LM3886 - regulator on the mute pin ?

[ash_dac]

Hi,

http://www.national.com/an/AN/AN-898.pdf

"Of the multiple ways to set the
mute current and utilize the mute function, the use of a
regulator can continuously control the amount of current out
of the mute pin. This regulation concept keeps the attenuation
level from dropping below 0 dB when the supply is
sagging. More information about mute circuit configurations
will be provided later in a future application note."

Has anyone tried this approach with regards to the mute pin?

[jackinnj]

You could use the LZ vresion of the popular LM317/337 series --- the mute pin only takes a few milliamps.

jack

[glennb]

Originally posted by ash_dac
Hi,
http://www.national.com/an/AN/AN-898.pdf
"Of the multiple ways to set the mute current and utilize the mute function, the use of a regulator can continuously control the amount of current out of the mute pin. This regulation concept keeps the attenuation level from dropping below 0 dB when the supply is sagging. More information about mute circuit configurations will be provided later in a future application note."

Has anyone tried this approach with regards to the mute pin?

I wouldn't bother with the complexity of a regulator for the mute pin. Just set the mute current to say 1mA when there is no load on the amplifier. Even with very low value caps on the power supply, it shouldn't sag low enough under load to drop the mute current below the lower limit (0.5mA) to keep the chip in 'play' mode.

If you are worried about sag, use a R-C-R setup with say C=100uF. This will prop up the mute current during transient sags.

[diy_audio_fo]

Originally posted by ash_dac
Hi,

http://www.national.com/an/AN/AN-898.pdf

"Of the multiple ways to set the
mute current and utilize the mute function, the use of a
regulator can continuously control the amount of current out
of the mute pin. This regulation concept keeps the attenuation
level from dropping below 0 dB when the supply is
sagging. "

Has anyone tried this approach with regards to the mute pin?

http://www.rane.com/pdf/ma3sch.pdf

An unusual way to control the current of the mute pin.

Can anyone explain how the circuit works?

Regards
Marco

[jackinnj]

Originally posted by glennb


I wouldn't bother with the complexity of a regulator for the mute pin. Just set the mute current to say 1mA when there is no load on the amplifier. Even with very low value caps on the power supply, it shouldn't sag low enough under load to drop the mute current below the lower limit (0.5mA) to keep the chip in 'play' mode.

If you are worried about sag, use a R-C-R setup with say C=100uF. This will prop up the mute current during transient sags.

a regulator is cheaper than a capacitor and takes less space.

[SpittinLLama]

You can pull more current from the Mute pin than 1mA. If you are really worried about it then set the mute resistor to 10K. Then the amp will ALWAYS be in PLAY mode because before the Mute current can drop low enough the under voltage will kick in and shut down the amp anyway. It won't hurt anything to pull out 3mA when fully on. Or pick a point in between, say 20K so supply has to drop well below +/-20V to lose enough current to affect anything.

-SL

[glennb]

Originally posted by SpittinLLama
You can pull more current from the Mute pin than 1mA. If you are really worried about it then set the mute resistor to 10K. Then the amp will ALWAYS be in PLAY mode because before the Mute current can drop low enough the under voltage will kick in and shut down the amp anyway. It won't hurt anything to pull out 3mA when fully on. Or pick a point in between, say 20K so supply has to drop well below +/-20V to lose enough current to affect anything.
-SL

Yep, agree entirely. There is no real need for a cap or regulator.

[glennb]

Originally posted by diy_audio_fo
http://www.rane.com/pdf/ma3sch.pdf
An unusual way to control the current of the mute pin.
Can anyone explain how the circuit works?
Regards
Marco

OK, I'll have a go. At switch on, as the power supply rails ramp up, it does a controlled increases of the mute current thru pin 8, from 0mA at Vs=20V linearly up to 1mA at Vs=32V. How?

Think of Q20 as a constant current source. D26 (a red LED) acts like a zener of about 1.6V. Subtract the Vbe~=0.6V of Q20, leaving 1V across R67. A constant current of 1V / 2Kohm = 2mA will flow through it.

R70 and R58 act as a voltage divider across the V- power supply, in the ratio of 4.42/(4.42+51.1) = 0.0796. When this divider reaches a voltage equal to the Vbe of Q19 plus the V across R67, then Q19 will begin to conduct thru the mute pin 8.

The Vs at which the conduction begins is Vs * 0.0796 = 0.6 + 1, therefore Vs = 20.1V. As Vs continues to rises to 32 volts, the voltage divider will rise to 32 * 0.0796 = 2.55V. Subtract the Vbe of Q19 giving 1.95V across R67. There is no longer sufficient Vbe on Q20 for it to conduct, so it no longer contributes any current through R67. At this stage the current thru R67 is 1.95V / 2Kohm = 0.975mA, which funnels up thru Q19 into the mute pin. C45 will slow down the current rise somewhat.

A neat circuit, but one simple resistor and maybe a capacitor can do a very similar job.

[glennb]

Originally posted by glennb
...Think of Q20 as a constant current source. D26 (a red LED) acts like a zener of about 1.6V. Subtract the Vbe~=0.6V of Q20, leaving 1V across R67. A constant current of 1V / 2Kohm = 2mA will flow through it...

Actually, Q20 is more like a voltage regulator, with R67 as its load. Before Q19 begins to conduct, Q20 regulates 1V across R67.
The overall circuit behaviour is the same even with this slightly different interpretation of Q20.