- Can anyone answer these questions?

For the QSD circuit in the Ensemble Rx II:

1) if the output side of the mixer is terminated in a 50 ohm resistive load (e.g. no opamp, no capacitor, just a 50 ohm non inductive resistor back to an appropriate common point) what is the impedance looking into the input of the mixer from either side of the input transformer secondary?

If I understand correctly, this should be 50 ohms plus the switch-on resistance (aprox 5 ohms according to the datasheet I think) as the input side is always connected to one of the 4 outputs. Is this correct?

2) Is it correct that the output of the mixer (e.g. as seen on a spectrum analyzer) consists of only two components, the sum and difference of the input RF and the LO? For example, if the LO is 14 mHz and the only RF is CW at 14.010 mHz, then the output is 10 kHz and 28.010 mHz.

3) If the mixer is now terminated with two lines in parallel a) a capacitor in series with a 50 ohm resistor and b) an rf choke in series with a 50 ohm resistor, with the capacitor presenting a low Z at 28 mHz and high Z at 10 kHz and the rf choke the reverse, then the impedance seen at the mixer input is unchanged, still 50 ohms plus the switch on resistance. Correct?

4) Now if the circuit is as in 3) but the resistor on the capacitor line is changed to 100 ohms and the resistor on the rf choke line is changed to zero ohms, what is the impedance seen at the mixer input? How does one analyze this to see what the correct answer is?

The answer to 4) will help me to try an experimental redesign of the circuit.

Thanks! - On 7/1/2014 12:43 PM, tharlam@... [softrock40] wrote:
Can anyone answer these questions?

Each secondary of the input transformer will "see" 50 ohms during two of the four quarter cycle switching intervals. If each winding on the input transformer has an equal number of turns, then the input to the transformer will be 50 ohms (plus switch on resistance) at frequencies where the self inductance of the windings is high relative to 50 ohms.

For the QSD circuit in the Ensemble Rx II:

1) if the output side of the mixer is terminated in a 50 ohm resistive load (e.g. no opamp, no capacitor, just a 50 ohm non inductive resistor back to an appropriate common point) what is the impedance looking into the input of the mixer from either side of the input transformer secondary?If I understand correctly, this should be 50 ohms plus the switch-on resistance (aprox 5 ohms according to the datasheet I think) as the input side is always connected to one of the 4 outputs. Is this correct?

Yes.2) Is it correct that the output of the mixer (e.g. as seen on a spectrum analyzer) consists of only two components, the sum and difference of the input RF and the LO? For example, if the LO is 14 mHz and the only RF is CW at 14.010 mHz, then the output is 10 kHz and 28.010 mHz.

No. The LO is a square wave. It is rich in odd harmonics. The output will consist of n*flo +/- 14.01MHz where n is 1,3,5 etc and flo is 14.00MHz. Therefore there will be output signals at 14.00+/-14.01, 42.00+/-14.01, 70.00+/-14.01 and so on.3) If the mixer is now terminated with two lines in parallel a) a capacitor in series with a 50 ohm resistor and b) an rf choke in series with a 50 ohm resistor, with the capacitor presenting a low Z at 28 mHz and high Z at 10 kHz and the rf choke the reverse, then the impedance seen at the mixer input is unchanged, still 50 ohms plus the switch on resistance. Correct?

Approximately true at the operating frequency. You would have to calculate the exact impedance at each frequency based on your chosen capacitor and inductor values.4) Now if the circuit is as in 3) but the resistor on the capacitor line is changed to 100 ohms and the resistor on the rf choke line is changed to zero ohms, what is the impedance seen at the mixer input? How does one analyze this to see what the correct answer is?

If the resistor in series with the inductor is reduced to zero, then the impedance of that leg of the circuit would approach zero ohms at low frequencies. This is different from the first scenario that you described, where the impedance of the RL leg would approach the resistor value at very low frequencies. The mathematics is a bit too complicated to describe. Your best bet to figure out how the circuit would behave, with the changes you describe, would be to use a circuit simulator that can handle switches and discrete components.The answer to 4) will help me to try an experimental redesign of the circuit.

Milt,

Thanks!

W8NUE

- You also may be interested in the following links that explain the Tayloe Mixer

www.norcalqrp.org/files/Tayloe_mixer_x3a.pdf

United States Patent: 6230000Dick K9IVB

- Thanks Dick and Milt for the helpful replies. A good simulator program should show what is happening nicely. I think I will start with the empirical approach and try to measure outputs with various configurations.