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walid
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The voltage gain of this amp = RC/RE, where Re not bypassed.
What would this gain be if RE bypassed?
thanks. -
Dear Audioguru,
This may be the last question to complete this circle of questions, but i expect u also want to finish this.
you said this statement: "A 10:1 ratio in a voltage divider doesn't allow the divided voltage to change much when a low gain transistor causes a 5:1 ratio or a high gain transistor causes a 20:1 ratio.
The resistors could also be 360k for R1 with 51k for R2."
1)I understand from this that you are usually take a 10:1 ratio but you may go slightly up or down. Can I take this as a rule in my own design?
2) You said also:"doesn't allow the divided voltage to change much when..."
Are you mean by this "divided voltage" that the VB which now = 1.2V doesn't change much by the applied AC signal, and WHY?
3) if R1=330k and R2=47K, then, the divided voltage VB = 10 * R2/(R1+R2)= 1.25v and ID = 10/(47+330)=26.5uA and the ratio is 11, where
if R1=360k and R2=51K, then, the divided voltage VB = 10 * R2/(R1+R2)= 1.24v and ID = 10/(51+360)=24.3uA and the ratio is 10.
and from all that we can concude again that the raio must be around 10.
I love you Audioguru I feel that i'm very close to the truth. -
Hi
If Zin=R1//R2//(hfe*(re+R4)), where re=25mV/IE, then
re = 25/0.55=45.45
Zin = 330k//47k//(230*(45.45+1000)
Zin = 330k//47k//240.45k = 35.13K as audioguru said 34.9K
Augioguru, why u think this slight difference. I tryed all combinations of 26 instead of 25 and 0.45mA instead of 0.55mA and the results are more difference. -
Hi audioguru
u told me before that IE= 0.55mA was wrong and IE actually = 0.45mA
please, make another figure like the first one marked "s transistor.png " with the wright numbers.
thank you. -
I'm sure u OK but:
1) tell me where the error in my calculations
2) What was the first the circuit with its resistors value known, or the whole circuit is not known.
I mean when you finished the 1st stage and move to this VCO stage, you sure put your assumptions, is VE=2.25 is one of them then u calculate the other values according to it.
why not to tell me the full story in numbers and I'll be very thankful for your favor.
Yours, Walid. -
To AUTIr
Autir:"why did you choose a ratio of 10"
I think that there is a rule: If you want a voltage gain say 100 don't make one stage to avoid distortion, make it two stages (10 * 10) or if you can more. -
Hi all my friends
Audioguru said: "4) I made the mic preamp sensitive enough to pickup sounds in a room. Try 5mV as its output (I didn't measure it) and the transmitter should have full modulation of 75kHz."
Walid: Two questions needing direct answers
1) How u made the mic preamp sensitive enough to pickup sounds in a room. I guess through R1, the resistor feeding Vcc to the MIC. You said once a day that the rule u use when choosing this R1 is that when Vcc = 9V, R1 = 9K, and When Vcc = 3V, R1 = 3K and so on.
I think that R1 is not the only thing you take into account to to make the Mic is sensitive! What you do?
2) When we suppose that 5mV AC is out of Mic, the transmitter should have full modulation of 75kHz. What if this 1mV and what the advantages and disadvantages by this?
Thank you. -
Hi all
1) I don't understand what prateek mean!
2) thank you Audioguru for your interest though I don't know what the lettet you whant to receive me?
3) all those replys are out of my question.
yours, walid -
All the time I ask one question and with your reply it become 100000 questions, I don't know why.
Other friends in this community just look leaving me alone with you, also i don't know whyyyyyyyyyy!!!
In your last reply You said:
Audioguru:"a)The emitter voltage is 2.25V (a guess)."
Walid: I understand b) to i), but guessing VE=2.25V not, lets calculate it:
i/p loop: Vcc-R6 IB-VBE-VE=0, VBE = 0.72V as you said,
5-4700 *IB-0.72-(hfe+1)*IB*R7=0
IB = 77.4 uA
IE = (hfe+1) * IB = 17.9mA, so,
VE = 17.9m * 220 = 3.9 Volt and not 2.25V it is far from it.
What would you say?
What I want to say is that I can't build my solution to a guess when someone asking me the same qestion. -
What do you want?
