# question regarding impedance matching on a ne602 with a transformer

Discussion in 'Mathematics and Physics' started by BkraM, Aug 9, 2015.

1. ### BkraMMember

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Hi all,

I want to experiment with a ne602 mixer, but a bit confused on the required impedance matching.
In the stup i have, the input to pins 1 and 2 comes from a source with a 50 ohm resistance, the resistance between pins 1 and 2 is 3 kohm.

I'm reading that to match the impedance the turns ratio of a transformer is the key.
\sqrt{3000/50} or 7.75:1 should give a good impedance match between the source and the load.

The question is, how important are the number of turns itself. I could use a transformer with 7.75:1 windings or 387:50 or 775:100, which one would be preferred?
My initial thought is that it would be important to achieve some voltage level at the input pins, and therefore require enough inductance (depending on frequency) to achieve this.
But looking though the internet, all I read about is the importance of the ratio. Am i missing something and are all of the above choices leading to the same result?

Thanks,

2. ### Tony StewartWell-Known MemberMost Helpful Member

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Edit... I now understand this an INPUT step up transformer from 50:3000 Ohms internally loaded

you are making this for the differential output and so will have bifilar wound primary with tuned cap for resonance. as in fig 4b

Added.. the bandwidth is determined by the Q where Fc/BW-3dB=Q and Q is determined by ratio of Real/Reactive impedance of either element .

Chip Input capacitance will be a few pF depending on layout with ground plane and chip.

Normally the broadband inductance of the XFMR if sufficiently higher than circuit drive and load impedance then the number of turns is irrelevant

But in this case it is being used as a tuned filter front end so turns ratio will affect tuning capacitance and Q or BW of filter. For Highest Q, it means smallest capacitance and inductance at resonance.

What is your centre freq. and BW?

I suspect for RF you be using much lower number of turns

Last edited: Aug 9, 2015
3. ### ronsimpsonWell-Known MemberMost Helpful Member

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I can't find them but somewhere there are application notes on the input LC circuit.
----------edit--------------
AN1993

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5. ### Tony StewartWell-Known MemberMost Helpful Member

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Note This is probably for >> 100MHz
This step up autotransfomer will work better than an isolated transformer with poorly controlled inductance.
Tap ratio will determine input impedance

Note this will need to be an helical air coil so that inductance value is stable and accurate with the tap accurately placed for impedance transformation and a Spectrum Analyzer or sweep generator to measure Return Loss is essential.

Last edited: Aug 10, 2015
6. ### BkraMMember

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Thanks for the replies!

the frequency info should have been in my initial post, as the low frequency is the reason for this question...
I want to mix a 30 kHz signal from the LO with a 10 kHz signal. All this to drive a 40 kHz ultrasound transducer.
At the receiving end the signal is concerted back to a 10Khz signal, which will be logged.
the bandwidth will be limited to about 1, maybe 2 kHz.

7. ### ronsimpsonWell-Known MemberMost Helpful Member

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First you don't have to have a transformer. Is your source sign wave? Do you need a LC filter on the input?
A transformer for 100mhz might have 3 turns but at 10khz you need many turns to get the inductance up.

8. ### BkraMMember

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source is a chirp with a small bandwidth, so more or less a sine wave.
I can insert the signal as directly, as unbalanced. But was wondering what i would have to do to make it a balanced input.
I expected this would take a lot of (handwound) turns, which seems to be the case.

so best i can do is use an unbalanced input as these low frequencies?

9. ### ronsimpsonWell-Known MemberMost Helpful Member

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I you look at some of the "pre-amps" that sit before the mixer, you can connect one input to ground with a cap and drive the other input with a signal.

10. ### Tony StewartWell-Known MemberMost Helpful Member

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Is you signal AM/FM or PM?
WHat dynamic range do you need?
What SNR , Harmonic distortion and image rejection?

These must be defined first.

11. ### BkraMMember

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10 kHz input chirp is a FM signal with a bandwidth of approx 1~2 kHz.
I've got the setup working with the cap to ground as ronsimpson mentioned.
The goal of using a balanced input would be to get rid of the input frequencies in the output signal.
Curious how i would get such a low input frequency balanced

I'm not sure what you mean by specifying image rejection and harmonic distortion.

Thanks,

12. ### JimBSuper ModeratorMost Helpful Member

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Why not try a phase splitter type circuit, as used in audio amplifiers in days gone by to drive the push-pull output stages.

A single transistor stage with the same value resistors in the emitter and collector circuits.
The signals at the emitter and collector will be equal amplitude and 180deg out of phase with each other.
Like this:

Cribbed from:
http://www.diystompboxes.com/smfforum/index.php?topic=74786.0

JimB

13. ### Tony StewartWell-Known MemberMost Helpful Member

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The questions are too specific to modulators and demodulators without a sense of purpose.
Why AM modulated 40kHz with FM modulated 10KHz?
This seems a poor way to transmit a signal with a potential loss in SNR and inefficient use of bandwidth.