AM Radio Design Log

#1

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Ok, now that I got some encouragement, I am going to start here. This is in 
spired by fitness forums where people post a training log to chart their pr 
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Reply to
M. Hamed
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On a sunny day (Mon, 15 Jul 2013 20:20:27 -0700 (PDT)) it happened "M. Hamed" wrote in :

I admire your step by step approach.

Remember the simplest way to _add_ 2 signals is with 2 resistors, cheap too.

As to mixer, it is _hard_ to find something that is 100% linear, so most things will mix (will multiply, are non linear), from a simple diode to nice little ring diode mixers, (I like those), balanced mixers, JFETs, dual gate MOSFETS (I like those too, RF in one gate and LO on the other, just play with gate2 bias for max non-linearity), or a good old non-biased transistor with RF on the base and LO on the emitter... etc etc..

As to Q, LC ratio and bandwidth, it may be worth it to run that in LTspice, and use small signal mode and sweep a couple of octaves, you may find that a bigger C may help sometimes. Using bigger values for C also means you can hang your scope probe on the coil and tune for maximum, while with C values if in the few pF range the probe capacitance will de-tune things, just a practical consideration.

Reply to
Jan Panteltje

Thanks!

Two resistors to sum voltages, wow, why didn't I think of that! It seems to me for this to work I'll need to make sure the input impedance of the next stage is much higher.

Thanks for the tips on L's and C's. Now to actually figure it out using LTSpice, wouldn't I need to know the unloaded Q of the inductors for the simulation to be any useful?

Reply to
M. Hamed

Oh my goodness. Are you designing a fixed frequency receiver or a tunable receiver? Two oscillators at fixed frequencies does not give you a tunable receiver without some serious DSP or FPGA components.

Are you designing a direct conversion receiver or a superhetrodyne receiver?

There seems to be some serious misconceptions here.

Please read a primer on basic receiver types and the principles that they work on.

?-)

Reply to
josephkk

Neither. The two fixed oscillators are not for the final receiver. They are just for testing purposes and experimenting with mixers and filters. The final receiver should be a superhet.

Reply to
M. Hamed

The way i understand his approach, is that initially, those questions are in the future of experimentation. I see it as: step 1 use 2 ARBITRARY frequencies and try various amplitudes, (linear) summing methods, and (nonlinear) multiplier/mixer methods. Document results, make notes one what works and how well it works/does not work. Fundamental groundwork for next step. I see absolutely NO "misconceptions here". I see a rather rigorous investigation of a well-known technology using scientific analysis. The approach is admirable.

Boil no water without pot or fire.

Reply to
Robert Baer

"Robert Bore"

** I see nothing else but misconceptions - too many to count.
** You on drugs ??

** Hamed's approach is patently absurd - nothing more than the day dreams of a bored code monkey.

... Phil

Reply to
Phil Allison

On a sunny day (Tue, 16 Jul 2013 15:49:53 -0700 (PDT)) it happened "M. Hamed" wrote in :

Actually no, you can drive it into a zero ohm inverting opamp input and it will still work. Or, if that eludes you, into a 1 Ohm resistor, try it in spice.

'Q' is a very vague thing, others may disagree, but what you probably really want is bandwidth. As I was mentioning resistors, you can make your own Q with a simple parallel or series resistor, and if you know what bandwidth you want, work out that resistor (and Q).

You can see me do that here:

formatting link

Note the two 2k2 resistors, bottom right QPSK modulator, and the LC, (actually L is 184 nH, not uH). I started with the required attenuation for the mirror frequency after the mixer, say you mix 25 MHz with 1 GHz, you get (when carrier is suppressed in a ring modulator) 1GHz + 25 MHz +- sidebands, and 1 GHz - 25 MHz +- sidebands. When not filtering the 25 MHz enough, then the sidebands (in this digital setup) will extend from zero to eeeeh Ferry Much. You want some real attenuation so the LO + 25 spectrum and LO - 25 spectrum do not overlap, as then receiver gets confused. (so it does not work then). So couple of MHz and at least some decent deebees at the other side is required, you start with BANDWIDTH. From the bandwidth the Q follows, and from the LC ratio AND that Q you can get the attenuation. I actually did try it in LT spice, although I already knew 220 pF is a good value for about 30 MHz (analog TV IF I designed in the past). But I ran the spice for a smaller C too and that sucked, just to make sure. The 2 2k2 SUM the signal, and in these case work into a VARIABLE impedance (parallel LC impedance is high in resonance), and gets SHORTED at other frequencies, there is a non-linear! load from the ring diode mixer too. And man does it work!

To get rid of the mirror I may add a Ghz or there about 25 MHz wide SAW filer, 10 for 5 dollar on ebay...

Well, maybe my way of designing is not conventional, but it always works and nothing goes phut for me, so I am doing something right.

Should not be writing this... LOL

PS LTpsice is actually easy, even I could use it, easy enough for me to use to decide not to write my own, something I sometimes do when a program's learning curve takes more time than writing my own version. But I use it just to test small sub-circuits, never a complete setup, that would not work likely, in the real world there are such things as parasitic capacitance and inductance, coupling between tracks (magnetic too), track impedances, you need a lot of experience making peeseebees or special RF software (microwave) for that. Wavelength is the keyword.. It is fun..

Reply to
Jan Panteltje

for this to work I'll need to make sure the input

What does the "zero ohm" part of "zero ohm inverting opamp" mean?

