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NE602 - Calculating input bandpass filter

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atferrari

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I intend to experiment the basics with an NE602 double balanced mixer. Frequency of interest 10 MHz.

From all circuits I found, there are two, more or less close to what I intend to build winding my own coils. No toroids available for the moment.

My questions:

a) The double tuned bandpass filter in this design is symetric: the secondaries are connected to the antenna pot and to the pins 1 & 2 of the NE602.

How is this possible since I am designing for a yet unknown antenna (thus unknown impedance) while having 1500 Ohms at the input of the NE602?

b) The other design (page 11) connects the secondary to the antenna and the primary of L2 straigth to pins 1 & 2.

I calculated Xl (= Xc) = 333 Ohms. Is this reasonable to match the 1500 Ohms input of the NE602? What about the secondary of L1 respect to the antenna?

c) I believe I will end winding two similar tuned transformers adjusted with trimmers.

Understanding that the secondaries will show an impedance proportional to their number of turns vis a vis that of the primary, what values should I use to calculate my coils?

d) Could I use a tap to the primary instead of a secondary to adapt impedances?

e) Today, revising my initial calculations for the possible values of L and C I found myself in doubt: in an LC tank at resonance, reactances are XC = XL, say 330 Ohms each.

What impedance should I actually measure? Provided that they cancel each other, only actual R is left. Am I right?
 
First of all, the DC receivers will work reasonably well, BUT with very little extra complexity, you can build a far better receiver with the 602/612.
It won't be long before you want to improve on it!!!

You need to treat the filter and the matching of the antenna as two separate things. The filter will work best with both ends matched (within reason), so a matching unit will be used to match the antenna to the front end of the filter.

Of course, the trimmers in the filter will help.

But then both circuits use a pot as an RF attenuator!

I found using toroids on 14 MHZ the bandpass filter was too narrow, and had to be re-tuned between the bottom and top of the band, so I replaced the toroids with the little inductors that look like 1/4 watt resistors, and the bandwidth was wide enough to cover the 350kcs.

The other design page 11 effectively doesn't really connect the filter direct to pins 1 and 2, it connects it to pins 1 and ground, as pin 2 is bypassed to ground with the 4n7 capacitor.

What do you intend using it for? (SSB, broadcast etc)
 
First of all, the DC receivers will work reasonably well, BUT with very little extra complexity, you can build a far better receiver with the 602/612.
It won't be long before you want to improve on it!!!

PLease, I used quite good equipment for years so I know what is good. I just want to build this myself.

You need to treat the filter and the matching of the antenna as two separate things. The filter will work best with both ends matched (within reason), so a matching unit will be used to match the antenna to the front end of the filter. Of course, the trimmers in the filter will help.

Have the idea that the maching stages are acting, per se, as filters. They are tuned.

But then both circuits use a pot as an RF attenuator!

So? Is that good, bad or nonsense? Many other circuits around there use them. Is that "matching"?

What do you intend using it for? (SSB, broadcast etc)
Time signals. Around 10 MHz as I said.

Please note that I am asking what values to have into account for calculating.

From all my ignorance, I understand that I need to adapt the eventual 50 or 75 or 300 Ohms of an unknow (yet) antenna to those 1500 Ohms of the input at pins 1&2.
 
Looking at the datasheet for the NE602 (OK SA602 on this sheet):
https://www.electro-tech-online.com/custompdfs/2013/02/SA602A.pdf
I dont think that the "matching" is critical for the antenna input circuit.

What I will say is that the circuit which connects the LC tuned circuit directly to the input will have a lower overall Q and hence a broader bandwidth than the one which uses the link winding to connect to the input.
Which of the two configurations is "better" is open to debate...
...one of those pointless debates which we see here on ETO and last for about 10 pages with no firm conclusions reached.

The two capacitively coupled tuned circuits are there to filter the required "band" of signals from the several volts of signals which will be present on the end of a wire antenna.
Without them the poor little 602 will be well overloaded.

And speaking of overloading, using a pot as an RF attenuator is an accepted practice on the front end of simple radio receivers.
It is not a calibrated attenuator, but then it does not need to be calibrated, it just provides attenuation if and when you need it.

JimB
 
I dont think that the "matching" is critical for the antenna input circuit.

Why? Even if no critical, I intend to have something less than marginal

The two capacitively coupled tuned circuits are there to filter the required "band" of signals from the several volts of signals which will be present on the end of a wire antenna. Without them the poor little 602 will be well overloaded.

