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Everything posted by walid
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Hi AN920 I thank you for your wonderful to deliver some of the concepts that are new, at least for me. I want to know from a practicle perspective, I want to know the importance of this subject. Where we need to calculate the -ve resistance? I did some search and get the following: http://www.st-andrews.ac.uk/~jcgl/Scots_Guide/RadCom/part5/page1.html I'll discuss with you: (1) From this i understand that Negative Resistance = superconducting. (2) I think that this is the best simplified definition of the concept of the Negative Resistance. Did agrees with me? (3)If you read the rest of the page and saw the pictures following, you'll know that there were three cases : a- R be greater than 0 . This is the factual situation that we know. b- R be equal to 0, a situation virtually impossible. c- Or to be less than 0, and here I ask how it can be made so? After receiving reply I'll discuss with u the contents of the next page : http://en.wikipedia.org/wiki/Negative_resistance thank u very much
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Hi AN920 What the meaning of: I am O+ who loves EE in a mainly <-O world. thanks
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Hi AN920 Thanks to the rapid and excellent reaction. I have only a digital multimeter Equipped with the function of measuring capacitance. i have not an oscilloscope or function generator. Therefore, I will not be able to know the parameters of any crystal using the methode mentioned above. The discussion has been more than impressive, and the information provided by AN920 was excellentand new, at least for me. I will stop discussion on this subject in the hope that I met with Mr. AN920 in another new subject. Thank you very much and goodbye.
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Hi AN920 I was very happy when I read your reply, I wish you a long and happy life, thank you very much. (1) (a) That is the same as: SQRT(RS* DC gain) = SQRT(50 * 100) = about 70 (correct me) (b) What this value? I mean what does that mean? (2) omega = 2 * pi * freq (yes it is) The rest is very clear, but I can use laws, although I do not understand how it came, and what it means. This by no means diminishes the value of the rare information you provided. Thank you very much. Now lets discuss about some relevant points : (1)The Crystal examiner that we are discuss about it. I built it to check for one type of crystal which is 455KHz. Here is its image: from: http://www.ersatzteile-onlineshop.de/gruppen/3853000000-1.htm All the remote controls that They were supposed to do repair containing this type. When i get its model data LS, CS and RS, I'll redesign the original circuit according to the method you provided me and put it here for comments and corrections. (2)Some sources have labeled them Resonator and others said crystals, Is there any difference? (3) I understand easily what resistor do in a circuit, also the cap and coils, but i can't imagine How crystal functioning in electronic circuit? (4) When I asked my first question on this topic, the circuit did not operate because of the error in the wiring but after correcting the error it worked well. I kept some of the damaged crystal because I noted that some of the damaged crystals back to work after a period. I noted that by coincidence. What I want to say here that I also noted that the testing device glows strongly with the good crystal quality and lighting weak with bad crystal. From all of the foregoing Can I say that the crystal does not destroy just like transistors (0 or 1) it weaken or its condition worsened. (5) The last question: I have heard repeatedly that the "remote control" pattern working with the carrier wave of 40 kHz, why use Crystal written on it 455 kHz? Thank you very much
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Thank you AN920, (1) (a)It was difficult to obtain any information from the seller in my country. How can measure these parameters by network analyzer? (b) Do you not see that the value of Ls = 0.08H is very large value? (2) Can you shed more light on this point and the relationship between Q and stability? (3) From datasheets i found: Yes we have more ft (200 MHz at 10mA) than at 0.5mA, and this is more clear with the following graph: Now, At Ic =10mA, DC hfe = 100 as shown in the following graph. Why you take the DC hfe not AC hfe? (4) In other words, you want to say that: Make VCE = 50% of V1? (5) Lets calculate it: Vcc=9V, hfe = 100, VE = 4.5V and Ic = 10mA ==> IB = 10m/100 = 100uA Vcc - RB*IB - VBE - VE =0 VBE at IC = 10mA (from one of the graphs in the datasheets) = approx. 0.68 V, so: 9 - 100u * RB - 0.68 - 4.5 = 0 RB = (9 - 0.68 - 4.5)/100u = 38.2 K ohm (far from 82 K)??? AND note that if IC = 10 mA then VE = 5.6 V not 4.5V. Please I want further clarification. Correction : Rs * 200 = 50 * 200 = 10 K , yes you are right 38.2 K > 10 K (6) How a low value of R2 can load the crystal, please Explain this point. I will complete the remaining questions tomorrow. Thank you very much.
