# Decibels Instrumentations   Whatsapp  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.

## Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250