) I ask u this question:"Why C2, it is strange, may be for negative feedback" and u answer:"C2 shorts the base to emitter at RF frequencies to prevent RF pickup at the mic from being rectified by the transistor which would upset its bias. " I'm not understand it as u expect, but I'll do my explaination and u correct me: At RF freq. like 100Mhz, Xc2= 16 ohm = short circuit At AF freq. like 2KHz, Xc2 = 800 Kohm = open circuit Now if the MIC picks up any RF power from the TX itself or from other Tx's, it give it to the base of Q1 which rectified them into DC and this DC will change the Q1's bias. I'm sure this is wrong wrong explaination, because this rectified RF power is Dc and C2 can't pass DC to ground! I don't know, u may tell me. Another point about this: I never see a like this cap btn B and E, how this idea come to u? thank u audioguru, I feel that your student in advance.... may be in one day have the ability to answer and help people like u do.
  7. Good night Audioguru, U said: You have used hFE (DC gain) of 230 in your calculation for Zin instead of Hfe (AC gain) of 100 so the actual impedance is less. walid: Why to use AC hfe, u never said that before? And how to find this AC hfe? And u said: Since the pre-emphasis capacitor affects only high frequencies then the correct value for input impedance occurs only at about 30kHz and higher. At lower frequencies the input impedance is much higher. walid: I can't understand anything from this.
  8. Hi audioguru, u said: The pre-emphasis at 15kHz in North America is about 7.2 times the gain at lower frequencies.With its lower frequencies gain of about 18 then the max gain of the transistor of about 130 results in a pre-emphasis which just makes it and sounds perfect. My questions: (1) What the meaning of The pre-emphasis from the view of electronic? (2) What kinds of audio at 15KHz? (3) Why u relate the 15kHz to North America region only?
  9. Hi audioguru; you said: My Mod-3 version of the circuit had an extra RE to raise VE so that different Vbe in different transistors wouldn't affect the VC too much. The question: Are u mean a bypassed RE added in series to a not bypassed RE? you said: But I didn't have space for it and a large bypass capacitor for it. walid: I don't know what do u mean!
  10. audioguru: "I didn't have 320k for R2 nor 60k for R3 so the divider current to base current ratio is 20:1 instead of 10:1 which better controls the effect of different current gain in different transistors." walid: who said that u use R2=320k and R3=60k, and if u do so the divider current to base current ratio will be 5:1 and not 20:1. And Why say that, I can't understand u!
  11. In fig3 seems to be that figure u use to determine hfe vs the operating IC. Figure 4 is to determine hie. tell me how exactly u use them, and if VCE or frequency is defferent from specified in the curve, how can I use them. thanx.
  12. Hi audioguru because I have 4 figures, I'll make this question into two. From What you said:"You should never assume Vbe in a transistor circuit that has a low emitter voltage. It is typically 0.62V on my datasheet." I can understand: 1) Are u tell me to go to datasheets only when determining VBE and never to assume it? You teach me before that it is not different in electronic calculations. What the problem if it 0.6, 0.7 or 0.8 volts if we know that resistors are 5% error. 2)From the beginning of this discussion till now you frequently said datasheets datasheets datasheets datasheets datasheets datasheets datasheets datasheets datasheets datasheets... I finally look at these datasheets of FAIRCHILD 2n3904 transistor. Fig.1 and 2 are the relation btn VBE and IC, one in saturation and the other in ON. What I can understand is that u choose the 25 degree curve. I dont know the difference btn them or from any of them u take your values.
  13. I understand from your reply that mine is true.
  14. It is ok, when we come to design such amp, and when choosing Av we must take other things into account at the same moment, that things are the IC value and the VBE value. example: u want Av=10, then u have many values of (RC/RE) which = 10, like 10K/1K, 4.7K/470, 1k/100 .. etc The difference in choosing one of these ratios depends on What Ic you want. Lets assume that your VCC=9V, and you choose 10k/1k ratio, then the saturated current Isat=9/(10K+1K)= 0.8mA, the best choise of IC=Isat/2=0.4mA to be at the center. Taking into account VCE=VCC/2=4.5V, then only 4.5v remain and it is divided between RC and RE according to there ratio, that is VC=10VE. If VE = 1K*0.4mA=0.4V then V(RC)=4V With the same VCC=9V and RC/Re=4.7K/470, then Ic= Vcc/(2*[RC+RE])= 0.87mA and VE= 0.87m*470=0.41V and so on.. In sammary, First: decide Av second: put several values of RC/RE = Av Third: Choose the RC/RE ratio that give you the desired Ic, then complete your calculations ...
  15. I want now to analyze the 1st stage of this Tx according to the discussion in autir's BJT biasing formulas: VCC=5V, RC=10K, RE=470, RB1=160K, RB2=30k The voltage gain Av=RC/(re+RE), but since RE bypassed ==> Av=RC/re, where re = 25m/IC. VB=5*RB2/(RB1+RB2) ==> VB=0.79v assume VBE=0.65V, so, VE=Vb-VBE=0.14v ==> IC=0.14/470=0.3mA If hfe= 230 ==> IB =0.3mA/230= 1.3uA Idiv = 5/(160k+30k)=26.3uA and the ratio = 26.3/1.3 = 20 not about 10 so i'm wrong.there are 2 possibilities: 1) hfe is lower than 230, it must be 1/2 230 to make that ratio ok 2)VBE is less than 0.65V, i don't know if it may be 0.51v If any of these is true, then we have IB=2.6uA and the ratio ia about 10:1. Back to Av; re=25/0.3= 83.3 ohm Av=10000/83.3 =120 it is a big gain!!! Zin = 160k//30k//(hfe*re)= 25.3k//(230*83.3)= about of 10.9K Zout= RC = 10K.
