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Audio amplifier relation of dB to watts

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deepakchikane

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Hi Technocrats,
Hope you are doing well in this Covid-19 pandemic situation.

I need your help for the Audio amplifier design.
Problem statement-to design audio amplifier which will able to produce 85dB outputs.

I have designed the Amplifier with TDA 1905 but i am not able to get the dB relation with respect to Watts.

Could someone help me with the relationship.
I have used Speaker with impedance 8 OHM and input to the speaker is 1.84VRMS.

So i am able to get the power of P=V^2/R=(1.84*1.84)/8=0.4232W

now i am able to get how much dB i am producing.
Please help out.

Regards,
DC
 

A dB rating is always referenced to something.

Let's say you choose 1 mW as your reference. Then an output of 1 mW would be 0 dBm.

Because we are talking power here, the equation to calculate how many dBm your example above is: 10*log(0.4232/0.001) = 26.26 dBm.
 

There's no relation of dB to Watts. dB is describing a ratio of voltage, power or sound pressure value to another value. As a relative value, it can e.g. describe a voltage gain. Absolute dB units are supplemented with a reference character, e.g. dBm (power relative to 1 mW), dBV (voltage relative to 1 V), dB(A), dB(C) and dB SPL (units of sound pressure).
 

You need to know the efficiency of the speaker, which is usually give in sound dB (SPL) per watt at 1 meter from the speaker such as:
SENSITIVITY 1W/1m 82 dB
so 1W of power to the speaker input would generate an 82dB SPL sound level at 1 meter on axis from the speaker cone.
SPL is the sound level as measured by a calibrated sound meter.
 

A dB rating is always referenced to something.

Let's say you choose 1 mW as your reference. Then an output of 1 mW would be 0 dBm.

Because we are talking power here, the equation to calculate how many dBm your example above is: 10*log(0.4232/0.001) = 26.26 dBm.

Hi Schmitt
Thanks for your reply.
i was referring to some documents from PUIAUDIO, where they have clearly mentioned about the same.
https://www.puiaudio.com/resources-white-papers-speaker-power.aspx

Now.. i want to understand if i want to get 75dB output from speaker-How i would achieve it?

- - - Updated - - -

You need to know the efficiency of the speaker, which is usually give in sound dB (SPL) per watt at 1 meter from the speaker such as:
SENSITIVITY 1W/1m 82 dB
so 1W of power to the speaker input would generate an 82dB SPL sound level at 1 meter on axis from the speaker cone.
SPL is the sound level as measured by a calibrated sound meter.

Hi Crutschow,
Could you please help me with More details.

What are the steps to achieve it?
What is the selection criteria?
What need to take care?
 

85 dB or 75 dB "output from speaker" is sloppy speech. Actually means dB SPL, see https://en.wikipedia.org/wiki/Sound_pressure

As discussed in the article, it's measured at a specific distance, strictly in an anechoic chamber (or free space).

A speaker with a sensitivity of 85 dB SPL @ 1 W + 1m needs 1W input power to generate 85 dB SPL. Or 0.1 W for 75 dB SPL. In so far your question can be answered quite easily.
 

Look at a table of decibel ratings for various audio sources, from rustling leaves all the way up to a jet plane. Say 85dB is volume level of... hmm, factory, restaurant, vacuum, child screaming.
So iIf your speaker sounds about as loud, then you're not too far off.

hearingsense.com.au/wp-content/uploads/2014/06/Noise-levels-image.gif

Whenever I use my analog VOM to measure AC voltage at a speaker, it's usually 1/2 to 1 volt at normal listening volume (and a bit higher when the bass thumps loud). The needle responds more correctly to low frequency sine waves such as mains AC. Moreover the AC volts range contains diodes which create a threshold voltage internally, so my meter gives a rough indicator of voltage.
 

The speaker sensitivity information is given by the speaker manufacturer.
Otherwise you would need to measure it with a sound meter.

My speaker datasheet states that..
SPL=99dB (+6dB/-2dB) at 1 meter (Anechoic) with 2.83Vrms input.

Does it means that at 1W it gives 99dB?
V=2.83V
R=8 OHM
P=V^2/R=(2.83^2)/R=1W

so for 0.5W it will give 99dB-3dB=96dB
0.25W=93dB


DC
 

My speaker datasheet states that..
SPL=99dB (+6dB/-2dB) at 1 meter (Anechoic) with 2.83Vrms input.

Does it means that at 1W it gives 99dB?
V=2.83V
R=8 OHM
P=V^2/R=(2.83^2)/R=1W

so for 0.5W it will give 99dB-3dB=96dB
0.25W=93dB
Yes, that's all correct.
But note that is at 1 meter distance in an anechoic chamber (no sound reflections from walls), so the actual intensity will vary with the type and size of room the speaker is in, its location in the room, and the distance from the speaker.
 
The TDA1905 amplifier IC is obsolete and is not made or available anymore. Then this thread must be homework because old teachers use obsolete old things.
 

Does it means that at 1W it gives 99dB?

Right, but you should use the error limits also.

I mean that for 1W input power the speaker can output 99+6 to 99-2 db SPL at 1m

When you go down to lower power levels, the error range stays the same (+6 to -2 db).

Also remember that the speaker was tested and specified at one power level without mentioning the frequency.
 
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