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Old 24th February 2013, 03:31 PM   #11
xrk971 is online now xrk971  United States
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Mark,
@Cervelorider has built many of these amps and may be able to assist if you PM him.
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Old 24th February 2013, 04:48 PM   #12
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Quote:
Originally Posted by Mark2727 View Post
Nigel,

I'm not understanding what you did. Could you give me a sketch and do you mean by "PIC", a printed integrated circuit?
Mark
I would try an RC on the shutdown pin to start with.

A PIC is a microcontroller.
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Old 24th February 2013, 05:24 PM   #13
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Hi,
I checked the schematic and it is not using the mute and also the shutdown pins. They are connected to ground means they are enable all the time.
I agreed with nigelwright7557 suggesting using a micro. You can use the Nano 8 from Basic Micro cost 2.98 and it is program in basic. You can control both pins Mute and shutdown easy with the micro. Just programming.
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Old 24th February 2013, 05:31 PM   #14
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Originally Posted by xrk971 View Post
Mark,
@Cervelorider has built many of these amps and may be able to assist if you PM him.
X

Tried him (D.S.) already. !!! He was writing me for awhile, then dropped off.

Mark
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Old 24th February 2013, 05:35 PM   #15
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Nigel/Tauro:

Are you fellows telling me there's a pin/leg on the Texas Instruments chip that is running to ground, and should be run through some sort of microprocessor instead? And that that device must be programmed in BASIC?

Interesting. But it may be beyond my technical skills.

Look...I'm just a beginner. So sorry for the dumb questions.

Nevertheless, I'm off to look at a pin out of that chip.

Wow, this gets crazier all the time. Thanks!

Mark
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Old 24th February 2013, 05:46 PM   #16
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Hmmm....pins 2 and 3.

So, we somehow run/connect those pins to this Nano 8 and program that to enable a soft start?

Mark
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Old 24th February 2013, 06:08 PM   #17
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Hi,
In the schematic the shutdown pin 2 it is connected to the + voltage means that the shutdown it is disable. Zero volts means enable. The mute pin 3 it is connected to ground. Zero means that the mute is is disable. 2 volts means that the mute it is enable. How much voltage your are using in you PS? I may come up with a schematic using the basic micro. I can make the circuit using the micro basic nano 8 if you are interesting. I will programming it for you. It is a simple program.
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Old 24th February 2013, 06:14 PM   #18
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I usually use a dropper resistor and zener to power the microcontroller as it needs very little power.
The micro is an internal oscillator so is easy to use.

I think the micro will drive those pins direct.

When I did my circuit for the IRS2092 I had to use an opto coupler to drive its reset pin but you probably shouldn't need to do thi9s.
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Old 24th February 2013, 06:39 PM   #19
xrk971 is online now xrk971  United States
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This problem must have an easier solution than the complicated schemes with external parts and programming. The application sheet says the following for the shutdown mode:

"The TPA3122D2 employs a shutdown mode of operation designed to reduce supply current (ICC) to the absolute minimum level during periods of non-use for power conservation. The SHUTDOWN input terminal should be held high (see specification table for trip point) during normal operation when the amplifier is in use. Pulling SHUTDOWN low causes the outputs to mute and the amplifier to enter a low-current state. Never leave SHUTDOWN unconnected, because amplifier operation would be unpredictable.
For the best power-up pop performance, place the amplifier in the shutdown or mute mode prior to applying the power supply voltage."

So place amplifier in shutdown mode prior to applying power. Use a 3 pole toggle switch that has the middle pole grounding an RC circuit tied to the shutdown pin with a suitable time constant to drain it prior to power on. The RC circuit takes a finite amount of time to charge and reach positive logic. The 3-pole switch then has to go to the third position ("on") to apply power to Vcc which then has to charge the RC circuit through a large R value so that the amp stabilizes before the shutdown is disabled. Assuming the amp stabilizes in 2 seconds and stated the logic value is 2 volts (p 3 of spec sheet), using the RC time constant equation: Time = 0.7 RC, set R=100k ohm, solve for C=2.0 sec/(0.7*10^5 ohm)=28 uF. Kind of a big cap, but easy with an electrolytic here.
Does this make sense?

You need a 3-pole switch, a 30 uF 35 V electrolytic cap, a 100 kOhm resistor. Problem should be solved. No programming or fancy external IC's.

EDIT: You can skip the 3-pole switch by putting a couple of 1 Meg ohm resistors in series to the positive electrolytic cap terminal to drain it to ground slower than the 100 k ohm charges it (20 x in this case). Then it is all passive.

Last edited by xrk971; 24th February 2013 at 06:42 PM.
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Old 24th February 2013, 06:44 PM   #20
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Why not just an RC as I suggested earlier ?
No need for switches.

I would also put a diode across the resisitor to discharge the capacitor on power down.
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Last edited by nigelwright7557; 24th February 2013 at 06:48 PM.
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