[AM reciever mixer]

Hi
the folloing fig is the front end of a simple AM radio reciver



It contains the AM ferrite antenna and two variable caps, one connected // to the primary coil of the antenna and the other // to the primary coil of the red (local oscillator) transformer.
AM sig freq from 530 KHZ to about 1600 KHz is tuned and go to the base of the transistor through 20n cap.
the local osc connected to the emitter through 10n cap. its freq is always higher than the AM sig by 455 KHz.
the o/p of this mixer at the transistor collector is 4 signals: the two signals and sum and the difference.
That was a simple summary of circuit. My question is that i need a simplified explanation of the work of this mixer?
thank you very much

[Are datasheets important?]

First I used the dictionary for translation so please forgive that there were some errors in language

In fact, he is not an Engineer

OK, I'm engineer in communication and control, 5 years in university after high school , Despite this, I feel myself as if I'm small and stupid pupil when discuss [glow=red,2,300]GURU[/glow].
I know many of Engineers, and dealt with through the hundreds of forums, and even in this forum did not find a better than [glow=red,2,300]GURU[/glow].
Obtaining a degree in engineering is required, but not enough. The world needs a genius like [glow=red,2,300]GURU[/glow], does not need papers written in it a certificate in engineering.
Why take advantage of the engineer does not understand anything and does not have something to offer to society.

For all these reasons, I find that even [glow=red,2,300]GURU[/glow] right to criticize any person manner it/he deems appropriate
when [glow=red,2,300]GURU[/glow] says that, I believe this because [glow=red,2,300]GURU[/glow] said that.
Do not think I crazy or naive, in the community that I live in, we have a saying heritage states : [glow=red,2,300]"Who taught me a characters I was a slave to him " [/glow]
Only [glow=red,2,300]GURU[/glow] taught me milions of characters and was patient on my often mindless questions.   
In the end, I say to all that [glow=red,2,300]GURU[/glow] precious wealth of scientific I hope to preserve and benefit from him

Ask God to extend his life And live happily with his family.
thank you [glow=red,2,300]GURU[/glow]
thank you MP
Thanks to all Forum members 

[Are datasheets important?]

Hi

It is my understanding that the author of the project is an R&D Engineer. I have seen him elsewhere on the web. You are throwing out rumor and speculation instead of fact. (theory?  ;D ) Are you claiming to be an R&D Engineer? You stated what R&D Engineers do, then you started talking about your selection. I was a little confused of the meaning behind that.

MP

I believe that Mr. [glow=red,2,300]AUDIOGURU[/glow] is a big  R&D Engineer, and he is the best one i meet in a forum
OK i'm not so good in electronics, but I can distinguish between person and person, and I know from the answer the person's abilities.
Therefore, I ask everyone to deal with Mr. AUDIOGURU courtesy and respect because he deserves it 
thanks

[Frequency response for a transistor amp.]

thank you guru


Summary of what guru said:
(1)Pre-emphasis: a process is being carried out at a transmitter to boost high audio frequencies
(2)De-emphasis: a process is being carried out at a reception side to cut the treble audio frequencies back down to normal and cut hiss at the same time.
(3)Why do they boost the treble audio frequencies at the transmitter? If a transmitter doesn't boost the treble audio frequencies then a radio will cut them and it will sound bad, like a stereo with its treble control turned all the way down.

I have a few questions:

(1)Why remained an American used the same pattern despite the tremendous progress of technology?
(2)Why they need to boost then cut, if they didn't cut at the reciever they haven't to boost at Tx?
(3)

The knee where the frequency response is +3dB is 2.1kHz (RC= 75us) for America, and is 3.2kHz (1khz (RC= 50us) for Europe.

(a) 2.1kHz: Is it from MATHCAD curve or from your design and calculations?
(b) RC = 75us: RE * C4 = 470 * 150n = 70.5us?
© 3.2kHz (1khz (RC= 50us) : RC = 50us is ok but what are these two frequencies?

