[sdiy] Equation for converting V PP to dBm?

[Tom Wiltshire]

Hi Justin,

I've been through this same process, so maybe I can help.

You know that:

	GdB = 20 log10(Vout/Vin)

So stating the gain as a multiplication factor ("gain of x2" rather than "gain of 6dB"):

	GdB/20 = log10(Vout/Vin)				(divide both sides by 20)

	pow10(Gdb/20) = Vout/Vin			( raise both sides to pow10)

	Gain = pow10(Gdb/20)

We also know from AD that

	Gdb = Vcv/-33 (when Vcv is in mV)

so 

	Gdb = Vcv * 1000 / -33 (for a CV in volts)

You can then stick this into the other equation to get a final answer that gives you the gain for a given CV:

	Gain = pow10( (Vcv * 1000 / -33) / 20 ) 					( messy, so lets cancel *1000/-33/20)

**********************************************
	Gain = pow10( Vcv * -1.5 )
**********************************************	

A quick check -  stick in a CV of 3.3 volts:
	Gain = pow10(3.3 * -1.5)
	Gain = pow10(-4.95) = 0.00001122, which sounds plausible for -100dB.
Or stick in a CV of 0V:
	Gain = pow10(0 * -1.5)
	Gain = pow10(0) = 1, which is also correct.

I went round and around with this when trying to work out the equation required to compensate for the log response. Ultimately it isn't that difficult, but I had a hell of a job keeping things straight in my head. Rusty maths. Maybe you'll do better.

Good luck!

Regards,
Tom


On 30 Dec 2011, at 13:18, Justin Owen wrote:
> 
> Thanks for all the input into this - much appreciated.
> 
> Based on the replies here and a reading of http://en.wikipedia.org/wiki/Decibel - this seems to do the job
> 
> GdB = 20log10(Vout/Vin)
> 
> 20log10(10/10) = 0dB and 20log10(10/5) = 6dB "...an increase of 6 dB corresponds to approximately four times the power and twice the voltage..." Right?
> 
> For the purpose of comparing the response of one VCA against another in the comfort of my own home - is there any reason why I can't call a 10V PP sine from a given output '0dB' (Vin) and plot any gain/attenuation as +/-dB (Vout)?
> 
> If the 2164 is capable of a maximum of -100dB then if I (am able to...) measure 20log10(0.0001/10) = -100dB - can I accept that is indeed -100dB as referenced in the data sheet?
> 
> Assuming this all makes sense and that GdB = 20log10(Vout/Vin) = x ...is a suitable equation for my current task - has anyone got any advice on resolving
> 
> GdB = 20log10(y / Vin) = x - where I know the value of 'x' and need to calculate the value of 'y' - e.g. GdB = 20log10(y / Vin) = 40 ?
> 
> J
> 
> P.S. - I'm not ignoring the thorough workings that some of you were kind enough to offer up - just need to work through this one step at a time. Ta!
> 
> 
> 
> 
> 
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