Yes. That's why I want them properly calculated.

And speaking of overloading, using a pot as an RF attenuator is an accepted practice on the front end of simple radio receivers.

OK.


Jim, risking to make this a never ending thread (I hate that) I will add these questions:

Those "secondaries" (link windings) could be also implemented as taps in the L part of the tanks? Would they work like an autotransformer?

Just using the data from one of those circuits L= 5.3 uH / C= 47 pF the inductive reactance is appr. 333 Ohms.

If I wind a coil of 30 turns, for the above value, a link winding of 10 turns, would it show and impedance of 333/30*10 = 111 Ohms?
 
Looking at the datasheet for the NE602 (OK SA602 on this sheet):
https://www.electro-tech-online.com/custompdfs/2013/02/SA602A.pdf
I dont think that the "matching" is critical for the antenna input circuit.

What I will say is that the circuit which connects the LC tuned circuit directly to the input will have a lower overall Q and hence a broader bandwidth than the one which uses the link winding to connect to the input.
Which of the two configurations is "better" is open to debate...
...one of those pointless debates which we see here on ETO and last for about 10 pages with no firm conclusions reached.

The two capacitively coupled tuned circuits are there to filter the required "band" of signals from the several volts of signals which will be present on the end of a wire antenna.
Without them the poor little 602 will be well overloaded.

And speaking of overloading, using a pot as an RF attenuator is an accepted practice on the front end of simple radio receivers.
It is not a calibrated attenuator, but then it does not need to be calibrated, it just provides attenuation if and when you need it.

JimB


I agree, and in practice it doesn't seem to mind that much, probably more useful to keep it lower than the 602/612's input.

The 602/612 does overload very easily, I have found that a good antenna "tuner" serves the purpose if a wide range of frequencies is used, a narrow bandpass filter no doubt will be better under difficult conditions.

I was not knocking the pot as an attenuator, but the main point of the questions I saw as impedance matching.

There are other practical points that can make the difference between a very good 602/612 based receiver and one that just works, but I better not go on, I no way meant to imply that the OP didn't know what good equipment was or that he was ignorant, my apologies if he read that into what I said, maybe due to my enthusiasm after building 9 602/612 based receivers since the beginning of last December, and learning a bit along the way.
 
after building 9 602/612 based receivers since the beginning of last December, and learning a bit along the way.

So, is it any way that you could give some hints on how to calculate what I intend to build?

Gracias.
 
Jim, risking to make this a never ending thread (I hate that) I will add these questions:
Never ending thread are fine if they actually lead somewhere.
Many are just like a bunch of drunks discussing nothing in particular.

Those "secondaries" (link windings) could be also implemented as taps in the L part of the tanks? Would they work like an autotransformer?
Yes, that is often done.
Sometimes you will see a tap on the tuned coil and a link winding taking the signal on to the next stage.

Just using the data from one of those circuits L= 5.3 uH / C= 47 pF the inductive reactance is appr. 333 Ohms.
Yes, that is correct.

If I wind a coil of 30 turns, for the above value, a link winding of 10 turns, would it show and impedance of 333/30*10 = 111 Ohms?
No, on several counts.
The inductance (and hence reactance) of a coil is proportional to the SQUARE of the number of turns. So the reactance of the link would be 333/9 = 37 Ohms.

However, if you are trying to determine the resistance which is transformed into the the link from the tuned winding, that will not be possible from the information given here.

The equivalent parallel resistance of a perfect tuned circuit in isolation is infinite.
The Q of the tuned circuit is infinite.

In practice this in not possible there will be losses which will be seen as a resistance.
Let us assume that the tuned circuit has a Q of 100, we can calculate the resistance as R = Q.X
(X is reactance).

Coming back to your original question about impedance matching from the unknown antenna to the 602 input, it may be better to consider the effect of the antenna and the 602 on the Q of the tuned circuits of the input filter.
The input impedance of the 602 will be reflected onto the tuned circuit via the link winding, or direct connection depending on which of the two circuit you intend to use.
What is certain is that the direct connection will put a greater load (lower resistance) across the tuned circuit that the connection via the link winding.
Hence the direct connection will make the Q of the tuned circuit lower and so it will have a wider bandwidth.

Enough of this rambling for now, it is late and I am going to bed!

JimB
 
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