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# AM reciever mixer

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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

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Hi Walid,
The mixer in an AM radio multiplies the input signal with the oscillator signal.
I think the input signal amplitude-modulates the oscillator signal, producing multiplication that results in the sum and difference signal frequencies.

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Hi guru

I think the input signal amplitude-modulates the oscillator signal

the input signal is already AM signal.
How AM signal amplitude-modulates the oscillator signal?
thank you guru
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Hi Walid,
As the AM signal increases in amplitude then the oscillator signal's amplitude is also increased in the mixer.
As the AM signal decreases in amplitude then the oscillator signal's amplitude is also decreased in the mixer.

An AM signal has sum and difference sidebands.
A mixer creates a sum frequency and a difference frequency the same as an AM signal except both frequencies are RF.

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http://www.electronics-tutorials.com/receivers/am-radio-receivers.htm

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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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I think the text says it all that it is a coupling capacitor to couple back the signal for the circuit to oscillate. You don't need high gain for the oscillator. The total loop gain only need to be >1 to overcome circuit losses and enough phase shift to feed back the signal in-phase. The value of the coupling capacitor is not very critical as long as it has low enough reactance (Xc ~ 10 Ohm) at the frequency of interest.

Oscillators with more than 6-8dB of loop gain is normally not a good thing, as it produces many other unwanted products. Think of an oscillator as an amplifier with A>1 with in-phase feedback.

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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)

{it has low enough reactance (Xc ~ 10 Ohm) at the frequency of interest.}

Xc = 16 Kohm at  1000 KHz, and Xc = 8 Kohm at 2000 KHz

(2)
{Oscillators with more than 6-8dB of loop gain is normally not a good thing, as it produces many other unwanted products. }

Please I want further explanation on that point.
(3)
{Think of an oscillator as an amplifier with A>1 with in-phase feedback.}

A>1 To what extent?
(4)
{in-phase feedback}

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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Hi Walid, i have a doubt on the schematic put up- there has to be another transformer to recover the 455Khz IF after the osc coil.-- in fact the primaries of Osc coil and part winding of 1Sst IF (yellow Generally) wiil be in series. the secondary of this yellow device will give you 455KHz--
other calculations you are discussing with AN920 , i shall be watching curiously Sir,

regards
Sarma

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Questions:
(1)

Xc = 16 Kohm at  1000 KHz, and Xc = 8 Kohm at 2000 KHz

No, Xc = 15.9 Ohm at 1MHz and 8 Ohm at 2MHz

(2)

Please I want further explanation on that point.

You don't want oscillator device to saturate causing unwanted harmonics or spurious signals.

(3)

A>1 To what extent?

Gain must be > 1 to satisfy one of the two oscillation criteria.

(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?

It is important to note what happens to phase at the point of resonance. Look at the attached picture  of gain(blue) vs. phase(red) and it will become clear. There is no active gain due to absence of the amplifier. Now if you add the amplifier, the additional phase shift will result in a total phase shift near 360 deg with gain above 0dB. Now all oscillation criteria are met and circuit will oscillate

Last picture shows the additional signal change through the amplifier circuit. It is clear that the amplifier circuit adds more phase-shift of its own and a total loop gain >1 (~ 5dB) and circuit can now oscillate.
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With all basic elements connected we can see that circuit starts to oscillate when power is applied.

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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)

{Oscillators with more than 6-8dB of loop gain is normally not a good thing, as it produces many other unwanted products. }

(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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Program is Multisim from National Instruments.

I indicated a total phase-shift of 360 degee which bring you back to 0 degree, correct.

In oscillator design the term loop-gain is used to indicate the total gain in the loop (gain of amplfier - any losses from tuned circuit and other elements). Initial design starts with an open-loop analysis and ends with closed-loop analysis.

Phase shift caused by tuned circuit (C1//L1+L2) is 180 degree, and amplifier adds another 180 degrees for a total of 360 degrees.

I have added the last diagram (wthout the tuned circuit) to show that any change in current into the emitter terminal will cause an output waveform shifted 180 degrees at the collector. I have over driven the amplifier so that the phase shift will be clearly noticeable.

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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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The program is not free. They used to have a student or demo version. I am not sure if that is still the case.

This link

http://www.campustech.com/c/campust/title.html?id=gDHxDQRf&mv_arg=NINT100&mv_pc=65

you can buy it for\$35

You should be able to use Linear's switchercad also which is free

Another free spice
http://www.5spice.com/download.htm

Here is the same result using SWcad

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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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Here is osc startup with SWcad. Probe at emitter.

Also current into emitter and in collector circuit

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