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how to find out the frequency of an amplifier


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It is the capacitors in an amplifier that limits its low frequency response. Old amplifiers had transformers that also limited their low frequency response. A DC coupled amplifier amplifies very low frequencies and DC just as well as any other frequency.

Calculate the reactance of the coupling capacitors with their loads at a frequency you select. Each reactance that equals its load drops the response -3dB. At about 5 times the -3dB frequency the response will be flat.

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Hi audioguru!

it looks like that perhaps u didnt get my question.

Well! my dear teacher! i want to ask you that how i will know the frequency with which my amplifier is working so that i can select the value for my capacitors.

U remeber that LM380 amplifier circuit. There u told me to use 1000uf capacitors. Then im asking you how u select this value.

Thanks in advance

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Hi Shaiq,
In my last post I said to calculate the reactance of a coupling cap with its load at the frequency you select. Our hearing extends down to 20Hz. Your speaker is 8 ohms. The reactance of a 1000uF cap is 8 ohms at 20 Hz. The frequency response will drop -3dB at 20 Hz and be flat down to about 100Hz. -3dB is half-power. ;D

post-1706-14279142607466_thumb.png

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I'm missing something here...................

If a capacitive coupled signal is driving a resistive load, this is like a high pass filter!!!!! Yes, you do apply the Maximum Power Transfer Theorem  and the corner frequency will be, f=1/(2* pi*R*C) (the -3dB point, a.k.a. half power point), but as the frequency increases from 20Hz up, the gain goes up, not down provided the resistance (inductive reactance of the voice coil) remains relatively constant over the bandwidth, because capacitive reactance is going down.

What does the "equivalent circuit" for a speaker look like, and what are some "real world" values??

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Hi Indulis,
At about 5 times the frequency where the capacitor's reactance equals the speaker's impedance, it is no longer a highpass filter and the small increase of the speaker's impedance with rising frequency due to its inductance is negegible.
There isn't gain in a capacitive coupled circuit. The amplifier's output imedance is extremely low. It uses max voltage transfer, not max power transfer. The amplifier's extremely low output impedance damps the speaker's resonance.

In the real world, an 8 ohm speaker has a low frequency resonance of about 24 ohms (cheap ones are even more) and is 8 ohms a couple octaves above resonance. At 20kHz the impedance is about 100 ohms. ;D

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OK............. let's call it a decrease in attenuation, which from a certain point of view is gain.

As you go below 20Hz  towards 0, the voltage across the resistor (in this case the speaker) will continue to decrease... this is basic "CR" circuit stuff, the transfer function is (sR1C1)/(1+sR1C1) and the corner frequency of the pole is f=1/(2*pi*R1*C1) (C1=coupling cap and R1=speaker "resistance").

You said:    "The frequency response will drop -3dB at 20 Hz and be flat down to about 100Hz. -3dB is half-power."


I don't understand!!!!! As frequency increases above 20Hz, the signal is not going to drop...... it's going to increase as it

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The frequency response will be flat (no attenuation) from 100Hz to a high frequency. The response below 100Hz will gradually drop until it is -3dB at 20Hz.

A rather large output cap and resulting very low cutoff frequency was chosen to allow the very low impedance of the amplifier's output to still have a low impedance of the capacitor in series with it to damp the speaker's resonance. If a smaller cap was used, most bass would be less but resonance would be more: Boom-box sound. :'( ;D

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