(4) Our region (Israel-Palestine) is closer to Europe. If I did my calculations to design the first stage pre amp And went to the selection of this C4 value, then I have to use (RE * C4 = 50us)? 
(5) The following Fig is drawn using a cwCAD III

I want to watch the freq response But I do not know how to do it, can u please tell me.
I want that to compare between Math calc and swCAD results.
(6) You told me once:

The frequency response of the preamp is flat from 60Hz to about 600Hz and is boosted above about 1.5kHz due to the selectable pre-emphasis capacitor.

How the freq response be flat if we know: Xc4 decrease when freq increase
XC4// RE  is also decrease with f
so Av increase linearly with f ?!
thank you audioguru

[Audioguru FM Tx]

Hi GURU

Quote from: awash on August 14, 2006, 10:16:14 AM
C1 and C3 = 330N I got only one 330N cab I replace the other one with like 470 n or should I have make the same capacitor for both please help me to solve shortage of one 330N cap.
They are just audio coupling capacitors. A higher value will pass lower audio frequencies. 330nF passes frequencies down to about 40Hz, 470nF will pass audio frequencies down to 28Hz and 220nF will pass audio frequencies down to 60Hz, with the impedances of the circuit. Their values don't need to be the same. Use 100nF if you don't like deep bass.

From the above values of C1 and C3 and the corresponding values of the freq they can pass, I deduce that u assumed or calculated the i/p impedance of the first stage and the second stage to be 12K.

For the 2nd stage: you did these calculations:

{The input impedance of the 2nd stage is determined when the emitter current is determined:
a) The emitter voltage is 2.25V (a guess).
b) Therefore the emitter current is 2.25V/220= 10.23mA.
c) The hFE is 230 so the base current is 44.5uA.
d) The voltage across the 47k base bias resistor is 44.5uA x 47k= 2.09V.
e) As a check, the voltage across the 47k resistor is 5V minus
[the emitter voltage of 2.25V plus the Vbe at 10mA of 0.72V]= 2.03V. Pretty close.
f) Part of the transistor's input impedance at 10mA= 500 ohms from its graph.
g) The input impedance of the emitter resistor times the hfe of 180 (from the graph)= 39.6k.
h) The total input impedance of the 2nd transistor = 40k.
i) The 47k bias resistor in parallel with the transistor's total input impedance= 21.6k.}

So u concluded that Zin = 21.6K
RC filter freq at that Zin = 1/(2 pi R C) = 22 Hz and not 40 as u said>

For the First stage u concluded that the i/p Z is about 10K and so the cutoff freq = 33Hz and not 40

NOTE: I am not looking for mistakes to say that this is a mistake, I feel the forest when I find a contradiction. I studying this discussion often without any comment and every time I understand more things

Thank u guru for everything u teached me

[Frequency response for a transistor amp.]

Finally, after a long discussion with Mr. Audioguru, i discover that the frequency response is a relation (curve) of freq vs voltage gain (Av).

The following Fig is the 1st stage of guru's FM Tx:

Av = RC/re, this is assumed at very high freq that makes Xc4 = 0 so shortened the RE
But at audio freq (30Hz to 15KHz), Xc4 changes from about 35k at 30Hz to about 70 ohm at 15KHz.
So, RE (=470 ohm) can not be considered as shorted by C4 at all freqs
RE//XC4 range from 61 ohm at 15kHz to 464 ohm at 30 ohm

Av = Rc/(re+[RE//XC4]) changes with freq

I wrote a MATHCAD commands to look at this effect of C4, her is it:

And her is the results:

the strange thing is that the freq response is look like this:

with some knees (3dB)
my curve has not a knee

From my MATHCAD curve, I can see that C4 bossts the hi freq signals more low freq signal

please comment and explain this difference for me

http://NOTE: I'm sorry i made MATHCAD for C4 = 100n and discuss for C4=150n
It'll take long time to correct this and to upload the new fig

thanks

[Problem on two parallel plate capacitor.]

