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Transmission

To understand power in RF transmission, we need to go back to high school physics, and do a little thought experiment.

Imagine a Dyson Sphere, an artificial hollow sphere of matter around a star. The energy (in Joules) that the encompassed star provides spreads out uniformly in all directions. The Luminosity, or Power the star puts out is based on Joules/second, or Watts. The Brightness, or power received at the surface of the Dyson-sphere received is based on the distance from the star and the surface of the inner surface of the sphere (and is measured in Watts/m^2). Suppose the Dyson-sphere has a radius of r, then the surface area of that sphere is is 4{pi}r^2.

If the total output power (or Luminosity) of the star is L Watts, and the surface area of that sphere is is 4{pi}r^2, So the energy flowing through each square meter of the sphere every second is:

B = L/{4{pi}r^2}

We call that the star’s brightness, B, and is measured in Watts/meter^2. This is known as a form of the “Inverse Square Law”, as the relationship is 1/{r^2} not 1/{r}

If the distance, r, is doubled then the brightness, B, (or received brightness) decreases by a factor of 1/{2^2} or a quarter the brightness.

If the distance, r, is tripled then the brightness, B, (or received brightness) decreases by a factor of 1/{2^3} or a ninth the brightness.

If you still are not sure, have a look at the video.

You can also look at this from the opposite way - if you need to read a book on the surface of the Dyson-sphere, what kind of output power do you need from the star, to get the appropriate amount of brightness? To double the brightness, you need to increase the power of the star by 4 times.

Free Space Path Loss

As energy is transmitted by any transceiver it becomes dispersed across space and absorbed by the surrounding environment. This is typically referred to as path loss, or loss or attenuation of energy over a distance. Although generally not accurate in most situations, the most common model for path loss is free-space path loss (FSPL). FSPL is the energy loss in an obstacle-free, line-of-sight path through free space. FSPL is proportional to the square of the distance:

L=({4 pi d}/lambda)^2 = ({4 pi d f}/c)^2,

where d is the distance (meters) from the transmitter, f is the signal or carrier frequency (hertz), lambda is the wavelength (meters), and c is the speed of light (meters per second). The equation above also assumes the transmitter is isotropic and is measured from the far field.

This equation is typically provided in dB which can be calculated as:

L_{dB}=10log_10(({4 pi d f}/c)^2) = 20log_10(({4 pi d f}/c)) = 20log_10(d) + 20log_10(f) + 20log_10({4 pi}/c).

Let us get some perspective on this equation from looking at the idea of receiver sensitivity (RS), which is the minimum power a signal can be received at and still recovered. For WiFi 802.11ac, this requirement is -93 dBm[FIX ME and ref]. Operating at 2.4 GHz and examining distance as a function of transmit power we can observe the limited range of Pluto alone in orange in the figure below. However, by adding a 20 dBm amplifier to Pluto we can substantially increase its range.

Practical things

Log and dB

The power PdBm in dBm is equal to 10 times the base 10 logarithm of the power PmW in milliwatts (mW) divided by 1 milliwatt (mW):

P_{dBm} = 10 log_{10}({P_{mW}/{1mW}})

Power in mW Power in dBm
0.1 mW -10 dBm
1 mW 0 dBm
10 mW 10 dBm
100 mW 20 dBm

A doubling of output power (from 1mW to 2mW) is only +3dBm. A gain of +20dBm, is output power increasing by a factor of 100 times in mW.

Peak to Average

P1dB

Normally, there is a direct relationship between input and output of an amplifier. As the input increases, the output increases the same amount. However, once the input reaches a certain level, the output signal begins to soft-limit, or compress. A parameter of interest here is the 1 dB compression point. This is the point where the output signal is compressed 1dB from an ideal input/output transfer function. This is shown in the figure within the region where the ideal slope = 1 line becomes dotted, and the actual response exhibits compression (solid).

Typically, the 1 dB compression point is a function of frequency, and as one would expect, the distortion is worse at higher frequencies.

Noise

Distortion

Power Supply Limits

The relationship between dBm (power with respect to 1mW) and Volts peak-peak for a sinusoidal signal in a 50-Ohm systems is:

dBm mW Voltsrms Voltspp
-10 0.1 70.711 mV 199.970 mV
0 1 223.607 mV 632.360 mV
5 3.16 0.398 V 1.125 V
10 10 0.707 V 2.000 V
15 31.6 1.257 V 3.556 V
20 100 2.236 V 6.324 V

The question is, how do we get +20dBm (6.324V peak-peak) out of a system, when the power supply is limited to 5V? The trick is in how we connect the output stage. The output stage (RFOUT) is connected to Vcc through the inductor L1. From a DC perspective, inductors become short circuits, and RFOUT is setting at 5.0V, allowing a 10Vpeak-peak swing from the amplifier. This is also why it is AC-coupled by the output capacitor before it attaches to the antenna.

Power Dissipation

Supply Current in the ADL5606 datasheet is specified for a max on 390mA at 5.0V (or 1.95W). Although the ADL5606 is packaged in a thermally efficient 4 mm × 4 mm, 16-lead LFCSP (thermal resistance from junction to air (θJA) is 52.9°C/W. This 1.95W dissipation times the θJA is a temperature increase of 103.2 °C. Given a maximum junction temperature of 150°C, we either only use the part in sub 46.845°C enviornments, or work harder on spreading the heat out of the package, onto the PCB.

Measurements

university/tools/pluto/users/amp.1524607213.txt.gz · Last modified: 25 Apr 2018 00:00 by Robin Getz