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Class D Switching Power Amplifiers and Power D/A conversion

class d amplifier current output
class d amplifier current output
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Old 29th January 2017, 02:49 PM   #11
franticDIY is offline franticDIY  Italy
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Yes.

Could you suggest me, or send in private, a possible scheme from witch to start with my studies? Do you have a tutorial or application notes on this subject?
At the moment the accuracy is not a problem.

Thanks
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Old 29th January 2017, 06:34 PM   #12
Pafi is offline Pafi  Hungary
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Quote:
Originally Posted by franticDIY View Post
Yes.

Could you suggest me, or send in private, a possible scheme from witch to start with my studies? Do you have a tutorial or application notes on this subject?
At the moment the accuracy is not a problem.

Thanks
I will try to find time for drawing a schematic. The basic is not much more than yours, only an RC feedback is missing from output voltage, series resistor on input, and an inductor would help in series with load.

Last edited by Pafi; 29th January 2017 at 06:47 PM.
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Old 29th January 2017, 08:20 PM   #13
franticDIY is offline franticDIY  Italy
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Ok thanks i also try to do something.
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Old 30th January 2017, 05:41 AM   #14
moschfet is offline moschfet  Germany
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Maybe is an old thread of me interesting for you:

INSOD-AMP
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Old 30th January 2017, 12:02 PM   #15
franticDIY is offline franticDIY  Italy
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Originally Posted by moschfet View Post
Maybe is an old thread of me interesting for you:

INSOD-AMP
It is very interesting and seems that the second amplifier has current and voltage control.
Do you have ltspice files?
Do you remember another useful information?
Seems that the current feedback and voltage feedback are connected at the same node. How do you calculate components values?
Thanks let me know.
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Old 30th January 2017, 05:55 PM   #16
Pafi is offline Pafi  Hungary
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I modified the schematic found in that topic a little.
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File Type: asc V-I_converter_model.asc (3.6 KB, 6 views)
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Old 30th January 2017, 05:59 PM   #17
Pafi is offline Pafi  Hungary
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2 limiter diodes on output should be added.
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Old 30th January 2017, 06:35 PM   #18
franticDIY is offline franticDIY  Italy
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Thanks a lot. I will try it as soon as possible.
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Old 31st January 2017, 10:01 AM   #19
Pafi is offline Pafi  Hungary
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Youre welcome!

But at second thought it would be better using current mode PWM control right in the first (or only?) control loop.

It is very straightforward.
Determining L: it should provide 1.4 Ap with some margin at 20 kHz on 50 V peak sinusoidal voltage. Therefore L=50V/2A/20kHz/2/pi=200 uH.
Feedback of PWM:
It should be PI type. (Type II in other terminology.). P component can be determined by the requirement that voltage slope must always be slower than the slope of triangle. Feedback is a current controlled voltage source. (How to make it in real life is a question left later.)
V/I_P<(2Vpp/2us)/(100V/200uH)=1V/us /
0.5A/us=2ohm, but in order to maintain margin I would choose 1 ohm.

To be continued...

Last edited by Pafi; 31st January 2017 at 10:04 AM.
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Old 31st January 2017, 10:33 AM   #20
Pafi is offline Pafi  Hungary
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Without any I(ntegrating) feedback component loop gain is
G(s)=RP*Gpwm/(L*j*s+Rl)
Where Gpwm=100V/2V=50, RP is the transresistance of P(roportional) component of feedback, and Rl is load resistance.

For maintaining 10% amplitude accuracy you need 2...9 of loop gain depending on the phase angle. I will try setting phase angle close to 90 degree, so maybe |G|=2 will be enough. (Very optimistic assumption.)

At 20 kHz lets solve |G|=2 for Rl, to find the highest possible load resistance that is still controllable!

To be continued...
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