MMR40 First Mixer Analysis
MMR40 First Mixer Analysis
The first Mixer is used to develop the 10 MHz IF in receive, or the 7 MHz final frequency in transmit.
On Transmit, the PTO output is mixed with the 10 Mhz Transmit IF signal to create the 7 MHz output signal.
On Receive, the PTO output is mixed with the incoming 7mhz Receive signal to create the 10 MHz receive intermediate frequency.
The first mixer in the MMR40 is a SA612 double-balanced mixer. The datasheet can be downloaded from the Phillips Site.
- Pin 1 and Pin 2 are signal inputs
- Pin 3 is ground
- Pin 4 and Pin 5 are mixer outputs
- Pin 6 and Pin 7 are the Oscillator inputs
- Pin 8 is VCC (4.5 to 8 vdc)
The 5 volts VCC voltage is supplied by U2, a 7805 3-terminal regulator. U2 also supplies the other SA612 mixer further down the line.
On receive, the 40-meter input signal is coupled through the DC blocking capacitor C16 to pin 1. The 0.001uf capacitor has about 22 ohms impedance at 7 MHz. The second input at pin 2 is effectively grounded via analog switch U7c and the 0.1uf (about 0.22 ohms impedance at 7 mhz) capacitor C40 on pin 4 of U7.
The PTO oscillator is coupled via dc blocking capacitor C6 to pin 6. C6 and C23 form a voltage divider that allows us to apply about 17% of the signal from the PTO to U6 pin 6 (maybe 15% when you take into account the 20K impedance of U6-6 in parallel with the 22pf cap. The datasheet states that the oscillator should be between 200mv and 300mv.
When the 40-meter (~ 7 MHz) input mixes with the PTO oscillator (~ 3 MHz), the mixer chip outputs the difference frequency (~ 4 MHz) and the sum frequency (~ 10 MHz). The sum frequency is the one that ends up getting used as the 10 MHz receive intermediate frequency. This signal is output from pin 4 to the analog switch U7b, which eventually feeds the 10mhz crystal filter to remove the unwanted frequencies.
On transmit, the 10mhz transmit IF signal out of the crystal filter is fed to pin 2 of the mixer from U7b, pin 12. After mixing with the 3 MHz PTO, the resulting sum and difference frequencies are output from pin 5. Transistor Q6 acts as a buffer. The relatively lower impedance of its emitter circuit matches the primary of T3 a little better. Tuned circuits at T3 and T2 select only the 40-meter frequencies for further amplification for our transmit signal.
Here is the PTO Oscillator signal feeding C6:
And here is the input on U6 pin 6:
It measures about 12% of the signal available at the other side of
C6. I attribute the difference to probe loading, and the fact that I
didn't take the ~ 20K impedance at pin 6 into account in the
calculation above. It's close enough for our purposes.
So, 0.8vpp / 2 = 0.4v pk
0.4v pk * 0.707 = 282mv rms, right in the range specified for input to U6-6.
The input Rx signal at pin 1 and the mixer Rx output at pin 4 are at such a low level, that I cannot measure them with my test equipment.
For Transmit, here is the 10 MHz IF signal on pin 2:
This is a pretty high impedance point. It looks to be at about 100K ohms on the base of Q6. This is based on my ltspice model, which shows about 32 ua of base current at 3.1v. So the scope probe may be loading it down a little.
Migrated Comments:
2 Responses to “MMR40 First Mixer Analysis”
August 1st, 2012 at 8:20 am
Hello,
The SA612 datasheet states that 0.2 to 0.3 V peak to peak should be applied to pin 6. You say you have 0.8 V peak to peak at pin 6 in your discussion above? This is much more than the SA612 requires. Hope to hear from you. Camil.
August 1st, 2012 at 6:35 pm
You are right, Camil. I saw some references that just said 200 – 300 mv, and did not specify peak-to-peak. So when I did my measurements, and applied the formulas, it looked like it all fit the design.
At any rate, the receiver is acceptable quality, and overloading is handled by turning down gain, so I guess I’ll live with the designer’s choice on this one.
73 de WB0SMX