I can answer that.thomasfw said:
Your equation is correct. To minimize the DC offset, you should try to make the resistance to each input pin the same (since the bias currents at each pin are in theory, the same), so:
R3 = (Rin * RF)/ (Rin + RF)
However, note that we are talking about DC offset, and therefore, we must use DC impedances for Rin and Rf, not necessarilly just the resistor values. Note that Rin has a capacitor in series with it, and that changes things. That means that at DC, the impedance (of this R-C network) is infinite (or nearly infinite). Then R3 = ( Rf in parallel with infinity ), which is R3 = Rf.
jaudio also recommends using a trimpot for R3. This is because the bias currents for each input pin are not exactly the same for any given chip. The trimpot lets you adjust the resistance to compensate for variations in bias currents for a particular chip.
It depends on the amount of gain,that you want. In inverted mode, an input resistor of 10k and a Rf resistor of 220k will give a gain of 22. So with 1volt in you get 22 out. In bridge mode your get(-22)+(+22)=44. With that amount of
gain get lots of noise. Lower gain would be better.
Just match dc offset with trimmer and match ballast resistors.
I hope this helps
gain get lots of noise. Lower gain would be better.
Just match dc offset with trimmer and match ballast resistors.
I hope this helps
Thank You For Your Fast Response !
Let me asked in detail :
Rf=200K, Rin=10K, Gain=20
Rf=20k, Rin=1K, Gain also=20
What is the different in between them ?
If I want the total Output=200W,8 ohm (after Paralell & Bridge)
and let the Vin= 1v.
What should I do with the Rf , Rin & Trim Pot ?
Many Thanks !
Thomas
Let me asked in detail :
Rf=200K, Rin=10K, Gain=20
Rf=20k, Rin=1K, Gain also=20
What is the different in between them ?
If I want the total Output=200W,8 ohm (after Paralell & Bridge)
and let the Vin= 1v.
What should I do with the Rf , Rin & Trim Pot ?
Many Thanks !
Thomas
20k and 1k will have less noise.
You would need a gain of about 24(+24 + -24)
If you plan on making your own preamp,I would incease the preamp gain to 2v output and go with (+12 + -12)
I would go for 4K 48k and 50k trimmer.
If you plan on parallel six use a input resistor with more than 1k resistance.
You would need a gain of about 24(+24 + -24)
If you plan on making your own preamp,I would incease the preamp gain to 2v output and go with (+12 + -12)
I would go for 4K 48k and 50k trimmer.
If you plan on parallel six use a input resistor with more than 1k resistance.
Same gain, but different input impedance.
A 1K input impedance can place too much of a load on say, a CD player line-level output. But note that National's BPA amplifier design has buffered the input with an opamp, which has no difficulty driving a 1K input impedance (in fact, it drives four of them).
If you need a higher input impedance, then you will also need a higher feedback resistor to achieve the same gain . Note that higher resistances create more thermal noise (a.k.a. "resistor noise"). So if you increase it too much, then it will begin to add extra noise into the system, which you can hear as background hiss. (Another way to get high gain, without a high feedback resistance and the noise it makes, is to use a T-network feedback, but that's another thread in itself).
Correct, that a bridged amp will double the voltage at the speaker terminals, which, by P=V*V/R, gives 4 times the power (P). Note also that the current flowing through each of the amps is also doubled, which is in effect, the same as decreasing the load (speaker) impedance by half. An 8 ohm speaker connected to a bridged amp loads each 'side' of the bridged amp with a 4 ohm load. The National app note describes this and what you need to do about it when designing an amp.
However, your calculation of power above may be wrong. Make sure not to use the peak voltage in your calculations. To calculate RMS power, you must use the RMS voltage, which is (for a sinewave), the peak voltage divided by sqrt(2), or Vrms=Vpeak/1.4142
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