Since attenuators are specified as giving an attenuation in db, is there a difference in calculating the resistor values when dealing with voltage and power attenuators? This is being asked since db for voltage equals 20log Vo/Vi and the db for power equals 10log Po/Pi. If there is a difference, how does one know if the attenuator is specified for voltage or power? I would like to know the formulas for calculating the resistor values.

 

Remember that if you double the voltage into a fixed load the power is four times. (P= V^2/R)

Les.

 

since db for voltage equals 20log Vo/Vi and the db for power equals 10log Po/Pi.

The are the same.
A dB is a dB.
The difference in calculation is because power is proportional to the square of the voltage.
So if an attenuator gives 10dB of attenuation it will reduce the output power by a factor of 10 and the voltage by a factor of the square-root of 10 = 3.162, assuming the load impedance matches the attenuation characteristic impedance.

 

Interestingly. Voltage is dB. But power is usually dBm.

 

Hello,

Voltage is dB. But power is usually dBm.

Not always. dB is a ratio.
dBm is also a ratio, but in relation to 1 mW.

There are many more relations in dB:
https://www.softwright.com/faq/engineering/FIELD INTENSITY UNITS.html

http://www.jimprice.com/prosound/db.htm

Bertus