# DBs Units

## dB

A logarithmic expression of the ratio of two power levels: $10 \times \log_{10}\left(\frac{P1}{P2}\right)$

P1P2$$\frac{P1}{P2}$$dB
1110
2123
1011010
0.510.5-3
7.21.64.56.5
100110020
10001100030

Since a dB is a ratio, it is not an absolute quantity, like a watt, etc.

To make it an absolute quantity, we must specify a reference.

## dB suffix

• dBm -> reference is milliwatt (mW)
• dbµ -> reference is microwatt (µW)
• dBmV -> reference is millivolt (mV)

The suffix can turn dB into an absolute quantity dbM.

### Example

10 MHz sinewave at 950mVpp into $$50\Omega$$.

$Vpp = 950mV$ $Vrms = \frac{Vpp}{2 \times \sqrt{2}} = \frac{950mV}{2 \times \sqrt{2}} = 336mV_{rms}$

Therefore, the power in the $$50\Omega$$ load is:

$P = \frac{\left(Vrms\right)^2}{R} = \frac{0.336^2}{50} = 2.256mW$

To express this in dBm, the reference is 1mW:

$dBm = 10\times\log_{10}\left(\frac{2.256mW}{1mW}\right) = 3.53dBm$

In dBµ, the reference is 1µW:

$dBµ = 10\times\log\left(\frac{2.256 \times 10^{-3}}{1 \times 10^{-6}}\right) = 33.53dBµ$

## Voltage ratios

The dB relates to power, so we have to calculate power.

$10\times\log\left[\frac{\frac{V1^2}{R}}{\frac{V2^2}{R}}\right]$

$10\times\log\left[\frac{V1^2}{R}\times\frac{R}{V2^2}\right]$

$10\times\log\left[\frac{V1^2}{V2^2}\right]$

$10\times\log\left[\left(\frac{V1}{V2}\right)^2\right]$

$2\times10\times\log\left[\frac{V1}{V2}\right]$

$20\times\log\left[\frac{V1}{V2}\right]$

Thus, our 950mVpp can be expressed in dBmv

$V = \sqrt{P \times R}$

$dBmv = 20\times\log\left(\frac{336mV}{1mV}\right) = 50.5dBmV$

• dBc means it is dB relative to some "carrier" power leev

• very common in RF applications