Decibels Instrumentations

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A number of modern VOMs and DMMs have a dB scale designed to provide an indication of power ratios referenced to a standard level of $ 1 \mathrm{~mW} $ at $ 600 \Omega $. Since the reading is accurate only if the load has a characteristic impedance of $ 600 \Omega $, the $ 1-\mathrm{mW}, 600 $ reference level is normally printed somewhere on the face of the meter, as shown in Fig. 1. The $ \mathrm{dB} $ scale is usually calibrated to the lowest ac scale of the meter.
Fig. 1:Defining the relationship between a $ d B $ scale referenced to $ 1 \mathrm{~mW}, 600 \Omega $ and a $3-V-rms$ voltage scale.
In other words, when making the $ \mathrm{dB} $ measurement, choose the lowest ac voltage scale, but read the $ \mathrm{dB} $ scale. If a higher voltage scale is chosen, a correction factor must be employed that is sometimes printed on the face of the meter but always available in the meter manual. If the impedance is other than $ 600 \Omega $ or not purely resistive, other correction factors must be used that are normally included in the meter manual. Using the basic power equation $$ P=V^{2} / R $$ will reveal that $ 1 \mathrm{~mW} $ across a $ 600-\Omega $ load is the same as applying $ 0.775 $ $$ \sqrt{(1 \mathrm{~mW})(600 \Omega)}=0.775 \mathrm{~V}$$ The result is that an analog display will have $ 0 \mathrm{~dB} $ defining the reference point of $ 1 \mathrm{~mW},$ $$\mathrm{~dB}=10 \log _{10} P_{2} / P_{1}= 10 \log _{10}(1 \mathrm{~mW} / 1 \mathrm{~mW}(\mathrm{ref})=0 \mathrm{~dB} $$ and $ 0.775 \mathrm{~V} $ rms on the same pointer projection, as shown in Fig. 1.
A voltage of $ 2.5 \mathrm{~V} $ across a $ 600-\Omega $ load would result in a $ \mathrm{dB} $ level of $$ \mathrm{dB}=20 \log _{10} V_{2} / V_{1}=20 \log _{10} 2.5 \mathrm{~V} / 0.775=10.17 \mathrm{~dB},$$ resulting in $ 2.5 \mathrm{~V} $ and $ 10.17 \mathrm{~dB} $ appearing along the same pointer projection.
A voltage of less than $ 0.775 \mathrm{~V} $, such as $ 0.5 \mathrm{~V} $, will result in a $ \mathrm{dB} $ level of $$ \mathrm{dB}=20 \log _{10} V_{2} / V_{1}=20 \log _{10} 0.5 \mathrm{~V} / 0.775 \mathrm{~V}=-3.8 \mathrm{~dB} ,$$ as is also shown on the scale of Fig. $1 $. Although a reading of $ 10 \mathrm{~dB} $ will reveal that the power level is 10 times the reference, don't assume that a reading of $ 5 \mathrm{~dB} $ means that the output level is $ 5 \mathrm{~mW} $. The $ 10: 1 $ ratio is a special one in logarithmic circles. For the $ 5-\mathrm{dB} $ level, the power level must be found using the antilogarithm (3.126), which reveals that the power level associated with $ 5 \mathrm{~dB} $ is about in the manual for such conversions.

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