Measurement of attenuation

After building an attenuator, the attenuation can be easily and accurately measured with a DC voltage.  Voltages can measured accurately with a digital voltmeter.

To get the wanted attenuation, the attenuator must always be terminated with a 50 ohms.
During the measurement, the attenuator must be terminated with an 50 ohm resistor, that is connected to the output.
Two resistors of 100 ohm in parallel form an accurate 50 ohm resistor.
After the measurement, this 50 ohm resistor, must be removed.

PA1B    Simple measurement of attenuation

Input voltage
An attenuator built with resistors of 1/4 watt, can dissipate slightly more 250 mW.
So for an attenuator with 1/4 watt resistors the voltage may not exceed 3.5 V

Apply a voltage to the input.
Measure the voltage at the input (V1) with the digital voltmeter.
Now connect the digital voltmeter to the output
Measure the voltage at the output (V2) with digital voltmeter

Calculation of the attenuation
A (in dB) = 20 * log (V2 /V1)

Example calculation
V1 is adjusted to to 3.2 V.
V2 is measured as 0.99 V
A (dB) = 20 * log (V2 / V1) = 20 * log ( 0.99 / 3.2)
A (dB) = 20 * log (0.309) = 20 * 0.5095 = -10.19 dB   (Thus an attenuation of 10,2 dB)
These values were measured at my 10 dB attenuator.

2 opmerkingen:

  1. got it, it's DC voltage
    73's de OE1MWW

    1. Hello Wolfgang, Yes, a DC voltage can be measured accurately with a digital meter. But an AC voltage can also be used. If the values are measured in the same way, both Vtt or both RMS, the ratio will be the same. Eg. 2Vtt/200mVtt = 10
      Thank you for the compliment in your first comment. I am very pleased. 73, Bert