Volume 33, Number 3, March, 1999
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and the Linear Design Seminar Notes.*
On that point, let's digress for a moment to consider such a device. Think about what would happen when an ac input signal crosses zero and goes negative. Remember, the mathematical function, log
However, as the figure shows, the inverse hyperbolic sine, sinh
In general, the principal application of log amps is to measure signal However, there are some applications where a log amp is used to demodulate a signal. The figure shows a received signal that has been modulated using
Now, as the signal progresses down the gain chain, it will at some stage get so big that it will begin to clip (the term
After the signal has gone into limiting in one of the stages (this happens at the output of the third stage in the figure), the limited signal continues down the signal chain, clipping at each stage and maintaining its 1 V peak amplitude as it goes. The signal at the output of each amplifier is also fed into a full wave rectifier. The outputs of these rectifiers are summed together as shown and applied to a low pass filter, which removes the ripple of the full-wave rectified signal. Note that the contributions of the earliest stages are so small as to be negligible. This yields an output (often referred to as the "video" output), which will be a steady-state quasi-logarithmic dc output for a steady-state ac input signal. The actual devices contain innovations in circuit design that shape the gain and limiting functions to produce smooth and accurate logarithmic behavior between the decade breaks, with the limiter output sum comparable to the To understand how this signal transformation yields the log of the input signal's envelope, consider what happens if the input signal is reduced by 20 dB. As it stands in the figure, the unfiltered output of the summer is about 4 V peak (from 3 stages that are limiting and a fourth that is just about to limit). If the input signal is reduced by a factor of 10, the output of one stage at the input end of the chain will become negligible, and there will be one less stage in limiting. Because of the voltage lost from this stage, the summed output will drop to approximately 3 V. If the input signal is reduced by a further 20 dB, the summed output will drop to about 2V. So the output is changing by 1V for each factor-of-10 (20-dB) amplitude change at the input. We can describe the log amp then as having a
As the input signal drops down below about -65 dBm, the response begins to flatten out at the bottom of the device's range (at around 0.5 V, in this case). However, if the linear part of the transfer function is extrapolated until it crosses the horizontal axis (0 V theoretical output), it passes through a point called the
Slope × (P - _{in}Intercept)
For example, if the input signal is -40 dBm the output voltage will be equal to 18 mV/dB × (-40 dBm - (-93 dBm) = 0.95 V It is worth noting that an increase in the intercept's value The figure also shows plots of deviations from the ideal, i.e.,
A good test is to ground both differential inputs of the log amp. Because log amps are generally ac-coupled, you should do this by connecting the inputs to ground through coupling capacitors. Solving the problem of noise pickup generally requires some kind of filtering. This is also achieved indirectly by using a matching network at the input. A narrow-band matching network will have a filter characteristic and will also provide some gain for the wanted signal. Matching networks are discussed in more detail in datasheets for the AD8307, AD8309, and AD8313.
*Not yet released. For information, phone 1-800-262-5643 or use "Feedback" e-mail to request information.
Now take a look at the lower figure. This shows you what will happen if the frequency of the input signal is lower than the corner frequency of the output filter. As might be expected, the full wave rectified signal appears unfiltered at the output. However this situation can easily be improved by adding additional low-pass filtering at the output.
In the figure, an external 1.37-kW shunt resistor has been added. Now, the overall load resistance is reduced to around 1.25 kW. This will You may also want to take a look at the Application Note AN-405. This shows how to improve the response time of the AD606.
The degree to which the phase of the output signal changes as the input level changes is called
The table shows the correction factors that should be applied to measure the rms signal strength of a various signal types with a logarithmic amplifier which has been characterized using a
Since power in watts is equal to the rms voltage squared, divided by the load impedance, we can also write this as
^{2} / R)/1 mW)
It follows that 0 dBm occurs at 1mW, 10 dBm corresponds to 10 mW, +30dBm corresponds to to 1 W, etc. Because impedance is a component of this equation, it is always necessary to specify load impedance when talking about dBm levels. Log amps, however fundamentally respond to voltage, not to power. The input to a log amp is usually terminated with an external 50-W resistor to give an overall input impedance of approximately 50 W, as shown in the figure (the log amp has a relatively high input impedance, typically in the 300W to 1000W range). If the log amp is driven with a 200-W signal and the input is terminated in 200 W, the output voltage of the log amp will be higher compared to the same amount of power from a 50-W input signal. As a result, it is more useful to work with the
However, there is disagreement in the industry as to whether the 1-V reference is 1 V peak (i.e., amplitude) or 1 V rms. Most lab instruments (e.g., signal generators, spectrum analyzers) use 1 V rms as their reference. Based upon this, dBV readings are converted to dBm by adding 13 dB. So -13 dBV is equal to 0 dBm. As a practical matter, the industry will continue to talk about input levels to log amps in terms of dBm power levels, with the implicit assumption that it is based on a 50 W impedance, even if it is not completely correct to do so. As a result it is prudent to provide specifications in The figure shows how mV, dBV, dBm and mW relate to each other for a load impedance of 50 W If the load impedance were 20 W , for example, the V(rms), V(p-p) and dBV scales will be shifted downward relative to the dBm and mW scales. Also, the V(p-p) scale will shift relative to the V(rms) scale if the peak to rms ratio (also called crest factor) is something other than Ö2 (the peak to rms ratio of a sine wave). |