I think but not sure the following:
You know that C = e0 (A / d)
where A is the area of the capacitor plates, and d is the distance between them.

e0 is the permittivity of free space (8.85X10-12)

a) Vo is still as lt is cause the relation did not affected by batt voltage. the capacitance of a capacitor will is a fix value when it is manufactured>

b) capacitance, C: as d increase C decrease by the same factor (10 times) cause the relation is inversly proporetional
  the following link ia a calculator to see by yourself
http://micro.magnet.fsu.edu/electromag/java/capacitance/

c) electric field, E: V = Ed = (F/q)*d = Fd/q= W/q,  since q (the charge) is at the demonator then E increased 10 times

d) electric flux density, D: I don't know may be later I can found somthing
e) charge, Q: from the relation (Q = CV ) we see Q is directly proportinal to C, so Q is decreased 10 times as C
f) charge density: is decreased 10 times as Q
g) energy stored, U: W =

[Ft of the transistor]

Hi
I read carefully Kevin Weddle's reply an Audioguru's reply, but sorry i can't understand anything, i don't know why. Also I can't ask questions because i don't know about what to ask!
Despite that will ask :

I simulated my last circuit without the 75 ohm antenna load. The collector swung up to +20V.

how 20v and the supply voltage is only9v????

[BJT biasing formulas]

Thank you guru, I think that this is a big subject
it is better to focus now on topics simple and and when finished turn to the more difficult topics.

I want to tell you that a great honor for me to debate with you, you great wealth of scientific.

[Ft of the transistor]

My FM oscillator uses the transistor's alpha, because it is a common-base amplifier.

(1)I know that it is a common-base amplifier because of C5 connected between base and GND and at FM freq it short that base to GND.
But I did not know about the transistor's alpha, can u please tell me something about it?

(2)I attach your FM Tx,


I look at Q3, it is a 2N3904 transistor and it deal with 100MHz signal to amplify it.
How it can do this with only hfe =6 at 100MHz?
thank you guru, I wish that I'm not disturbing you by these frequently questions.

[BJT biasing formulas]

2) When the output capacitance of the transistor and stray capacitance combines with the collector resistor's value to form a lowpass filter. A lower value of the resistor would require a higher collector current for a higher frequency response.

Please I want further clarification on this point, I hope to support that with examples and illustrations
thanks

[What happen when downloading the previous stage?]

A very good example
thank you guru

[infrared help]

Hi

modulates the beam at about 40kHz

I understand how FM, AM and PM modulation takes place
If allowed to give us an idea of IR modulation
thanks

[What happen when downloading the previous stage?]

thank you guru

from what you said, if the o/p impedance of the first stage = 9K
and the i/p impedance of the next stage = 1k
and the signal is 1V
only (1/10) of that signal reach to the next stage
that is 0.1 volt.

If so, the word dowenload be inappropriate because I understand that the characteristics of the first stage change!

[BJT biasing formulas]

Hi
I'll back to the first design; C_E without CE.
I think in the past that increasing the value of Ic have better property to get higher o/p power,
and I was prefering -according to this- to design circuits with higher Ic values
GURU explained that this was a mistake and from my discussions with him I found that the
disadvantages of this selection:
1-consumption of the battery rapidly
2- more Ic = less hfe
3-i/p impedance to less with more Ic This leads to download the previous stage
My questions :
(1) Are there any other disadvantages?
(2) when I have to choose higher values for Ic?

