Look at this graph. Can someone tell me why is the amplitude measured in Volts?
http://img524.imageshack.us/img524/2628/...
Answer:
The first thing to understand is that the dB is not a unit of power but of power _ratio_. It tells you how many times more powerful your signal is than some reference signal.
And the second thing is that it uses a logarithmic scale to compress the enormous ratios found in practice into handy numbers which you can multiply together simply by adding.
The power in dB is 10 times the log (base 10) of the power of what you are measuring divided by the power of your reference:
D = 10 log (P1 / P0).
OK, in your case you haven't got a power as such, you've got a voltage. This is because oscilloscopes always work in volts, which is what they actually measure. However, because power is voltage squared divided by resistance, you can rewrite the above as
D = 10 log (V1^2 R / V0^2 R) = 10 log (V1^2 / V0^2) = 20 log (V1 / V0)
provided that both voltages are across the same resistance R (a point that is sadly often ignored in practice).
So all you need is a suitable reference voltage. The usual reference for small-signal circuits is the millivolt, although for radio work the microvolt is more usual – the reference is simply chosen according to the size of signal you are expecting, just to get convenient numbers.
The peak value of your signal is about 0.08V = 80mV. So, using the formula above, this becomes
D = 20 log (80) = 38dBmV
Notice the "mV" at the end. This shows that you are referencing your measurement to 1 millivolt (and also shows that this is a voltage measurement converted to dB).
And this is where it all gets a bit messy. Decibels are a measure of power ratio, so the reference should really be a power. However, the actual power of a waveform such as yours is quite difficult to measure, and in small-signal circuits is often unimportant. So engineers cheat and express voltages in dB as if P = V^2 / R applied to any old waveform, not just a sine wave. It gives them the convenient maths of dB without having to actually measure power. A typical engineering bodge, but it usually works. Unfortunately, it also causes a lot of confusion!
You can tell when it's a genuine power ratio because it will be expressed in a unit such as dBmW (or sometimes just dBm) to indicate that the reference is a power (1 mW in this case).
So, to summarise: to convert voltage to dB, take 20 times the log (base 10) of the voltage in millivolts and express it as so many dBmV. (Or use volts and dBV or microvolts and dBuV if appropriate.)
You may find in practice that engineers working in a particular field, where everyone knows what unit they are using, may drop the "mV" or whatever and just use plain dB. They do this just to make life easy for themselves and difficult for newcomers!
I'm afraid I've no idea what THT is, but if you could explain I will try to help on this.
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try this dude!
http://www.sengpielaudio.com/calculator-...
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