To: File 430mpxr1.wp6.wpd

From: Jon Hagen

Date: 11-24-99 REVISED 7-1-03

Subject: 430 Monoplexer - Theory of Operation

/4 stub for diode dc return

The monoplexer is shown above in simplified form. This device is put at the input to the receiver. When the diodes are forward biased, their resulting low impedance (about 1/10 ohm) shorts the inner and outer conductors of the coaxial feed line. This short reflects the leakage power from the turnstile (about 1 kW) to protect the receiver from burnout.

There are two diode "stations" along a 50-ohm line, separated by a distance D. Each station has 4 diodes to provide a better short. If a single diode station provided, say, 50 dB of isolation, then adding a second station a quarter wave from the first would increase the isolation to 2x50 + 6 = 116 dB. A quarter-wave separation provides the maximum isolation.

For receiving, the diodes are reverse biased. Their low resistance

disappears but they still present a capacitance of about 2.1pF.

Four diodes in parallel increase this to 8.4 pF which, at 430 MHz

is a reactance of 44 ohms. This reactance, shunted across the

50-ohm produces a reflection coefficient shown as point

A in the Smith Chart. Moving along the transmission line from

point A, we arrive at point B (the mirror image of point A about

the x-axis) where an identical station will bring us back to the center

of the chart, i.e. eliminate the reflection. If the capacitance had been very small, point A would

have been very near the y-axis. The separation on the chart would have been 180 degrees and

the physical separation would have been 90 degrees - the separation that gives the best isolation when the diodes are in the transmit state. No compromise is made; the separation is determined by the angle between A and B so that there will be no mismatch on receive. For our 8-diode monoplexer, the separation is only 31 degrees.

We can analyze the circuit to determine the isolation as a function of the diode ON resistance

and the distance between the diode stations. The equivalent circuit is shown below at the left.

A simplified version is shown at the right.

The source is made to be 2 Volts so that the output voltage, Vo, will be 1 Volt when the resistors R are infinite in value (no attenuation). Since the R will be much less than 50, the circuitry to the left of the transmission line has a Thévinen equivalent voltage source of 2R/50 and an impedance of R. This Thévinen equivalent is shown in the simplified circuit. Again, since R<<50, we can ignore the 50 ohm load (the receiver) in calculating the output voltage. The output voltage can be calculated as follows: The cable and the resistor at the right-hand end present an impedance given by Z = 50(R +50 j tan )/ (50 +R j tan ). The current flowing into Z is given by I= (2R/50)/(Z+R). The power delivered to Z is given by Pwr = |I|2 Re(Z). This power is equal to |Vo/R|2 Re(R). Solving for |Vo|2 we find

|Vo|2 = |(2R/50)/(Z+R)|2 |R|2 Re(Z)/Re(R)

where Z = 50(R +50 j tan )/ (50 +R j tan ). In this formula, the value of R can be complex, to include diode reactance. The attenuation is given simply by 10log(|Vo|2) and is plotted below in dB vs. diode station separation for three values of R.

When the diode impedance is taken as 0.1/4 + j0 (four diodes of 0.1 +j0 ohms in parallel, as at each end of our 8-diode monoplexers), the theoretical isolation reaches 120dB. If we add 0.6 ohms of reactance to each diode, corresponding to the 630pF bypass capacitors (metal segment plate with mica insulation), the isolation drops to 88dB. The measured value, however, is about 66 dB. As shown in the third curve, this is the value we would expect to measure if each diode had a reactance of 2.3 ohms. This corresponds to an inductance of only 0.85nH, which could be attributable to the 1/4" length of the stud-mounted diodes. It could also be attributable to the fact that the diodes don't directly contact the center conductor of the coaxial line. Instead, at each diode station, the center conductor becomes a thin disc, i.e. a radial transmission line. The diodes contact the disk at a radius that is about twice that of the center conductor. This section of radial transmission line might account for the apparent reactance.

We have no documents describing the original monoplexer design, which was done at Sylvania in Waltham Ma. Did the designer take into account the radial transmission line and attempt to resonate out its impedance by carefully selecting the thickness of the mica insulation in the bypass capacitors?

Monoplexer Driver

The driver circuit applies a positive voltage (+135 is used) to put the monoplexer into its receive state, and a negative voltage to put the monoplexer into its protect state with a current of 500mA in each diode. The positive voltage need not supply current, except while the monoplexer is being pulled up. But, since this pull-up must be fast enough (microseconds) if short range targets are to be detected, the pull-up current can be appreciable, both to drain the stored charge from the diodes and to discharge the 1000pF feedthrough decoupling capacitors (one per diode). Thus a substantial pull-up transistor is needed. The negative voltage needn't be applied quickly (except that extremely slow application would require that the IPP be increased) but, since considerable current is required from the negative supply (the PIN diode current), the pull down transistor must also be substantial.