Getting back to resistor mixers, here's something unintuitive (at least for me). One of my circuits uses two 10K resistors as a stereo to mono mixer to drive the 600 ohm input of a Valcom paging/music control unit. Theoretically one (ie me) expects the exact opposite. That is, for 600 ohms to drive 10K.

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Don Kuenz
Reply to
Don Kuenz

On a sunny day (Wed, 17 Jul 2013 10:00:08 -0500) it happened Don Kuenz wrote in :

It means that whatever current you send into it (the input), the output will feedback so the input stays at the same level, so no voltage will be generated at the input:

----- R1 ---- | | - in | opamp out -- |- + in

say you feed 1 mA into the - input in this example, then the output will go negative until the - input is at ground level again, so Uout / R1 is -1 mA. This is for an ideal opamp of course with infinite gain and input impedance.

Beware, when audio people say 'mixer' they actually mean ADDITION, a linear process, in fact audio people try normally to stay clear of non-linear processes.

When RF people talk about mixers they are talking about multiplication of signals, basically a non-linear process.

Audio people often have no clue, do not terminate cables with the correct impedance, maybe based on the idea that at these low frequencies it is not needed, but that is their idea. I rather talk to a video man about signals than to an audio one. And do not get me wrong, I worked in film audio for many years. And I learned a lot there too from them.

Reply to
Jan Panteltje

Hi Don

The inverting opamp is playing I->V converter. The inverting opamp will via the the negative feedback maintain the inverting input at zero volt (ideally).

/Glenn

Reply to
Glenn

Thank you. I was a bit worried there. Design some Hartleys, some Colpitts, and a couple other types then. Each one has different valuable properties.

?-)

Reply to
josephkk

OK. That's just a plain inverting opamp to me.

Unless one forces it to become a linear process by using Fourier transforms to move things to the frequency domain. ;)

At any rate, you busted me fair and square. What on earth made me think of audio mixers in a radio thread?

"Let he who is without sin cast the first stone." LOL.

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Don Kuenz
Reply to
Don Kuenz

On a sunny day (Wed, 17 Jul 2013 22:53:38 -0500) it happened Don Kuenz wrote in :

It shows the '-' input is zero ohm, behaves as a virtual ground.

Yea, trying some escapist moves?

I should have added (to 'usuallly do not terminate with the correct impedance') that that '600 Ohm' input of yours is probably not that, but a lot higher (could be diff opamp with 2 10 k resistors). It is easy to check, scope the signal before and after your 10K resistors, (600 / (10000 + 600) ) * Vin (I like those ')' ), anyways I am not betting but think it will be more volts at that imput than that predicts.

Reply to
Jan Panteltje

#2

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Toroids arrived. First target is the 550KHz oscillator using the first Hart 
ley oscillator shown in Exp. Methods. Using a JFET, diode, 1MOhm resistor,  
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Reply to
M. Hamed

About 5 percent off in frequency, looks like parts tolerances. How is the wave shape for being clean? Can you digitize it and do a FFT?

?-)

Reply to
josephkk

#3

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Wired up the second oscillator. The inductor value should come to about 16  
uH. Tried 5 turns on FT50-43 with a center tap but this didn't work. My gue 
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Reply to
M. Hamed

The wave looks very clean and nice. It would be nice if I had a spectrum analyzer to look at the spectral purity :)

At the moment I have no way to digitize it and FFT it. I may be able to bring the circuit to work and do that. I have access to a high end scope there.

Reply to
M. Hamed

s fixed at about 0 V and the power supply shows it's drawing milliamps of c urrent. As I increase my supply voltage, the signal at the diode starts inc reasing and looking more sinusoidal while the sine wave center point starts moving more negative. This is not clicking yet and still mysterious.

I forgot to mention that as this happens, the current drawn from the supply goes less and less as I increase the voltage. A bit counter-intuitive.

Reply to
M. Hamed

On a sunny day (Sat, 20 Jul 2013 09:00:04 -0700 (PDT)) it happened "M. Hamed" wrote in :

Those oscillators can be simplified:

  • |---------- | | |--- d === C3 --------------->| | | | |-- s /// L === C1 | | |------------| /// === [ ] R1 | C2 | /// ///

The 'L' provides the grounding of the gate for DC. The ratio of C2 to C1 is the 'gain', normally I use C2 = 2 * C1 There is no tap needed on the L, and it requires fewer components. For frequency calculation C1 and C2 are in series. You do not need R5 (10k to ground) in your circuit either, the output impedance is about R3 and R4 in parallel.

Decoupling is always required, voltage sources are never ideal, batteries age, other circuits may be affected by the ripple you create, and vice versa, etc.

Because it is a strange circuit, in the above example R1 takes care of auto-bias (source will be positive relative to the gate).

In the above circuit you can, within reason, connect almost any inductor and it will oscillate. I have used it with 220 pF and 100 pF for C2 and C1 from kHz to many MHz.

.

Input capacitance of FET, and also diode capacitance, depends on applied voltage (look up varicap). So making the anode of the diode negative should decrease capacitance, and increase frequency. Your Chinese meter (good you have one with a counter) probably has a lot of input capacitance, this will lower the frequency if in a way parallel to the tuned circuit, note your hands capacitance to the test leads too, and you to ground, in series. It is important to notice that if you take signal from the L, then it will also swing _negative_, that may matter if you connect it to a next stage, that could need a DC blocking capacitor.

Reply to
Jan Panteltje

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