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Old 24th March 2012, 02:49 PM   #21
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Quote:
Originally Posted by Roushon View Post
As I look at the application sheet AN-1192 I find that
in parallel of two LM3886 the feedback resistors are 20K,
on the other hand the same is about 47 and 51K in the
bridge of two LM3886. Why is this? My circuit is attached
in the parallel version. I thought I just have to exchange
the Amp input and the output of the AD8620 connection
to one of the LM3886. Do I have to change the feedback
resistors also.

Please let me know.

Thanks
Roushon.
Thanks Roushon
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Old 24th March 2012, 03:13 PM   #22
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Quote:
Originally Posted by Roushon View Post
I requested for a suggestion in my old following thread. But it got the
wrong thread name. That is why started this new thread. The request is
here:

PCB design of parallel of two inverted LM3886 with buffer and dc-servo.

For a 8Ohm speaker which sounds better: parallel or bridge of two LM3886.

Thanks
Roushon
I don't recommend to build an audio amplifier with two LM3886 in parallel. There are a lot of factors that must follow to obtain an appropriate quality level. Solution is used if you wish to use LM3886 the very low impedance (less than 4 ohms). Distortions obtained are quite high and the circuit implementation is even more difficult make as we are forced to find two LM3886 as more identical.
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Old 26th March 2012, 11:44 AM   #23
Roushon is offline Roushon  India
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Default Modified circuit

More trouble than I thought. Have to cut several tracks
and need one resistor and a capacitor more for the
non-inverted amp. The circuit is attached.

Please have a look and tell me if any correction is needed
before I start cutting tracks. It will be cumbersome. The
buffer may or may not be there at the input.The attenuator
has input and output impedances both 10K.

N.B. Note that the pin numbering of the AD8610 is not correct.
It was AD8620, I changed it but forgot to the change the pin nos.

Thanks and regards
Roushon
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Last edited by Roushon; 26th March 2012 at 11:47 AM.
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Old 26th March 2012, 11:50 AM   #24
AndrewT is online now AndrewT  Scotland
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and repeat the additions for the other half of the signal.

what do you mean "10k attenuator"?
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Old 26th March 2012, 11:58 AM   #25
Roushon is offline Roushon  India
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Quote:
Originally Posted by AndrewT View Post
and repeat the additions for the other half of the signal.

what do you mean "10k attenuator"?
The attenuator circuit is attached. I do not understand
the first line. Sorry!

Regards
Roushon
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Old 26th March 2012, 12:17 PM   #26
AndrewT is online now AndrewT  Scotland
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The attenuator shown does not have a 10k output impedance.

To give accurate dB attenuations for each step and combination of steps the attenuator needs to see a load impedance of 10k. That is very different.

With all the relays pulled in to bypass all the series resistors, the output will see the Source Impedance.
Let the last (rightmost relay contact) trip over to the 6r31 connected to Signal Ground.

The load now sees 6r31//[10k+Rs]. This is very definitely not =10k.
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Old 26th March 2012, 12:20 PM   #27
AndrewT is online now AndrewT  Scotland
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You have added 0u47F & 470k to the lower amplifier DC servo circuit.
The upper amplifier does not show these additions. Why?

You have shown the extra 0u47F connected to -IN PIN. I think it should be connected to the other side of the lower 470k, i.e. to Signal Ground.

Note that the whole of the lower DC servo circuit is in parallel to R14.
The DC servo circuit is an alternative route for the feedback signal.
That must be repeated for the upper amplifier.
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Last edited by AndrewT; 26th March 2012 at 12:23 PM.
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Old 26th March 2012, 02:28 PM   #28
Roushon is offline Roushon  India
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Thanks Andrew for the clarification.

I am following (may be blindly) National Application notes AN-1192 for the amp.

1. in the lower dc-servo the 0.47 will be connected to SG, this was a mistake.
2. should I repeat the same for the dc-servo for the non-inverted amp too?
that is 0.47 between -IN of AD8610 to SG and 470K between +IN and SG (instead of directly to SG). This is not done in AN-1192.

Regards
Roushon.
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Old 26th March 2012, 03:13 PM   #29
Roushon is offline Roushon  India
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Just looked at the application notes again. It is clearly mentioned (page 13, 3rd
para) that the non-inverting type servo can be used for the inverting amp
(connected parallel to Rf), in that case we will need the extra 0.47 and 470K.
But this extra RC can be avoided if an inverting dc-servo is used with the
inverting amp.

I hope it is OK.

Regards
Roushon
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Old 26th March 2012, 04:55 PM   #30
AndrewT is online now AndrewT  Scotland
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No.
Are you using inverting DC servos or non-inverting DC servos?
Do you know which is which?
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