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I forgot 2 things
(1) is Vcc to the 1st stage is 5V after the regulator
(2) Why C2 it is strange, may be negative feedback!
thanks -
Hi all my friends,
I write these commands for Pspice sim.:
Audioguru FM Tx 1st stage (Audio preamp), V- divider configuration
VCC 1 0 5V
Vmic 8 0 SIN( 0 XXX XXX 0 0 0 )
Rmic 8 7 XXX
R2 1 2 150K
R3 2 0 39K
Q1 3 2 4 2N3804
R4 1 3 10K
R5 4 0 470
RL 6 0 XXX
C1 7 2 330N
C2 2 4 100P
C3 3 6 330N
C4 4 0 100N
.MODEL 2N3804 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 5us
.OP
.probe
.end
As you can see there are some xxx inside these commands:
The first two xxx are the amplitude and frequency out of the mic if we consider the MIC as a signal generator with Vs and internal R.
The frequency, i think, is between 300 and 3400 Hz (the human voice) and the amplitude may be few tens of mV. I can't make a decision, please help me in that.
The 3d xxx represents the Mic impedance taking into acount the supply resistance R1 in the above schematic and that the mic have typically 5k ohm so if i still remember Rmic = R1//5k = 10k//5k = 3.3k, correct me.
The 4th xxx represents the load impedance, that is the i/p Z of the second stage in the figure above. please calculate it for me to complete the commands.
when I finish it i'll show you many thing about this subcircuit.
Note: the trasistor model is found in its datasheet.
thank you all
walid. -
Hi AUDIOGURU
Audioguru:"The 1st thing was to choose a voltage gain of 10"
Walid: Voltage gain Av = Vo/Vin, how this affect my calculations of resistors, tell me please.
Audioguru:"so leave the collector resistor as 10k."
Walid:" Why you insistent to take it 10K, 8.2K is a close value to not to recalculate it from the beginning!"
Audioguru:"I didn't use a reverse approach. The gain of 10 and requirement for fairly low current demanded that R3= 10k and R4= 1k. Since idle current through R4 would raise the emitter voltage, I knew that the operating point must be with the collector voltage about 5.5V for max output swing. HFE and Ohm's Law calculated the remaining parts."
Walid:" Please Audioguru I don't understand this. You can calculate it as I do, save your time. The best way to make someone easily understand is to show him in form of calculated example, so why they do this in any technical book. Numbers solve the problem, thank you."
please Audioguru I want not to repeat my questions back and forth, with numbers and examples you make me understand fast and saving your time to others who need your experience.
yours, walid. -
To Alun, I mean it, I want all the readers and you to see and share the discussion with Mr. Audioguru.
To Mr. Audioguru:
Audioguru: "Do you have questions about it?"
Walid: Sure I have, I have a dozen of questions, lets start:
FM Tx is one of the important subjects that interests me and many of people, so I start to disinter it hopping that finishing it with the ability to make our special designs or at least to discuss with others any design we may faced.
(1)(2)(3)(4).......(10000) I'd ask you about every element value (resistors caps ...others) in your circuit, so, why not to tell us a brief summary about these elements.
At the meantime I'll test the stages using Pspice to enter deeply into the design.
I know i add weight to you but hope you bear us.
thank you very much.
yours, walid. -
Hi my best friend Alun,
I'm never ignore your replies, I read it carefully every time and I'm waiting for your next instalment that you will add Re in it to the previous common emitter amp and then finally the resistor below the base and you will explain everything, I wait for this tell now.
The problems are that:
(1) Your answers are direct, simple and easily understood compared to Audioguru's
(2) Your answers in most are similar to what I already read in many textbooks. It is true that there are some of helpful additives.
(3) It seems to me that the more hot discussion is that with Audioguru, and you instead of helping me to stand together face to face with Audioguru, you a rise subjects (only to read) not pour into the point of the hot subject.
Please don't admonish at (or to or from) me. I looked at this as there are many points in Audioguru's replies were hard to understand and I expect from you and others to help me to over this, then if any of you have an addition welcome, but not to ignore yours. I respect you and others from all my heart. I feel you are my electronic family and if I meet any of you someday in a road, I'll be very pleasure and I'll help him as I could.
Yours, Walid -
Look at this site:
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/flipflop.html -
Dear Audioguru, please can you lell me where can I find a full text about your FM Tx whose circuit is attached below.
thank you.
