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Chip Amps Amplifiers based on integrated circuits 

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13th December 2012, 12:17 AM  #1 
diyAudio Member
Join Date: Dec 2012

What is Decibel Scale
Guts I'm really confused on this decibel scale numbers.
In my earlier AVtuner it has volume control numerical values marked on the front panel display.It will go down from 0 to 32 levels.The numbers got bigger as you increased volume & when you reduced the volume, the numbers got smaller. Very easy to understand. But newer amplifiers consist with decibel figures!!! I really confused.It will go from 25dB to +18dB.What does it means? If I have a home built amplifier ex:LM1875 how can you scale it to decibel scale? I'm new to these decibel figures 
13th December 2012, 03:11 AM  #2 
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Join Date: Oct 2008

db=20*log(voltage_gain)
if db<0, it means a gain of less than 1...for example: 12 dB => gain of 1/4 6 dB => gain of 1/2 0 dB=> gain of 1 +6 dB=> gain of 2 +12 dB=> gain of 4 +18 dB=> gain of 8 Akitika GT101 
13th December 2012, 04:05 AM  #3 
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13th December 2012, 04:39 AM  #4 
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In its roots the decibel(1/10th of a bel) is just an expression of ratio in logarithmic scale. As posted above, in case of voltage the expression is :
db = 20* log(U1/U2) You can measure any ratios you want with it, for example a distance or oil price compared to something. It was selected as a common unit for sounds because our hearing works in more like logarithmic way. Two times more sound pressure is not heard as two times "louder". Regards, Lukas. 
13th December 2012, 05:40 AM  #5 
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Join Date: Jul 2012
Location: Wichita, KS

When your tuner is at 0 dB, the signal out is at the same level as what is coming in. When it is at +10 dB it should sound about "twice as loud", 10 dB half.
If you want to put a log scale on something with a volume knob, it can be tricky. Unless you know the total gain of the amplifier, you need to use an AC volt meter. (Most inexpensive ones are only accurate with a 60 hz signal.) Measure both the input and output voltage, then just use the formula db = 20* log(voltage out/voltage in). 
13th December 2012, 11:35 PM  #6  
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Join Date: Oct 2012

Quote:
20*log(2) = 6.02 20*log(1/2) = 6.02 Btw, power is measured as 10* log(P1/P2) as power involves a voltage squared for a given load, and square rooting inside a log is the same as halving (hence 10 not 20) outside it Last edited by Robert Kesh; 13th December 2012 at 11:38 PM. 

15th December 2012, 11:39 AM  #7 
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Join Date: Dec 2012

Hi guys sorry for the delay that I have been outside for a while.
After reading many times I slightly understood what is decibel.I'll do some calculations & make some practice with it. Thanks for the support. 
16th December 2012, 09:11 AM  #8 
diyAudio Moderator

Put some numbers in as you suggest.
Lets take the LM1875 here. R3 and R4 determine the voltage gain at AC. Numerically the gain is (R4 + R3)/R3 which is 21. That means that 100 millivolts RMS input would deliver 2100 mv RMS output. If we talked of 300 mv peak input we would get 6300 mv peak output and so on. So whats the voltage gain in db. The formula is Gain = 20Log(VOut/Vin) So using either of the above two sets of numbers we get 26.4 db using either set of numbers. Also you'll notice if you swap Vin and Vout around in the formula that the answer is the same except with a minus sign in front of it. So if your amplifier really did give 100 mv output for 2100 mv input (an attenuator) then the figure of 26.4db would indicate that. So thats voltage gain. For that amplifier it is 26.4db at AC in the "midband". Why midband ? Because the input coupling capacitor rolls the gain off to zero at DC. The capacitor C2 also rolls the gain "of the amplifier itself" down to 1 or unity at DC. If we shorted those caps out (don't ) the amplifier would maintain its gain down to DC. At high frequencies the gain falls off again due to limitations of the IC. Now imagine its running into an 8 ohm load. Using the same figures we can say that the output of 2100 mv RMS produces a power of 0.55 watts RMS (W= V squared/R). into that load impedance. We can also say that the 100mv input would produce a power of 0.00125 watts into 8 ohms if it were applied across the load (same watts equals v squared over R) In db that ratio is 10Log(Pout/Pin) which is 26.4db (the same figure as the voltage gain) Quote, But newer amplifiers consist with decibel figures!!! I really confused.It will go from 25dB to +18dB.What does it means? Figure like that can be a bit meaningless because they are not quoted with "respect to anlything". Maybe it means 0db equals some notional figure such as 1 watt into 8 ohms. So perhaps it can go 18db above that (70 watts).
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16th December 2012, 11:04 AM  #9  
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Join Date: Aug 2008
Location: High Wycombe

Quote:
Note that those numbers say nothing about maximum output, only the range of gains available within the systems linear region. The key thing to remember about the dB scale is that it is (in theory) a ratio of power levels, hence the 10 log thing for power but 20 log for voltage (power is the square of voltage). Note also that in much modern audio usage we do not actually use it as a power ratio as we tend to use voltage transfer rather then power transfer on our lines, usually our output impedances are very much smaller then the loads, hence we tend to use dB assuming equal impedances in and out which is not actually the case. The RF guys generally do use power transfer so use the dB in the technically correct way. Be very careful about the differences between voltage, power and volume, the first two have a square law relationship, the last has a somewhat complex, non linear, frequency dependent relationship to power being a physiological/psycological response, a very rough rule of thumb is that below 'temporary' threshold shift, double the volume takes 10dB more power (10 times the power). Google the term 'Equal Loudness Contour" for more on volume vs power. Regards, Dan. 

16th December 2012, 11:32 AM  #10  
diyAudio Moderator

Quote:
I suppose what I was wondering is, why quote a variable gain amplifier (like an integrated amp) as having a gain below and above a notional 0db point. Is some reference value implied in the 0db figure. I don't know the answer to that
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