When input and output impedances are different, why is 20log(voltage-out/voltage-in) still valid?

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kj4saf

Joined Oct 16, 2011
2
In accepting as definition: dB=10log(Power-out/Power-in), we can derive an equivalent expression for voltage as vdB=20log(voltage-out/voltage-in). Indeed, it is always(?) seen that way. But, in order to derive the dB voltage expression from the dB power expression, one must(?) assume resistance remains constant. But in the case of amplifier gain, for example, there is no requirement that input and output impedances are the same. Is the expression invalidated in the case where input and output impedances don't remain the same?
 

ronsimpson

Joined Oct 7, 2019
2,054
For RF amplifiers it is common for the impedance to be 50 ohms.
For audio the input might be 100k and output 4 ohms. But who measures input power?
For audio a speaker's impedance changes with frequency so there is another layer of complications.

For power you should not just measure voltage but power.
dB=20log(voltage-out/voltage-in)
I think this is a simple version of the formula that assumes you know the impedance(s) and they are the same.
You can use a more complicated formula that has Zin and Zout inserted.
 

Papabravo

Joined Feb 24, 2006
18,400
The input and output impedance set the dB level for DC and low frequency. For example, when the source impedance is 50 Ω and the load impedance is 50 Ω the output will be 1/2 of the input and that will result in the attenuation being 20*log(1/2) = -6 dB. Let us say the source impedance is 100 mΩ and now the load impedance is 300 Ω, the output will be 0.99967 times the input, and 20*log(0.99967) ≈ -2.9 mdB ≈ 0dB. From this we conclude that at low frequency the output is slightly attenuated or essentially equal to the input.
 

crutschow

Joined Mar 14, 2008
29,755
Technically 20log(Vout/Vin) only applies for equal input and output impedances, but it is commonly used to indicate voltage gain even when the two impedances aren't.
 
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