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I write the following prog to simulate the circuit shown in Fig.C-E below:
common E Amp freq response
VCC 7 0 DC 12
VS 1 0 AC 1
Q1 4 3 5 Q_2N2222A_N
RS 1 2 1K
C1 2 3 2uF
R1 3 0 10K
R2 7 3 30K
RC 7 4 4.3K
RE 5 0 1.3K
C3 5 0 10uF
C2 4 6 0.1uF
R3 6 0 100K
.MODEL Q_2N2222A_N NPN( IS=11.9F NF=1 NR=1 RE=649M RC=1
+ RB=10 VAF=83.5 VAR=41.7 ISE=350F ISC=350F
+ NE=1.58 NC=1.58 BF=204 BR=5 IKF=149M
+ IKR=149M CJC=13.1P CJE=30P VJC=2.87 VJE=678M
+ MJC=330M MJE=330M TF=531P TR=69N EG=1.11
+ KF=0 AF=1 )
.op
.end
The .op command results are:
NAME Q1
MODEL Q_2N2222A_N
IB 1.25E-05
IC 1.71E-03
VBE 6.66E-01
VBC -1.74E+00
VCE 2.40E+00
BETADC 1.37E+02
GM 6.53E-02
RPI 2.35E+03
RX 1.00E+01
RO 4.90E+04
CBE 8.44E-11
CBC 1.12E-11
CJS 0.00E+00
BETAAC 1.54E+02
CBX 0.00E+00
FT 1.09E+08
And I need to discuss some of these results with you:
(1) How far or close these results to the real life, can i depend on it initially?
(2) RPI is the i/p impedance looking into the base of Q1, so, the total i/p impedance is: R1//R2//RPI = 1.8K ohm. so we can look at the i/p portion of the circuit as: Vs connected in series with Rs then to C1 then connected inseries to that 1.8k.
if we look at Vs as a MIC and Rs as the impedance of the MIC and C1&Rtot as a high pass filter, is the value of C1 suitable and why?
(3) What exactly these parameters mean: GM, RX, RO, CJS, CBX.
(4) FT is (as I remember) is the freq at which the gain =1, that is at this freq there is no amplification, Is this OK?
(5) In the MODEL Q_2N2222A_N, BF = 204 is this the upper value that this Q can reach?
(6) In the MODEL Q_2N2222A_N, what these mean: NF=1 NR=1 NE=1.58 NC=1.58 IKF=149M IKR=149M VJC=2.87 VJE=678M MJC=330M MJE=330M TF=531P TR=69N EG=1.11 KF=0 AF=1
thats all
thank you in advance
walid
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Dear Audioguru
I take 4 peices of paper, write every hard aspect u tell me in a descrete papare. then I read all this document from the beginning, when reach a hard point i look to the related paper, finally i understand many points (but not all) and i descover that u are a genius, yes you are really genius and we all must profit from your experience.
At least for me it take at average an hour to understand one paragraph of your answers. You talk from a real world and this strikes the info (from theoritical textbooks) in my brain, this is why i can't understand you at once!!
Each time I read it i disvover somthing new deserve to ask about, but before this i want to redisign your V-divide Amp and ask you to correct me:
Any design needs three points:
1) assumptions
2) Rules
3) calculations and corrections
1) Assumptions:
here you must choose the components and the intial conditions of your circuit as follows:
Audioguru choose the famous 2N3904 NPN transistor whose typical hfe=230 and minimum hfe=100
Also he choose VCC=10v and IC to be 0.55mA and Vc = 5.5VDC fom max signal swing.
2) RULES:
a) Ohm's law; V=IR
b) VBE = 0.65 V approx.
c) hfe = IC/IB -
Hi all my friends in this community especially Audioguru
I'm sorry for the many mistakes in my last reply.
Audioguru is true I'm false
(1) hfe is 230 and not 20.3 as i do, because it is 0.55m/2.4u and not 0.55m/027u
(2) I mean a fixed bias and not a self-bias sorry I'm confused
(3) All other thing I told is not correct if Audioguru commented it.
I'll back from the beginning and ask these questions, it may seem boring but I'm insistent.
AUDIOGURU: " A transistor's collector is a high impedance current sink, so the output impedance is its collector load resistor. "
Walid: I need more explanation about this.
AUDIOGURU: "5.5V was chosen as its idle collector voltage so it can swing equally up and down."
Walid: Why 5.5V why not 1/2 Vcc thats 5.
AUDIOGURU: "Ohm's Law calculated the resistors."
Walid: I want now try to calculate them:
IC = (10 - 5.5)/10k = 0.45 mA not 0.55mA confused at the start
AUDIOGURU:"A 10:1 ratio in a voltage divider doesn't allow the divided voltage to change much when a low gain transistor causes a 5:1 ratio or a high gain transistor causes a 20:1 ratio.
The resistors could also be 360k for R1 with 51k for R2."
WALID: this is the main problem for me to understand
AUDIOGURU: Each part number and each sample of transistor has a different current gain.
WALID: Its OK
AUDIOGURU: Transistors have very reduced gain at high currents.
WALID: u mean if IC =100mA beta reduced, WHY?
