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TDA2030 VS. LM383

Integrator741

Jun 16, 2013
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Hello everyone,

I am trying to build an amplifier with a flat gain between 36Hz - 18kHz that can be driven from a microphone and have an output of around 7 watts. The amplifier must be capable of being powered from a 12 volts car battery.

I found these two ICs and I don't know which one to use... TDA2030 and LM383. Can anyone please help me to choose between them two. which one would be better? Why??

Thanks!
 
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KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
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Well, both of them seem to be obsolete, which is never a good sign!

There's also a problem with your supply voltage, if you use a single-ended output stage. If you assume that the output stage can swing fully to both rails (they can't), the maximum it can generate is 12V peak to peak. Divide that by 2 * sqrt(2) to get the RMS voltage, 4.2V RMS. The amount of power, in watts, that this can produce depends on the speaker impedance and can be calculated using P = V^2 / R.

8 ohms: 4.2*4.2/8 = 2.2 watts
4 ohms: 4.2*4.2/4 = 4.4 watts
2 ohms: 4.2*4.2/4 = 8.8 watts.

So you would need an ideal output stage and a 2 ohm speaker to get 7 watts of output power.

This is borne out by the graphs in the data sheets that show you the output power versus the power supply voltage:

attachment.php


Luckily there's a trick called "bridge-tied load" (BTL) that helps a lot. The speaker is connected between TWO output stages, which move in opposite directions. This can produce an output voltage, as seen by the speaker, of +12V to -12V, i.e. 24V peak to peak - assuming the outputs can swing all the way to the rails.

This arrangement also removes the need for an output coupling capacitor, which saves space and cost, and gives a flatter frequency response at the bottom end. So really, what's not to like? :)

Let's assume that the outputs can swing to within 1V of each rail. That means the voltage across the speaker will be +10V to -10V or 20V peak to peak. Divide by 2 root 2 to get 7V RMS and recalculate the output power using P = V^2 / R.

8 ohms: 7*7/8 = 6.25 watts
4 ohms: 7*7/4 = 12.5 watts
2 ohms: 7*7/2 = 25 watts

With a BTL configuration, each output stage has to deliver twice the amount of current that a single-ended output stage delivers, for a given load resistance. So there's more stress on the amplifier.

A great way to see the available options is using Digikey's parts selection filter. They have a specific section for audio amplifiers: http://www.digikey.com/product-sear...s-ics/linear-amplifiers-audio/2556583?stock=1

I've had a look and it seems there is only one option for power amps in through-hole packages that will operate at 12V and deliver 5+ watts:

STMicro TDA2005: http://www.digikey.com/product-detail/en/TDA2005R/497-11749-5-ND/2214669

In BTL configuration, it can deliver typically 20W into a 4 ohm load at 14.4V supply voltage. At 12.0V supply instead of 14.4V, this will drop to about 10.7W. With an 8 ohm load, the maximum power will be slightly more than half this; say 5.5W.

So as you can see, the supply voltage is very important - a reduction of just 2.4V can halve the output power! Also the load impedance is very important; I think you'll want to use a 4 ohm speaker (or two smaller 8 ohm speakers in parallel).
 

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