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Old 26th May 2012, 09:03 AM   #1
Vostro is offline Vostro  South Africa
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Default Needed: Amplifier schematic, 4.5V to 9V supply

Hi Im looking for a working schematic of a small amplifier than can run off
3x 1.5V cells = 4.5V, or one that can run from a 9V battery.

Ive googled ofc but prefer to get a circuit here from someone I know would give a nice one that works.

I even saw some on google search that have errors that make it unusable,
And here I thought that all things on the internet are true, .

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Old 26th May 2012, 09:24 AM   #2
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Your spec is akin to saying you've got this 20 liter fuel tank and want to build a motorized vehicle with it.

It might help to specify how many wheels you need, how far you want it to go and how much you'd want to transport with it.

Amplifiers come in all sizes and shapes. At the very minimum, it would be necessary to know what you intend to drive with said amplifier - which impedance and sensitivity. And number of channels, if applicable.
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Old 26th May 2012, 09:27 AM   #3
wahab is offline wahab  Algeria
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Most efficient would be to use a STMicro TEA2025
as it work from 3 to 12V and can be loaded with 4R loads
plus it is a stereo amp that is eventually bridgeable.

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Old 26th May 2012, 10:25 AM   #4
Vostro is offline Vostro  South Africa
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thanks wahad , i forgot to say it must be a discrete amp.
as for specs, nothing special, just 4.5v into 8 ohm spkr. not headphones.

might use the ic you sugested, its fantastically cheap here, if i cant get discrete amp

Last edited by Vostro; 26th May 2012 at 10:28 AM.
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Old 26th May 2012, 12:54 PM   #5
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The TEA2025 looks like a pretty decent part as far as cheap, oldschool IC amps go. Not even all that noisy at 1.5 V typical (input-referred). Then again, minimum recommended gain still is relatively high at 36 dB (Rf = 100 ohm), which still is more than you'll typically need.

Note that the datasheet is far from error-free. Figure 12 shows R4 incorrectly connected (see Figure 2 for how it should be), and even if the bridge gain formula shouldn't look suspect to you on paper yet, punching it into the calculator with the values given shows that it cannot be correct.

So let's try and derive a correct formula.

The voltage tugging at one end of the speaker is that same as in stereo operation. Assuming we're looking at a frequency high enough for C1 to be neglected, we obtain:
V(out+) = [ 1 + (R1 / (R2 + Rf)) ] * V(in1)

The other amplifier is run in inverting mode and fed from the positive output via the R3 - R4 voltage divider.

Output level at the Bridge pin is
Vout(1) = V(out+) * R4 / (R3 + R4) ~= V(out+) * R4 / R3
with an output resistance of
Rout(1) = R4 || R3 ~= R4

Inverting amplifier input resistance at pin 6 is
Rin(6) = R2

Inverting amplifier output thus becomes
V(out-) = [ -R1 / (R4||R3 + R2) ] * Vout(1)
V(out-) = [ -R1 / (R4||R3 + R2) ] * V(out+) * R4 / (R3 + R4)
V(out-) ~= [ -R1 / (R4 + R2) ] * R4 / R3 * V(out+)
V(out-) ~= - [ R1 * R4 / ((R4 + R2) * R3) ] * V(out+)

Note that for the resistor values given,
V(out-) ~= -V(out+)

Total V(speaker) thus becomes

V(speaker) = V(out+) - V(out-)
V(speaker) ~= V(in1) * [ 1 + (R1 / (R2 + Rf)) ] * [1 + [ R4/R3 * R1 / (R4 + R2) ]]
or with the finding above,
V(speaker) ~= 2 * V(in1) * [ 1 + (R1 / (R2 + Rf)) ] = V(out+) + 6 dB

Looks like the datasheet formula is almost correct, except they mistakenly wrote R3/R4 as opposed to R4/R3.

If you were after minimum noise, it would be possible to do so by both reducing Bridge pin attenuation factor and inverting amplifier gain (now 40 dB each, with a noise gain of 101, so you're not too far from 36 dB already). With an extra series resistor between pins 1/Bridge and 6 called Rf2 and another in parallel to R3 (between 15/out+ and 1/Bridge) called Rp3, the above formulas can be reused with the following replacements:
R4 + R2 --> R4 + Rf2 + R2
R3 --> R3 || Rp3

Looks like I'm finally "getting" this BTL stuff.

Last edited by sgrossklass; 26th May 2012 at 01:02 PM.
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