  16. Hi audioguru, What did you mean by the following: 1) Odd resistor values 2) Select the RC and RE resistors first AND why u audioguru tend to make VE very small like 2mA*100ohm=0.2V. I see many circuits in that the VE is greater than 1 volt. WHEN IC is big (10mA), we use a transformer power supply rather than batt. audioguru: "3) The lack of adding a bypassed emitter resistor to raise the emitter and base voltages means that the DC operating point will change too much with different transistors." Walid: (1) u all the time want VE very small and now you want to raise it (2) u now introduce a new aspect (a bypassed emitter resistor) we never use in the last design. I think u like to dispersed my mind. (3) Please make good work for your friends and design it as you want and show it to us. I beleive u never do so, u want to always ask u...... (4) Why now, the DC operating point will change too much with different transistors. Why it does not change in your first design. I wait.... but please, we now study the autir' first design without any bypass
  17. Hi lets do it again, I'll design a new one close to autir's: Decisions: Vcc=9v, 2N3904 transistor of hfe=230 typically and VBE=0.65v Av=20, VCE=4.5V, IC=10mA Calculations: IB=10m/230=43.5uA, Idiv=10*IB=0.435mA Rc=20RE, VCC-VCE=IC*(RC+RE) ====> RC=428.6ohm and RE=21.5ohm VB=VBE+IC*RE = 0.864V R1+R2=VCC/Idiv, VB=VCC*R2/(R1+R2) ====> R1=18.71K, R2=2K. please correct me or tell me it is agood work yours, walid.
  18. yes autir, Arab food has these "side-effects" but appear after 40 years, so be aware to not have any dog in your house. I hope u enjoy with arab food and thank u for these feelings. I'm from Gaza-Palestine
  19. I read carefully autir's design and audioguru's reply and my comment is: If I disregard the low values of R3&R4 for a moment I can say that autir do a good job in organizing this subject and make it more direct. I'll continued autir's design after these lines in the .pdf file (age 2): By Kirchoff ’ s second law,the voltage drop across R2 will be VBB=VBE+IC * R4 ==> VBB=0.87 Volt . then, according to audioguru's theorem: Idiv= around (10 * IB) ==> Idiv = 10 *25uA =0.25mA so, R1+R2 = 9V/0.25mA = 36K .....(1) VB = 0.87 V (from autir) ==> 0.87 = 9R2/(R1+R2) .....(2) solving these two equations we get: R1 = 32.52 K and R2 = 3.48 K , you must round them to a practicle Values. Now, autir choose Av=20 and IC=600*25uA = 15mA. I think it is an o/p stage because of this high gain and current, isn't so? Values of R3&R4 are not good as audioguru said and autir may choose 2k and 100ohm. to do this with keeping Av=20 and Ic=15mA, VB must increased: VE= 100*15mA=1.5V, so VB = 1.5 + 0.66 = 2.16V, use this value instead of 0.87 in eq(2) above you get: R2=8.64K and R1 = 27.36K and you still keeping the 10:1 divider and your initial conditions. Note: I'm not trying to teach you, simply i repeat what u teach me, to be more rightful, i repeat what audioguru teach me, hope u correct me if there is some mistake. thank you all. note (2): Audioguru, you teach your dog somethings wrong.
  20. IF i have a class a amp with o/p z = RC = 2K ohm and there is a second stage connected to it what the best Zin value of this second stage to not loaded the 1st one is it true Zin2 >= Zo1
  21. You don't answer this question that i asked before; The voltage gain of this amp = RC/RE, where Re not bypassed. What would this gain be if RE bypassed? thanks. Staigen gains 289 posts by sending 289 Hehe....hoho haha hehe isn't funny!
  22. Hi guys I learned English through reading Electronic books and articles, so I can talk in this subject. I other sides of English language I'm very weak and depend totally on my dectionary. I want to tell Audioguru something about his new PC like WAW but the language limit my feeling. I want to tell Alun that i hope him to be very well tomorrow I want to tell Alun that I thank him for all the efforts changing .doc to .pdf for me If it is in arabic i can tell you a good words but I hope you understand me and I'm sure you do. Alun:"What operating system are you running" Walid: windows95 Alun:"How much RAM do you have?" Walid: 48MB, it was 16MB and later I add 32MB anyway i live with this PC tell i buy anew one pentium 4 thank you Alun. A final note: I don't know till now what the difference between Autir and autir, please tell me to be aware. For kissing, In Arab socities, it is a habit to kiss one man another and this as a welcome, anyway i still learn from my best friends. Tell me audioguru what should I say to tell you that I respect u instead of what i said. I'm going to study the class A amp of autir to see what we can do with it.
  23. So if I want to build this Tx using a mic from a radio/cassitte, then i must put R1 = 4.7K and not 10K, or what would you say?
  24. Hi Alun I miss u .... I this have petium 1 133MHz PC since march 1997, it is 1300$ that days, today it is less than 100$. I can't buy new one pentium 4, not only for money but because this PC satisfies all my requirements with less problems. Audioguru, have 486 PC So i agree with u that AUTIR replace his file with .pdf thank you.
  25. hi all Audioguru: "Half the supply voltage across the mic determines the resistor's value." Walid: lets calculate it: R1 = 0.5VCC/0.5mA = 2.5V/0.5mA = 5K, not 10k why?
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