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Hi AN920 Thank you for the wonderful and important answer. Questions: (a) The feedback ratio is the loop-gain = 150/680 = 0.2 < 1, are u mean that? (b) The remot control's crystal is 455KHz, can u please tell me the appropriate values for those caps, What the role? © How can influence? (d) I know that the FET have a very large i/p impedance so not loading the 32KHz crystal. But the thing that I do not know is why a high freq crystal can work fine with BJT While a low freq crystal can't? Thank you very much.
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Thank you AN920, It is very clear Questions: (a) After considering all of the above: Can we consider that C1 play a impedance matching between the antenna and the tuned circuit? (b) Another point is still unclear to me is : if we consider that the length of one meter of the wire antenna is of 50 Ohm resistance. So 10 meters have a 500 ohm and 2m have 100 ohm resistance. Now, with 10 meter (500 ohm) the recommended C1 is about 5pF, please look at the total Z: note 31.8 Kohm with 2 meter (100 ohm) the recommended C1 is about 50pF, please look at the total Z note, 3.18 Kohm Big difference ten times??!! © As seen above, the writer said: Why did not say internal inductance also? (d) Regarding the last point you mentioned about a switchable narrow-wide FM detector in a measuring instrument, I think you mean C16 (100nF) that big value cause a wide BW. I noteced some strange thing, the collector of Q1 is not connected to any supply VCC. Thanks
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Hi AN920 I have read with interest all that came in your last reply, so I read it more than five times, I want to further shed light on the following points : (1) Are u mean the series impedance of the antenna? And if so, What can the antenna be considered from this point? (2) Are u mean that with C1 it is will be like connecting a high value resistance in parallel with the tuned circuit. (3) I want to offer my interpretation of this phrase and you should correct me if I was wrong: Connecting this low value resistance in parallel with the tuned circuit make the total i/p impedance is always low for every signal of any freq so there is no selectivity at all. (4) Why u consider the antenna as alow resistance? I have heard that it is 50, 75 300 ohm, why antennas are low value resistance? (5) Why antenna is considered to be // to the tuned circuit, why not inseries? [glow=red,2,300]NOTE: [/glow] Please reply to questions one after another, not all at once so I understand very well not have to repeat the question in different ways. [glow=red,2,300]I forgot to tell you that your answer above, consisting of 4 lines better than 2 pages in the book. [/glow] thank u very much
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I am sorry if continued reading found the answer which is: But if you have any useful addition, welcome
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Hi all These days I read a book on radio receivers from the site : http://www.mikroe.com/en/books/rrbook/rrbook.htm During the reading, sometimes I find it difficult to understand some of the points, so I ask you to help me understand. Thank you very much. Question #001 (from sec: 3.1.4. Other Components) The writer said: this is the figure he refered to: as you see, few pF (3 - 7 pF)for 10 meters antenna and few dozens (30-70 pF) for 2 meter antenna. note C1 value is In inverse proportion with the antenna length. (a) Why this inverse proportion? (b) What would happen if we removed this C1? [glow=red,2,300]Thanks alot.[/glow]
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Dear AN920, I do not find the right words to express to you my deepest thanks and respect. thank you very much. questions: (1)The transformer Comprising L1, L2 and L3 have DOT at the bottom of each coil. I remember I studied something of this kind before, but now I forgot its purpose, can u please tell me, and what happen if we reverse one of them? (2)I see a command line: K1 L1 L2 L31, i know K1 is the mutual inductance and L31 is the inductance affect L3 due to current pass through L1. but how to write this command and how to choose those values? NOTE: There is a small problem in the program which is the nodes numbers that appear in the graph didnot shown in the circuit, and one can't know some graph belong any node. Thanhs
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Hi AN920 You are good person , I benefited greatly from your help, thank you from the bottom of my heart . What if I want to make a center tapped transformer at one side, how to simulate it.