AUDIOGURU: Therefore I calculate the current in the voltage divider to have about 10 times the transistor's base current..
WALID: how you calculate it and is it a rule to be this ratio? Please teach us.
AUDIOGURU: The transistor I selected has a typical gain of 230 and a minimum gain of about 100 with my chosen 0.55mA emitter current. Therefore its max base current is about 5.5uA.
WALID: OK
AUDIOGURU: So the minimum ratio of divider current to max base current is actually 4.82:1, not 5:1 as I said.
Walid: 4.82:1 ratio between what?
AUDIOGURU: The divider could be 330k and 47k or 360k and 51k and the results would be the same.
Walid: What the rule?
Finally, dear AUDIOGURU to save your efforts and to answer all question now and in the future about this subject tell us the story step by step like this:
I want Vcc = x and I choose a transistor x (you actually said this) IC = x and....
We have these rules.... Then calculate it to make Walid understand
Thank you
Yours
Walid -
Hi autir
Audioguru's design not a holy, there are many schools in designing a transistor amplifier. One of the simplest ways you can found in this site:
http://www.electronics-tutorials.com/amplifiers/small-signal-amplifiers.htm
it is very good for you to understand what you want, read it carefully then ask.
90% of Audioguru answers are obscure at least for me. Take a look at these examples:
(1) autir ask: How did you choose the Ib (2.4uA) to I(R1R2) (27uA) ratio?
Audioguru's answer: A 10:1 ratio in a voltage divider doesn't allow the divided voltage to change much when a low gain transistor causes a 5:1 ratio or a high gain transistor causes a 20:1 ratio.
The resistors could also be 360k for R1 with 51k for R2.
I try to understand it fore an hour but could not.
If i want to answer this i say: assume that Ib = 0, that is IR1 = IR2 =I
this I must be 1/5 t0 1/10 the IE
I = 10V/(330K+47K)=26.5 uA close to 27uA
If we choose I = 1/5 IE ==> IE = 0.27 ma, Audioguru IE = 0.55mA the double, that is Audioguru choose I =1/20 IE , many schools
Why Ib =0 because it is a very small value and we can negect it in our calculations.
In the link above you can read the full story it is direct and simple.
(2) autir ask: The Ib/Ic ratio in your circuit is proof of the hfe value of the transistor you used. Why, then, does it work the same with different hfe's? Does this mean that the Ib/Ic ratio can be decided by me without any consenquences regarding the circuit's function?
Audioguru's answer: I selected a transistor whose hFE is never less than half its typical, nor more than double. The circuit would need more current in the R1/R2 divider if a transistor with a lower minimum hFE was used.
walid: hfe in Audioguru's design = IC/IB = 0.55mA/27uA = 20.37 and this is less than half its typical.
(3) autir ask: The gain factor is calculated by the R3/R4 ratio, if I'm not mistaken?
Audioguru's answer: it is long and far from a direct answer
Walid: this is the way audioguru answer 90% of the questions, indirect and obscure. His answer must be either yes or no with explenation to why yes and why no. there are many questions must answered by yes or no at the beginning.
If I try to answer i say: NO NO NO this formula you use suitable for emitter degeneration (a sublink in the same site above) look at that site and compare.
(4) In his reply to Alun, Audioguru surprised that ther is no RE and no R2, why i don't know.
I tell him that this circuit called a self biased and is found as the first circuit in a small sig amp in any electronic book and this circuit is used in realty, i see it many times in areal designs.
Finally I hope from all my heart that Audioguru accept this criticism and still my big friend.
yours
walid -
I write the following prog to simulate the circuit shown in Fig.C-E below:
common E Amp freq response
VCC 7 0 DC 12
VS 1 0 AC 1
Q1 4 3 5 Q_2N2222A_N
RS 1 2 1K
C1 2 3 2uF
R1 3 0 10K
R2 7 3 30K
RC 7 4 4.3K
RE 5 0 1.3K
C3 5 0 10uF
C2 4 6 0.1uF
R3 6 0 100K
.MODEL Q_2N2222A_N NPN( IS=11.9F NF=1 NR=1 RE=649M RC=1
+ RB=10 VAF=83.5 VAR=41.7 ISE=350F ISC=350F
+ NE=1.58 NC=1.58 BF=204 BR=5 IKF=149M
+ IKR=149M CJC=13.1P CJE=30P VJC=2.87 VJE=678M
+ MJC=330M MJE=330M TF=531P TR=69N EG=1.11
+ KF=0 AF=1 )
.op
.end
The .op command results are:
NAME Q1
MODEL Q_2N2222A_N
IB 1.25E-05
IC 1.71E-03
VBE 6.66E-01
VBC -1.74E+00
VCE 2.40E+00
BETADC 1.37E+02
GM 6.53E-02
RPI 2.35E+03
RX 1.00E+01
RO 4.90E+04
CBE 8.44E-11
CBC 1.12E-11
CJS 0.00E+00
BETAAC 1.54E+02
CBX 0.00E+00
FT 1.09E+08
And I need to discuss some of these results with you:
(1) How far or close these results to the real life, can i depend on it initially?
