Doing a class D Amp project using TL494

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circuitcity

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Joined 2009
Ok Thanks..

circuitcity said:
By increasing the gain of the error amplifier. I increase the sensitivity of the error amplifier.

So for a gain of 10, my sensitivity is 0.3 (Since my Sawtooth is between 0-3V). Whereas for a gain of 20, my sensitivity is 0.15?

So clipping occurs much faster at a gain of 20, then with a gain of 10?

So loudness will increase but the disadvantage is clipping will occur faster and give you distorted music as a result.
Click to expand...

Also is this correct?
 
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circuitcity

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Joined 2009
Teacher told me to look into closed loop instead of the type of carrier I am using. So I'll just do that.


Looking at Ledmania's Feedback circuit, he uses no low pass filtering, but just a voltage divider to bring the output to 2V for the error amplifier.

Most document say to low pass filter the feedback signal.

From my understanding so far, the feedback can be pre-filter or post-filter, which then has to be low pass filtered and since I am using full-bridge I have to use a Differential Amplifier (This is what I saw in IEEE documents).

This is what I saw in one document where they take pre-filter as feedback. It is low pass filtered via R1 and C1.

http://img407.imageshack.us/img407/2756/controlloop1.png

http://img515.imageshack.us/img515/2866/87062818.png

That's what I am thinking of doing and the output of this feedback signal gets summed with the music signal. My question is how low should you take the cut-off freq to in the feedback low pass filter?

Secondly its reasonable to want a output voltage of the diff amp to be 2V right? Or should I use a voltage divider at the output of the diff amp?

Thing that's confusing me is that, this voltage isn't fixed but charges up and down (From simulation observation). You want a fixed 2V for the error amplifier.

What about the gain of this feedback path? Unity Gain a good place to start of with?

Note: Don't think I'll be physically building this (just simulate it for the next couple of days).
 
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circuitcity

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Joined 2009
Ok what I have done so far

http://img8.imageshack.us/img8/7384/feedbackz.png

The Graphs

This graph has the switched output from each phase leg and then the low pass filtered output from the RC filter.
http://img230.imageshack.us/img230/7363/65394680.png

In this graph, it's Vout (Third graph). Which takes almost a second to reach 2V (well approx) but stays there afterwards.
http://img61.imageshack.us/img61/9070/85926960.png

So 2V (approx) enters the summing junction with the music signal.


Have I done this correctly?
 
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Pafi

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Joined 2004
snyaudio.hu/spafi/
Have I done this correctly?
Click to expand...

Of course not.
- V2out doesn't go anywhere.
- Filtered stages are overloaded by the input of the subtractor, and what is even worse: in different measure.
- 100 nF makes it too slow. Settling time is not 2 second ( :yikes: ), only 2 milliseconds, but 1: you shouldn't specify settling time to 99.8%, but tau (63%) wich is about 0.3 ms, 2: 0.3 ms is also too high.
- The basic topology is bad. Haven't you heard about PI controllers, and generally about control theory? I quess so.
- you shouldn't have to adjust DC voltage this way. Vout would be 0 if you hadn't ruin the subtractor.
- the end is missing from your scheme.
 
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circuitcity

Member
Joined 2009
Pafi said:
Of course not.
- V2out doesn't go anywhere.
- Filtered stages are overloaded by the input of the subtractor, and what is even worse: in different measure.
- 100 nF makes it too slow. Settling time is not 2 second ( :yikes: ), only 2 milliseconds, but 1: you shouldn't specify settling time to 99.8%, but tau (63%) wich is about 0.3 ms, 2: 0.3 ms is also too high.
- The basic topology is bad. Haven't you heard about PI controllers, and generally about control theory? I quess so.
- you shouldn't have to adjust DC voltage this way. Vout would be 0 if you hadn't ruin the subtractor.
- the end is missing from your scheme.
Click to expand...

oops. drew the circuit wrong and hooked the two ground together.

The subtractor I assume your talking about the differential amplifier. I just followed the electronic book said and let R1 = R3 (10k) and R2 = R4 (20K)

So your saying these resistances are too high?

Isn't the diff Amp, the proportional block part in a PI controller and a integrator circuit is for the integral block. End scheme your taking about is the integral block part?


I'll progress more slowly to understand the concept better. Again its too late for me to even attempt to build a control loop. This is just for education purpose now to try and simulate a control loop as close as possible to suit the amplifier I have built.

I've re-drawn my circuit with just the RC Low pass filter and diff amp and resistance reduced.

http://img207.imageshack.us/img207/9370/65992647.png


This is the graph i see now from it.

http://img260.imageshack.us/img260/6166/37395032.png

same thing, zoomed in

http://img21.imageshack.us/img21/3484/75105238.png
 
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Pafi

Member
Joined 2004
snyaudio.hu/spafi/
Again its too late for me to even attempt to build a control loop.
Click to expand...

You will build one, the question is will it be good or bad. A good control loop takes about 4 resistors and 2 capacitors.

So your saying these resistances are too high?
Click to expand...

Absolutely not.

Isn't the diff Amp, the proportional block part in a PI controller and a integrator circuit is for the integral block.
Click to expand...

Maybe. It can be done some different ways.

End scheme your talking about is the integral block part?
Click to expand...

I don't know what it is, since you haven't finished.
 
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circuitcity

Member
Joined 2009
I'm sorry. I should say the diff amp is more like a voltage sensor (with a gain of 1) which takes the two input to give me a single output and that then goes into a PI controller.


I saw this

http://www.engr.panam.edu/~hvasquez/Mechatronics/Lab 4_PI _controller.pdf

So they use a diff amp and the output from the diff amp goes into a proportional block and a integral block which is then summed through a summer. And I take the output from the summer into the error amplifier.


Wow. A good control loop just with 4 resistors and 2 caps (This includes the feedback path low pass filter as well?) :eek:
 
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circuitcity

Member
Joined 2009
Ok I see what you mean now.

The control book had circuits comprised of Op-amps as a PI controller and they used 4 resistors and 1 capacitor for a PI controller.

Well PI controller + Inverter..

My question is what's in the feedback path then after the RC low pass filter because I am confused now. I have a differential output when I need a single ended output. So what sort of circuit do i need to do this.

Diff amp like before?

This how I see the control loop as
http://img213.imageshack.us/img213/8120/10047089.png

And my PI controller as well
http://img122.imageshack.us/img122/1127/pi1.png


So I am clear on my PI controller now.. But the feedback path now I am confused now.
 
darkshy

darkshy

Banned
Joined 2011
phase_accurate said:
These tricks with one output transistor as common emitter and the other one as emitter-follower will not work. This IC is designed for push-pull or half-brideg SMPS operation. There is a toggle- FF inside this IC that serves to steer one transistoer during one phase and the other one during the other one. In order to be used for class-d audio they would have to be combined in wired-or fashion. There is absolutely no way (at least not known to me) around that. So your only choice is to have an additional IC that acts as an inverter between the TL 494 and an IR2110.

Regards

Charles
Click to expand...

81bas said:
TL494 CANNOT produce complementary PWM (ie, cannot generate inverted and non-inverted PWM signals). It can produce only usual PWM if you use open collector outputs mode and connect collectors together (with single pullup resistor), as phase_accurate said. You need to use XOR gates then to get complementary signals for mosfets driver (it is the best way to do it, and Eva knows it too ;)).
Click to expand...

Is this essential to use XOR gates instead "These tricks" ?

Also why not invert the square with only a single BS170 or a BC547 and a resistor ?
 
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