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ECE 166 Microwave Circuits

Homework #5 Due Thursday November 16, 2006

(Temperature and Noise Figure)

Problem 1: (Noise Diodes)

Noise diodes are basically avalanche diodes which are either biased at 0 V or at -28 V (for the breakdown +

Avalanche mode). At 0 V, they appear as nearly an open circuit (Rs=3 , Cj=0.5 pF, Rj=inf), and at breakdown,

they appear as a short circuit (Rs=3 , Cj=0.5 pF and Rj=5 ). The equivalent noise temperature at the diode

terminals is 300 K at 0 V, and 1,000,000 K when biased at -28 V (they produce a lot of noise!).

In order to get a matched diode with reasonable amount of noise, a 20 dB attenuator matched to 50 ohm is placed in

series with the diode. Calculate the impedance (and reflection coefficient) and the noise temperature seen looking at

the diode at 0 V and -28 V, at a frequency of 3 GHz. This is how we build noise diodes!

Problem 2: (MM-wave Cameras)

As mentioned in class, we all transmit P=KT W/Hz at microwave and mm-wave frequencies (we actually transmit

much more in the infrared, but at 1-200 GHz, it is P=KT). If we build a detector which can measure this noise, then

we can build a microwave camera. However, if the frequency is 3 or 10 GHz, then the wavelength is 10 cm (or 3

cm) and cameras can only give a resolution of one-wavelength at their focal plane (actually 0.5

with someprocessing, but never more) so it is going to be quite a bad camera. However, if we use 94 GHz, then the

wavelength is 3.2 mm and we can have quite a good resolution (2-3 mm). Since mm-wave frequencies (80-120

GHz) can pass through cloth quite easily, we can see below clothing and to the human skin and we can determine

(or see) if someone is carrying contraband material. If you do not believe it, look at the picture below taken with

a mm-wave camera and it clearly shows a concealed plastic/ceramic gun under a persons clothes. The reason we

can see it is that it is at a different temperature than the body (which is at 300K) and our receiver is so sensitive

that it can measure different powers (or temperatures) accurately!

Let us analyze if we can measure a small temperature difference or not (simple analysis). The antenna receives the

signal from the focused spot, amplifies it using a 94 GHz LNA (these are not cheap!), passes it by a bandpass filter,

a diode detector which transfers the RF energy in a DC level (see below), and then a low-noise op-amp.

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The amplifier chain (all three amplifiers) has a gain of 53 dB at 92-97 GHz and a noise figure of 7 dB. The

bandpass filter bandwidth is 5 GHz with 3 dB loss. The diode detector produces 200 V for every Watt it receives

and we call this theDiode Responsivity (200 V/W). Of course, the diode will receive nW-uW and will produce uV-

mV of voltage, but Responsivity is quoted in V/W. Finally, there is the op-amp which has a gain of 100 (simple low

frequency gain).

a) For Ta=300 K (spot focused on body), calculate the RF power at the diode andthe DC voltage at the output of the op-amp.

b) For Ta= 250 K (spot focused on the ceramic material), calculate the RF powerat the diode and the DC voltage at the output of the op-amp.

c) For Ta=299 K (spot focused mostly on body and a small part of the spot is onthe ceramic material), calculate the RF power at the diode and the DC voltage

at the output of the op-amp.

d) Assume that the output of the op-amp has a total noise of 0.1 mV over theintegration bandwidth, calculate the minimum resolving temperature that can

be seen by the mm-wave camera. That is, calculate the Ta which will result in

0.1 mV at the output of the op-amp.

Now, you can see that mm-wave cameras are a reality. Of course, they do not have a

lot of resolution compared to infrared or visible cameras, but they can detect throughthick clothing, jackets, etc. and they can also detect ceramic materials, etc. basically,

they detect everything which is at a different temperature than the background!

Problem 3: (Receive systems and effect of transmit noise)

a) Calculate the equivalent noise figure and noise temperature of the following receive network (neglect thePA and the transmit path for now). Assume that the receive frequency is 1960 MHz.

b) For Ta=300K, calculate the total noise power at the output of the receive network in a 1 MHz bandwidth.c) For full-duplex systems, that is when the power amplifier is always On even when we are receiving very

weak signals, we have a problem of the broadband noise from the transmit path. The power amplifiertransmit signal at 1900 MHz can be filtered out using the diplexers and the other filters, but the power

amplifier also sends a lot of broadband noise including noise at 1960 MHz! This noise just passes by the

receive system just as a signal, and the only circuit which can filter it out is the isolation between port 3 and

port 2 in the diplexer, and in this case, it is 50 dB. Calculate the noise power due to the transmit PA and

compare it with the noise power in (b).