[What happen when downloading the previous stage?]

hi
Today, I sat thinking at this point, what happens when the i/p impedance for the subsequent stage is less than the o/p impedance of the past stage? Please clarify the idea
Thank you

[Ft of the transistor]

Hi

Ft is the freq at which beta (hfe) = approx 1
for 2N3904 transistor I found that at 300MHz, hfe = 1



frome the same datasheets i found the following curve:

look at the red curve
at 300MHz, hfe = 5 dB
20 log (hfe) = 5
hfe = 1.8

Now at FM freq (say 100 MHz) and from the red curve, hfe = 15dB
that is hfe = 5.6 (very low)

In the past, when I was looking for a FT, and I find it = 300MHz, I think that hfe would be a big
value at FM freq = 100 MHz but it seems that the curve will be worth only about 6
My question: Is this enough when using this transistors as oscillator at 100MHz?
gURU use that transistor in his FM TX.
thanks

[The relationship of Amplifier vs Power vs Frequency]

Hi guru
I can use SWCAD III to see the freq responce of a C-E transistor amplifier circuit,
But I want to know how are calculated or how is expected during the design process
Can you please explain to me
thanks

[Salt-taster]

thank you guru it is a complete answer

[Salt-taster]

Thank you guru
I understand you
The author said that the voltage drop on a red led is 1.6v
Is this standard value
I measure it , it was 2v and sometimes 2.2 volt
thanks

[Salt-taster]

Hi
I feel I am alone with this question.
But in any case, I built the circuit and worked excellently.
The question remains puzzled me is the importance of R1 (=470 ohm)
when i tried to remove it, the red led is illuminate
It remains as to whether or not there is a salt solution
can anyone explain to me the function of this R1
thanks 

[Salt-taster]

Now I have a Specific question which is:
comparing the folloing figure:

with this figure (salt0001):

Noting that the voltage at pin 1 (Vo) = (e2-e1)(R2/R1)
I note that e1 = 1.8 V
e2 must be Slightly lower than e1, say e2 = 1.7V
so Vo = (1.7-1.8)(220K/10K) = - 2.2 V
If the solution resistance is increased, then e2 will be lower, say, 1.5 V
so Vo = (1.5-1.8)(220K/10K) = - 6.6 V
In all cases the o/p Vo will = 0 V because we supply LM324 with a single power supply.

Either the equation (Vo) = (e2-e1)(R2/R1) is wrong and must be (Vo) = (e1-e2)(R2/R1)
Or the design is wrong?????

[A very Urgent Question]

I am sorry I forgot to thank elix for his interest and involvement Thank you very much

[A very Urgent Question]

Hi dear guru

Hi Walid,
I think the very high supply current spikes from the 555 are interfering with the IR receiver.
A 555 creates current spikes of 400mA on the power supply. The datasheet for an LM555 warns that the current spikes can affect other ICs on the same power supply and recommends using two supply bypass capacitors at the 555.
The datasheet for the TSOP IR receiver shows a resistor that feeds it supply current which is missing in your circuit and is very important for it to avoid interference.
The datasheet for the 7805 shows a capacitor at its input and another capacitor at its output.

The circuit shows an NPN transistor that is being destroyed by high voltage spikes from the relay coil. The coil should have a diode across ir to arrest the spikes.

you are absulutely true
You are wonderful,
you are the better man i have dealt with him in the electronics
you our great teacher,
thank you very much
Yes, using two supply bypass capacitors at the 555 fixed the problem
Whatever I said I will not be able to express my deep gratitude to you

[A very Urgent Question]

Hi guru
Yes I forgot to attach the schematic of the "person detector" that drives the LED
The schematic is attached below from: http://www.electronic-circuits-diagrams.com/alarmsimages/alarmsckt19.shtml



As u see, it is a Tx and Rx
I used one IR LED and not two
I put the two units in front of each other and not beside as the designer said
My module is on until some one pass so I replace the NPN transistor by PNP one
I did not use a relay, the emitter of PNP connected directly +9V and collector connected to a RED LED through a 680 ohm resistor
The main o/p is taken from the collector point to the pic
Now, I am not interested in pic (microcontroller)
When I walk in between the two units and stop for a moment  (This possibility can happen in practice
) the LED (and so the o/p) is Intermittent lighting glows irregular
What  can  I do to change this o/p to be one short 5v pulse
Thank you guru very much