Relation between dBm and dBV with 50 Ohm Impedance

Relationship Between dBm and dBV with 50 Ohm Impedance

To derive the relationship between dBm and dBV with a 50-ohm impedance, let's start by understanding what these units represent:

Definitions

dBm: Power in decibels relative to 1 milliwatt (mW).

P_dBm = 10 log₁₀(P / 1 mW)

dBV: Voltage in decibels relative to 1 volt (V).

V_dBV = 20 log₁₀(V / 1 V)

Relationship Between Power and Voltage

For a resistive load (here, R = 50 Ω), the relationship between power and voltage is:

P = V² / R

Rearranging to express V in terms of P:

V = √(P × R)

Substituting Into dBV Equation

Substituting the expression for voltage into the formula for dBV:

V_dBV = 20 log₁₀(√(P × R))

Simplifying:

V_dBV = 10 log₁₀(P × R)

Using the expression for P_dBm, where:

P = 10^{P_dBm / 10} × 10⁻³ (in watts)

Substitute P into the equation:

V_dBV = 10 log₁₀(10^{P_dBm / 10} × 10⁻³ × R)

Splitting the logarithm:

V_dBV = P_dBm + 10 log₁₀(10⁻³ × R)

For 50 Ohm Impedance

Substitute R = 50 Ω:

10 log₁₀(10⁻³ × 50) = 10 log₁₀(0.05) ≈ -13.01

Thus, the relationship becomes:

V_dBV = P_dBm - 13.01

Final Formula

The relationship between dBm and dBV for a 50-ohm impedance is:

V_dBV = P_dBm - 13.01