### Table of Contents

# HF dummy load power meter theory

## Summary formulas

Formulas for a 50 Ω dummy load power meter.

- `P` is dissipated power in Watts
- `Uo` is measured DC output voltage in Volts
- `Uk` is the knee voltage of the diode(s) used in Volts
- `Ur` is maximum reverse voltage diode(s) in Volts

`P = ((Uo + Uk)/10)^2`

`Uo = 10sqrtP - Uk`

`Ur = 20*sqrt(P) + Uk`

`Pmax = ((Ur - Uk)/20)^2`

For high values of `Uo` and `Ur`, the influence of `Uk` is negligible and `Uk` may be omitted, simplifying the formulas.

`P = ((Uo)/10)^2`

`Uo = 10sqrtP`

`Ur = 20*sqrt(P)`

`Pmax = ((Ur)/20)^2`

## Introduction

The Dummy load power meter consists of a dummy load with the same impedance as the source, a single phase rectifier with smoothing and a measuring instrument to accurately measure the DC output voltage.

### Dummy load power meter

This page contains the theory and derivation of the formulas for this measurement principle. On HF dummy load power meter homebrew you will find dummy load power meters I built, including schematics and practical instructions for component selection.

### Basic scheme dummy load power meter

## Preconditions

Important boundary conditions for proper operation.

- The dummy load used has an almost perfect SWR of 1:1 at the frequency used, so the entire power supplied by the transceiver is dissipated in the dummy load. This eliminates the need to take reflected waves into account.
- The power absorbed in the dummy load is almost equal to the power delivered by the transceiver, provided the coaxial cable used to connect the dummy load has little loss. This is the case when using a short coaxial cable and coax suitable for the frequency used.
- The DC voltage from the rectifier is an accurate measure of the power absorbed in the dummy load. This DC voltage can be accurately measured with a high-impedance digital multimeter, without the meter noticeably stressing the rectifier.

## Limit values

Important boundary conditions for the practical design of the dummy load power meter.

### Maximum measurable power

The maximum measurable power is determined by:

- the maximum allowable power of dummy load used
- the thermal resistance of the cooling fin used
- the maximum repetitive peak reverse voltage of the diode(s) used in the rectifier

### Minimum measurable power

The minimum still reasonably measurable power is determined by:

- the knee voltage of the diode(s) used
- the number of diodes used

### Frequency range

The frequency range of this measurement method is determined by:

- the maximum frequency range of the dummy load
- the maximum frequency range of the diode(s) used
- The HF connector
- a construction with short connections

## Derivation formulae

The dummy load power meter with only phase rectification gives an output voltage that is a function of power. The rectifier consists of a diode with sufficiently high switching speed with a capacitor behind it. The output of the circuit is loaded at high impedance, making the output voltage ripple negligible. Below is the derivation of the formulas to calculate the power from the output voltage.

### Definitions

- `P` is dissipated power in Watts
- `Uo` is measured DC output voltage in volts
- `Uk` is the knee voltage of the used diode(s) in volts
- `Ueff` is effective value in volts
- `Up` is peak value in volts
- `Ur` is maximum reverse voltage diode

### Power measurement with an ideal diode

The following formula gives the power generated in a dummy load.

`P = Ueff * Ieff`

'eff' indicates that this is the effective value of the AC voltage and current, also called RMS. For a definition of RMS value see: Root Mean Square

We replace `Ieff` with:

`Ieff = (Ueff)/R`

Which gives the following formula.

`P = (Ueff)^2 /R`

Using an ideal diode, DC output voltage `Uu` corresponds to the peak value of the AC input voltage. We replace `Ueff` with the peak value. For a definition of peak value see: Amplitude

`Ueff = (Up)/sqrt2`

Which gives the following formula.

`P = (Up)^2/((sqrt2)^2*R) = (Up)^2/(2*R)`

We use a 50 Ohm dummy load in this design, so we may substitute R for 50, this yields the following formula.

`P = (Up)^2/100`

To easily calculate the power, I bring the partial factor `100` under the square sign.

`P = ((Up)/10)^2`

### Power measurement with a non-ideal diode

In practice, a diode is not ideal but has a knee voltage. This reduces the output voltage relative to the calculated peak voltage. For output voltage `Uo`, the following formula applies.

`Uo = Up - Uk`

We can transform this formula to.

`Up = Uo + Uk`

We replace `Up` in the power formula with this formula, yielding the following formula.

`P = ((Uo + Uk)/10)^2`

With this formula, after measuring `Uo`, we can accurately calculate the power output in the dummy load.

Note: At higher values of `Uo`, `Uk` plays an increasingly smaller role and `Uk` can be neglected.

### Calculating Uo at a desired power

We can transform the above formula to calculate `Uo` at a desired power. Useful for checking or calibrating other meters or adjusting the output power of a transceiver.

`((Uo + Uk)/10)^2 = P`

Both sides take the square root.

`((Uo + Uk)/10) = sqrtP`

Multiply both sides by 10

`Uo + Uk = 10sqrtP`

Bring `Uk` to the other side of the is equals sign

`Uo = 10sqrtP - Uk`

### The maximal diode reverse voltage

The maximum reverse voltage across the diode is reached at the negative peak of the HF input sine wave. The maximum voltage is equal to the maximum output voltage plus the voltage of the negative peak.

`Ur = Uo + Up`

We replace `Uo` with

`Up = Uo - Uk` ⇒ `Uo = Up +Uk`

Which gives the following formula

`Ur = 2Up + Uk`

We replace `Up` with

`Up = sqrt(2)*Ueff `

Which gives the following formula

`Ur = 2*sqrt(2)*Ueff + Uk`

We replace `Ueff` with the previously derived formula

`P = (Ueff)^2 /R` ⇒ `Ueff = sqrt(P*R)`

Which gives the following formula

`Ur = 2*sqrt(2)*sqrt(P*R) + Uk`

For R = 50 Ohm

`Ur = 2*sqrt(2)*sqrt(P*50) + Uk`

We simplify the formula in a number of steps

`Ur = 2*sqrt(2)*sqrt(50)*sqrt(P) + Uk`

`Ur = 2*sqrt(100)*sqrt(P) + Uk`

`Ur = 20*sqrt(P) + Uk`

For powers above 10 watts, the influence of Uk is negligible and we can simplify the formula to:

`Ur = 20*sqrt(P)`

Using this formula, we can work out what the minimum Ur of the diode in use should be based on the maximum power to be measured.

To calculate `P` at a given `Ur`

`sqrt(P) = ((Ur - Uk)/20)`

`P = ((Ur - Uk)/20)^2`

For a high value of Ur, the influence of Uk is negligible and we can simplify the formula to:

`P = ((Ur)/20)^2`

Using these formulas, we can work out what the maximum allowable measurable power is at the give `Ur` of the diode used. When using multiple diodes, multiply the `Ur` by the number of diodes.

### Using several diodes in series

Several diodes can be connected in series to increase the total Ur, Use diodes of the same type and from the same production batch, the properties of the didoes are then close to each other so that the total the reverse voltage is distributed proportionally among the diodes used. It is wise to keep a margin between the calculated- and the sum of the diode reverse voltage here.