velko
10+ year member
Member
I asked my friend to help me figure out the voltage on the speaker outs in order to set the gains. The problem arose that I didn't have a 0db tone but only a -12 tone and he tried to set me straight. I didn't have to do this sh!t with my home stereo!
Below is HyperGeek's response.
dB = 10 log y/x (where log is log to the base 10 -- it is NOT "ln," which is the natural log (log to the base e))
So, dividing both sides by 10, you get:
dB/10 = log y/x (where y is the output voltage and x is the input voltage)
In your case, your dB's are more accurately described as dBv's (decibels relative to 1 volt), so the x is presumed to be 1 volt, so you don't need to worry about the x, just the y, which simplifies the equation to:
dB/10 = log y
So, on a calculator, you would just do the following:
enter # of dB's
hit divide button
enter 10
hit "=" button
hit 10^x button (or if the calculator only has a y^x button, write down the # you got when you hit "=" then hit the clear button, enter 10, hit y^x, enter the number you wrote down, then hit "=")
That converts the dB value into volts (which isn't really an accurate description, as it's really volts relative to something, but, for your purposes, let's just think of it as converting decibels into volts).
You can then use that voltage value to adjust the expected output voltage to compensate for using a different input dB level. For example, if the input signal is at -12 dB, the above equation converts that 12 dB loss into a voltage divisor of 15.849. So, if you're looking for an expected voltage output of 17.89 volts from a 0 dB input, you would divide that 17.89 volts by your 15.849 divisor, giving you an expected output voltage of 1.129 volts for the -12 dB input signal.
You can use the same equation to convert volts back into dB's to know how many dB's you need to adjust the gain by to get the desired output level. For example, if you expect an output voltage of 1.129 volts, but get an output voltage of 0.800 volts, the gain of the amp (or some other part of the system such that you change the gain of the system as a whole) relative to the desired gain is:
0.800/1.129 = 0.709, then dB = 10 log 0.709 = -1.496 dB, which means you need to turn up the gain by about 1.5 dB.
Whenever dealing with decibels, there are only two things that make decibels confusing. (1) Decibels are on a logarithmic scale, not a linear scale, and (2) decibels are a ratio of one value to another value, not an absolute measurement of anything. #1 is why you have to use the log button and the 10^x or y^x button, and #2 is why decibels are properly described as being relative to some reference level, such as dBm's (decibels relative to a milliwatt), dBv's (decibels relative to a volt), dB SPL (decibels of sound pressure level). Since your sound system is converting a CD recording in dBm's and converting into an output voltage, you can conveniently ignore #2, but you still have to pay attention to #1, converting between linear and logarithmic scales using the log and 10^x or y^x buttons.
Below is HyperGeek's response.
dB = 10 log y/x (where log is log to the base 10 -- it is NOT "ln," which is the natural log (log to the base e))
So, dividing both sides by 10, you get:
dB/10 = log y/x (where y is the output voltage and x is the input voltage)
In your case, your dB's are more accurately described as dBv's (decibels relative to 1 volt), so the x is presumed to be 1 volt, so you don't need to worry about the x, just the y, which simplifies the equation to:
dB/10 = log y
So, on a calculator, you would just do the following:
enter # of dB's
hit divide button
enter 10
hit "=" button
hit 10^x button (or if the calculator only has a y^x button, write down the # you got when you hit "=" then hit the clear button, enter 10, hit y^x, enter the number you wrote down, then hit "=")
That converts the dB value into volts (which isn't really an accurate description, as it's really volts relative to something, but, for your purposes, let's just think of it as converting decibels into volts).
You can then use that voltage value to adjust the expected output voltage to compensate for using a different input dB level. For example, if the input signal is at -12 dB, the above equation converts that 12 dB loss into a voltage divisor of 15.849. So, if you're looking for an expected voltage output of 17.89 volts from a 0 dB input, you would divide that 17.89 volts by your 15.849 divisor, giving you an expected output voltage of 1.129 volts for the -12 dB input signal.
You can use the same equation to convert volts back into dB's to know how many dB's you need to adjust the gain by to get the desired output level. For example, if you expect an output voltage of 1.129 volts, but get an output voltage of 0.800 volts, the gain of the amp (or some other part of the system such that you change the gain of the system as a whole) relative to the desired gain is:
0.800/1.129 = 0.709, then dB = 10 log 0.709 = -1.496 dB, which means you need to turn up the gain by about 1.5 dB.
Whenever dealing with decibels, there are only two things that make decibels confusing. (1) Decibels are on a logarithmic scale, not a linear scale, and (2) decibels are a ratio of one value to another value, not an absolute measurement of anything. #1 is why you have to use the log button and the 10^x or y^x button, and #2 is why decibels are properly described as being relative to some reference level, such as dBm's (decibels relative to a milliwatt), dBv's (decibels relative to a volt), dB SPL (decibels of sound pressure level). Since your sound system is converting a CD recording in dBm's and converting into an output voltage, you can conveniently ignore #2, but you still have to pay attention to #1, converting between linear and logarithmic scales using the log and 10^x or y^x buttons.
