In this post we present a tool to convert Peak-to-Peak Voltage to Power in both dBm and Watt. The post also includes background information, example calculations and how the formula is derived.
Contents
Calculator
to use this enter:
- Vpeak-to-peak or Vpp and select the appropriate units Volt/millivolt/microvolt
- Impedance (Z) specified in ohm (Ω)
Vpp to dBm Formula
The formula to convert peak-to-peak voltage to an equivalent Power level in dBm is
PdBm = 10*Log10((1/Z)*(Vpp/(2√2))2)
As well, the peak voltage is given by
Vpp= 2*Vp
for a sine wave with DC = 0 Volt.
and the RMS voltage,
Vpp = (2√2)*VRMS
Vpp to dBm formula derivation
To do this let’s consider two scenarios 1) sine waves and 2) other waveforms.
Sine Waves
The power in dBm is related to the root-mean-square voltage level. This is usually referred to as voltage. In the above equations, we have considered a sinusoidal waveform shown in the picture below.
The peak to peak voltage indicated by (2) in the picture of a sine wave above. If the maximum and minimum voltage are Vp and -Vp , respectively. The peak-to-peak voltage Vpp = Vp – (-Vp) = 2*Vp
This is first converted to the RMS equivalent: VRMS = Vpp/(2√2)
The power is related to VRMS as per the following equation:
PdBm = 10*Log10((1/Z)*(VRMS)2)
A sinusoid is the most common signal type used in RF systems. For example the default output from a signal generator is typically a sine wave.
Other Waveforms
🤔 What about non sinusoidal waveforms? How to convert the peak-to-peak voltage of a square wave or triangle wave to its dBm equivalent?
In this case we have to find the RMS voltage from the peak voltage. The equations for a sine wave do not apply in general.
Then use this calculator to convert VRMS to dBm. It uses the same formula:
PdBm = 10*Log10((1/Z)*(VRMS)2)
Additionally in the case of arbitrary waveforms, the RMS voltage can be calculated from any number of time domain data points.
Where is this calculator used?
In many applications there is a requirement to convert Vpp to power in dBm. For example, the output level of an RF signal generator is typically specified in dBm. This might be used to drive an analog circuit with input limits specified in terms of peak-to-peak voltage. Exceeding this limit can damage the circuit so it’s important to keep dBm output level within the stated limit.
Example Calculations
Power output from analog signal generator
A clock reference circuit might state that the minimum level requirement is 1 Vpp. Use this tool to calculate the output power from the signal generator in dBm. Make sure you specify the input impedance which defaults to 50 ohm in the calculator above.
For a 50 Ω system, if the desired input level is 1 Volt peak-to-peak, the tool is used to determine: signal generator output should be set to +4 dBm. It is assumed that a CW signal generator is to be used and not a square wave for instance. A square wave with 1 VPP output will have higher power relative to a sine wave with the same peak to peak level.
Signal into an Analog-to-Digital converter
This ADC has an input range of 1.8 Vpp. Input signal levels in excess of this range will damage the ADC. The analog Input as seen on the plot is actually -1 dBFS (which means it is 1 dB lower than the Full scale value)
Using the calculator on this page, a Continuous Wave (CW) RF signal generator would have to be set to a level of +3 dBm to get this value with an input impedance to the ADC equal to 200 ohm.
Convert Differential Vpp to dBm
Vpp represents the difference between the min and max voltage of a single line. In the case of a differential pair the difference between two identical lines is given by VDiff = 2*Vpp
In this case, the dBm value can be calculated using the following relationship:
PdBm = 10*Log10((1/Z)*(VDiff/(2√2))2)
Enter
- Vpp for a single line
- Impedance Z in ohm (Ω)
Related Calculators
The dBFS to dBm calculator finds the highest spur level in dBm from the SFDR and input peak to peak voltage specification.