(2) RPI is the i/p impedance looking into the base of Q1, so, the total i/p impedance is: R1//R2//RPI = 1.8K ohm. so we can look at the i/p portion of the circuit as: Vs connected in series with Rs then to C1 then connected inseries to that 1.8k.
if we look at Vs as a MIC and Rs as the impedance of the MIC and C1&Rtot as a high pass filter, is the value of C1 suitable and why?
(3) What exactly these parameters mean: GM, RX, RO, CJS, CBX.
(4) FT is (as I remember) is the freq at which the gain =1, that is at this freq there is no amplification, Is this OK?
(5) In the MODEL Q_2N2222A_N, BF = 204 is this the upper value that this Q can reach?
(6) In the MODEL Q_2N2222A_N, what these mean: NF=1 NR=1 NE=1.58 NC=1.58 IKF=149M IKR=149M VJC=2.87 VJE=678M MJC=330M MJE=330M TF=531P TR=69N EG=1.11 KF=0 AF=1
thats all
thank you in advance
walid
-
Clear, he is student and his teacher asking him for a project and he want many idias to cgoose one. Alun understand this and directed him to our Project Index
thats all
is it clear ECET0purdue
thanks -
The following answer from Ante is very very very good:
""If you put a smoothing cap after the rectifier you will sabotage the function of this circuit. The circuit uses the half cycles (pulsing DC) to make it possible for the SCR to turn off, if it didn’t turn off each half cycle there would be to much current flowing!""
At first I'm not understand it, but when read it again I discover the GOLD inside it, How this? read on
For long time I know that thyristors when connected in circuits and applying a sufficient voltage to gate it will conduct and remain in conduction (latch) even if we disconnect the gate.
The only two ways to make the thyristor to be OFF are:
1) disconnect its anode from the supply
2) sorry forget it (perhaps to reverse the applied voltage on A and K)
Look at the figure above, Q1 is the main thy. and it controls the charging process, if you suppose wrongly (as I do) that its Anode is connected to purely (constant)DC it is 100% will not be OFF when we disconnect its gate. But if the DC applied is of the form of pulsing DC (ON OFF) the thy. will cuttig the charging current if the gate voltage is zero.
THIS IS THE MAIN RESON TO OMIT THE SMOOTHING CAP AFTER THE RECTIFIER BRIDGE, THIS CAP will sabotage the function of this circuit.
So we can say that ANTE is a good person, thank you Ante.
Still I have more questions:
(1) I asked before this question:"The GR1= 50V 6A Bridge Rectifier and the transformer is 4A also the Ammeter is 0 to 5A. What making the designer sure that the batt not demand mor current like 10A, what happen if this occur?"
the answer by tnk2k was:"The battery voltage builds ups to limit the current. In case the battery shorted out, it will blow the 5A fuse."
And the answer by Ante was:"If the circuit is built correctly there are no problems with the current."
Both of them don't understand me. What I mean is that: are all batteries demand only 4 to 5 A of current at the beginning of the charging process OR is this current is the upper limit that any car battery want at first. Why not to be 10A, the capacity of my car battery is 60A.h and if it is full discharged it may demand more than 5A at first.
(2) Also I asked:" Is it necessary to connect the EARTH to the -VE terminal of the supply as shown in the schematic, EXPLAIN?"
The answers from Ante and tnk2k were about risk and safety, and again the don't understad me.
I know the importance of the earth in safty, but I mean: connect the earth to the transformer body and leave the -ve terminal as all circuits do?
I see many many circuits don't use the earth terminal at all and depend on the main earth in the home like the radio/recorder power supply.
(3) For D1 i think it is for maintaining a 0.7V across the gate!!!
BJT biasing formulas
in Theory articles
Posted
Hi Autir
I downloded this ZIP file and when opening it in WORD97, the document looks like this:
EMBED Equation.3
14,28 Ohm.
R3=20
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
285 Ohm.
By Kirchoff's second law, the voltage drop across R2 will be VBB=VBE+IC
EMBED Equation.3
please help me to view his document.
thanx.