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Chip Amps Amplifiers based on integrated circuits 

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12th May 2006, 01:55 AM  #1 
diyAudio Member

Simple gain question
Hey people,
I have a LM4780 amp with a gain of 33.5. I want to decrease the output by 6dB. This is where I get a little confused... Does that mean I want a gain of 29.5? +6dB is like is like doubling the signal twice so I subtract 4.. Or does it mean I want a gain of 8.375, ie divide it by 4... Or maybe I'm just completely wrong?! 
12th May 2006, 03:02 AM  #2 
diyAudio Member
Join Date: Nov 2005

VOLTAGE gain in dB = 20 * LOG ( GAIN )
So you want to lose 6 dB... that's a minus 6 now. Gain = 10 to the ( 6 /20 )...... (this is the inverse log) Multiply that times your original 33.5 and you have it... OR each 3 dB = 0.707 SO... 6 dB = 0.707 to the 2nd power ~= 0.5 
12th May 2006, 03:35 AM  #3 
diyAudio Member

Dang, that's complex!
So I subtract 0.5 from 33.5 to get my gain that achieves 6dB? ie, 33.0? 
12th May 2006, 05:11 AM  #4 
diyAudio Member
Join Date: Nov 2005

Nope... you multiply the 0.5 times 33.5 to get a final desired gain of 16.75.
Here's the trick when you work with dB; you simply add or subtract gains. When you work with ratios... straight numbers, you multiply. dB is a logarithmic way of workng with things. It got started long before we had calculators... the "Bell" in deciBell stands for Alexander Graham Bell. By working out dB charts for standard problems, math problems could be solved with simple addition and subtraction rather than mutliplication and division. Strictly speaking, because of calculators, we don't need dB anymore. But it does make sense for alot of things and people are used to it. It works well for "signals" and radio... especially when long lines and distances are concerned. Heck... for that matter it would have made more sense for our number system to be base 8 instead of base 10. People insisted that we got our thumbs into the act as well. Here's the formula's: Voltage Gain (in dB) = 20 * Log (output / input) & the reverse output/input = 10 to the (gain in dB / 20) Power Gain (in dB) = 10 * Log (output / input) & the reverse output/input = 10 to the (gain in dB / 10) "output/input" denotes "straight" mathematical gain. Play with the numbers... 0 dB equals 1... 20 dB = 10... 20 dB = 1/10... it will make sense after awhile. 
12th May 2006, 05:53 AM  #5 
diyAudio Member

Excellent, thanks for taking the time Poobah, much appreciated

12th May 2006, 06:28 AM  #6 
diyAudio Member
Join Date: May 2006

dB represents a ratio between two things.
In this case, it's the ratio between the power that you have and the power that you want. 10 ^ (db / 20) = ratio that you want. plugging in 6, you get... 10 ^ (6 / 20) is about .5 so 6 dB means you want half as much gain as you have now. 
12th May 2006, 01:42 PM  #7 
diyAudio Member
Join Date: Nov 2003
Location: Brighton UK

Hi,
you are probably not aware there are hidden dangers in reducing power amplifier gain. A power amplifier has a gain margin, the gain margin (in dB) is the ratio of the gain at which it oscillates to the gain it is used. By reducing gain you are reducing the gain margin, making the amplifier more likely to oscillate, e.g. instability during clipping. Generally speaking to reduce gain you should attenuate the input. /sreten. 
12th May 2006, 11:03 PM  #8 
diyAudio Member

yep yep, I know this.
From the LM4780 datasheet: "Gain settings below 10 may experience instability.." So a gain of 16 should be fine. The reason I wanted it lower is because they will be powered by DRV134s, which add 6dB gain. 
12th May 2006, 11:05 PM  #9 
diyAudio Member
Join Date: Nov 2005

Good tip sreten and good work max! The data sheet is your friend.

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