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Thank you very very much, I have SWcad III, Guru give it to me it is a good news that SWcad can do it i'll try it with different values to see somethings then back to you to discuss thank u again AN920
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hi guru not at all, for example there is no variable cap and no transformer ..... how to get these parts? To Master_Ianick , thank you very much To Steve_hi , thank you very much
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Hi AN920 i put "Multisim National Instruments" in google but did not know how to download it please give me the link thank you AN920
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I am not expert like Mr. guru, I will try to answer using my simple information : Oscillators can be built using opamps or transistors or may be some special IC's. opamps can not handle high frequency but transistors can if u want to build a function generator try to search one in google or tell me to find one for u That is all I know regarding your question, and I hope that others help me in the answer Do not hesitate to ask about everything that comes to you
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Hello AN920, in fact, I thank you from the bottom of my heart to respond quickly. It is clear you made an effort to explain the idea but unfortunately I did not understand much, lets see: In your circuit, L = 4.7+4.7 = 9.4 uH C = 560p So the resonance freq fo is about 2.2 MHz. At fo i see from your graph that gain = 0 dB (Because of the absence of amplifier, as u said) and phase = 0 degree. when using amplifier, we get mor gain (5 dB as u said) and 360 degree phase angle and all these coditions are needed to maintain oscillation==> OK My questions: (1)First, before I forget, Can you please give me the name of the program which is used in the simulation to help me in analyzing some vague points. (2)360 deg is the same as 0 deg, Why you considered it as 360 and not 0? (3) (a) why you said "loop gain" and not simply "gain"? (b) Why Oscillators with more than 6-8dB of loop gain is normally not a good thing? (4)If I want to calculate the phase angle to the folowing circuit: using jwL and 1/jwC, can I look at caps and coils as: [(L1 + L2)//C1] +C2 or what? (5) Phase angle before using amplifier = 0 and after using amplifier = 360. What is the role of amplifier in this change. Can you explain this point. thank you AN920
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Hi please I want the program to simulate in electronics to perform some tests on the transmitter and receiver thanks
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Hi The statements which is not true is: d. An oscillator must have a negative feedback circuit. because An oscillator must have a positive feedback circuit to maintain oscillation. negative feedback is good for amplifiers to get more stability.
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Hi AN920, First, thank you for clarification and assistance, there are still some questions: Before that we have to agree on some things following: this is an AM radio reciever antenna, local osc and mixer AM band is from 530 KHz up to 1600 KHz the local osc' freq is always above that value of 455 KHz, so it can be set from (530+455) to (1600+455), that is from 985 to 2055 KHz. Questions: (1) Xc = 16 Kohm at 1000 KHz, and Xc = 8 Kohm at 2000 KHz (2) Please I want further explanation on that point. (3) A>1 To what extent? (4) I want to discuss this point in some details, Lets assume that there is some harmonic sig at the collector of 1000 KHz and 1mV rms. and Lets assume that the local osc transformer have the following turns numbers: the coil connected to the collector is of 10 turns (Let us call the primary coil). the other coil (secondary) is of 30 turns and tapped at 10 turns from the bottom. NOW: if the sig at the primary is 1000 KHz and 1mV rms, it will be of 1000 KHz and 3mV rms at the secondary also at phase with the original signal. At the tap we get the same sig with reduced amplitude to 1mV rms, when this sig pass through the 10n cap it will be 90 degree out of phase, so the it is not in-phase feedback??!! What is your reply to this talk?
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Thank you AN920, I saw this link previously, commentary good, but not enough. The most important thing in that link : my question: in my drawing: Can I consider the 10n cap connected to the emitter as a bypass cap and thus get a high voltage gain: I know that one purpose of using that cap is to block the DC from going to ground through the transformer coil. that cap have an reactance of about 16k at 1 MHz and about 8 k at 2 MHz. is the 10n value is choosen carefully? I thank everyone help me to understand the idea.
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Gang specifications used in FM receiver.
walid replied to electronics_shamim's topic in Theory articles
Hi shamim in addition to what guru said, look at the following link: http://www.mikroe.com/en/books/rrbook/chapter3/chapter3b.htm hope it help u -
Hi guru the input signal is already AM signal. How AM signal amplitude-modulates the oscillator signal? thank you guru