Audio amplifier (part 2)

Hello, it is me again ...

After "jouling" some transistors, I decide to move to LTSpice. Which is funny as I am doing electronics as hobby to change from being in front of computers all day long !

I have some questions on this schema:

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Simulation is here:
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I have only 2N3055 at disposal so I use a Sziklai pair (at least I think!)...

BC546 and 556 will be replaced by bigger ones

Questions:

1/ Any big mistakes ?

2/ To move from A to AB, I think I need to change the value of R1 so Q1 conducts more or less. By doing that I will have DC on the load, so I have to put a cap no ?

3/ What can be improved before building it ?

Thanks for your help,

Olivier

Reply to
Olivier Scalbert
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I don't think so. It looks like you have a basic understanding of what you are doing. Of course, that strongly depends on what you are trying to do.

You have two separate but interconnected problems. How to change the bias point for the two output transistors, and how to produce a zero signal output of zero volts.

If you change R1 and R2 at the same time, you should be able to change the bias point and still maintain the desired output voltage. But this circuit depends strongly on component matching and power supply voltage stability. More practical versions use negative feedback to correct for variations in components and supply voltages.

First, you should make a functional list of "improvements" (what would define a functional improvement in the operation). Then you are in a better position to consider circuit changes and how they interact with those functions.

First, I think you should explore frequency response. Your test waveform is 100 Hz. Change that to 10,000 Hz and I think one of the weaknesses of this circuit will start to show up. Both of your output devices have lots of turn on drive, but no turn off drive, so once they are on, they will be very slow to turn off. This will produce increasing distortion and higher bias current as frequency rises.

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Regards,

John Popelish
Reply to
John Popelish

You *really* need to add resistors across the base/emitter of the output transistors, otherwise speed will be an issue, especially once you put this thing in an overall feedback loop.

I would tend to make them on the real low side, considering driving 3055's to thier max needs 1A. Losing say, 100ma in these resister would be, 0.8/.1 = 8 ohms. Say, 10 ohms. However, this gives a tad of power in the drivers, so you may want to consider anything from 20 to 100ohms. I wouldn't go over

100 ohms.

Kevin Aylward snipped-for-privacy@kevinaylward.co.uk

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Reply to
Kevin Aylward

Thanks John for your help.

Do you think it is possible to change the design in order to have only one potentiometer to change the bias of the transistors, instead of changing R1 and R2 ?

For the negative feedback, I imagine that I have to measure the DC level at the load resistor, compare it with 0, negate it and re inject it somewhere in the input bias, but I do not see how ...

I have explored the frequency response. It is still ok at 10k and even more . Pehaps the spice model of my components are too ideal. I should put some parasitic inductances and capacitors ... or find other models ...

Olivier

Reply to
Olivier Scalbert

Thanks Kevin. Yes you are right ! see:

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Is it because there are too many charges into the base ? And with the resistors, it is easier for the charges to go out of the transistor ?

Olivier

Reply to
Olivier Scalbert

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Not necessarily.

As John Popelish noted, using negative feedback and providing a little
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Reply to
John Fields

Yes, though it might produce other problems. The most common way to produce the approximately 3 diode drop bias voltage between the driver bases is to multiply the base emitter drop of an additional transistor, with the multiplication ratio set by a potentiometer (instead of connecting 3 diodes in series).

The Vbe multiplier is described in this reference:

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You add a potentiometer in series with R1 and R2 to make the ratio variable over a small range to set your output device bias current. You choose the resistor value to make the total current through the resistor divider to be about

1/10th of the current through the transistor, so that the divider current times the current gain of the transistor dominates the transistor conduction. The potentiometer value compared to the fixed resistors should allow only enough multiplier voltage variation to cover all the part tolerances and a bit more. You can simulate the potentiometer as two resistors in series that add up to the pot total, but making the extreme value no lower than about 1 ohm (to keep the simulator happy, since it doesn't like zero ohm resistors, and this is pretty close to the contact resistance of many small trim pots).

Unfortunately, this bias generator eliminates the feed point you are using for the amplifier that best balances the input offset from zero. You could get his back by using the adjustable Vbe multiplier to only replace the pair of diodes on one side of the feed point, making only the drop of that pair adjustable.

Or connect the output load in series with a capacitor so that it is not possible for the amplifier to pass DC through the load. If the load does not need to have a connection to the supply ground, you might use a pair of capacitors, in series across the total supply, and reconnect the grounded end of the load to the common point in that series string. This places a predictable DC bias across he capacitors, and they help filter the real supply.

Since your amplifier stage has a gain very close to 1, just adding more negative feedback would not do much good. You also need to add gain to that operation. Say an opamp connected as an inverting integrator. But normally, this additional gain and inversion is produced by additional transistors that make up a gain block ahead of this unity gain output current booster.

Including the transistor collector currents? I think you should start to see some distortion in the current wave forms with this configuration.

Are you using the models that came with LTspice?

A couple suggestions that would make the circuit more practical:

Connect the lower transistor pair without the 0.5 ohm resistor between them. keep that resistor between them and the output node, so it is more similar to the upper darlington.

Add a pair of 10 to 100 ohm resistors between base and emitter of the two output transistors to make them turn off faster.

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Regards,

John Popelish
Reply to
John Popelish

Thanks John, I need some times to digest all you told me !!!

Olivier

Reply to
Olivier Scalbert

"Too many charges" is not quite the right concept.

It is necessary to store charge in the base emitter junction to turn the transistor on. It is necessary to remove that charge to make it turn off quickly, after it has been on. The junction will consume that charge, eventually, if you do not drain it out, but that is a component and temperature dependent process and most applications cannot tolerate the long and variable self discharge time.

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Regards,

John Popelish
Reply to
John Popelish

To cool ! Thanks John.

Some points:

- I am stupid as I focus on extracting DC from output .. instead of comparing output with input !

- I have never played before yesterday with LTSpice. It is incredible. You give me a circuit in ascii, and I can run it, modify it ! Wow !

- I do not know how Mosfet works. In fact they afraid me a little. Your amplifier as a gain of 5. Is it R6/R4 ?

- does it "sound" good ?

Olivier

Reply to
Olivier Scalbert

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They are, in effect, voltage-controlled resistors with the
drain-to-source resistance being lowered more and more as the gate
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Reply to
John Fields

--- ^^^^^ Oops... source

JF

Reply to
John Fields

Basically yes. Essentially, there is a big cap. The driver transistor can charge it up, but without a turn off it, can't discharge much.

Personally, I prefer full class Ab drive all the way, that way you can source and sink large currents to turn the thing off!.

Kevin Aylward

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- SuperSpice

Reply to
Kevin Aylward

Hi John,

For the FFT, just right click in the simulation screen and select FFT in the popup menu.

Olivier

Reply to
Olivier Scalbert

Also increase stop time in the simulation box to have more points. BTW how to convert harmonic levels in FFT to distorsion in percentage ?

Olivier

Reply to
Olivier Scalbert

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Thanks! :-)

JF
Reply to
John Fields

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OK, thanks.
Reply to
John Fields

Ok thanks ! Olivier

Reply to
Olivier Scalbert

Hi,

In fact, how do you compute R5 and R8 ? Olivier

Reply to
Olivier Scalbert

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Empirically. ;)

In the simulator, what I did was (with 0V into R1) arbitrarily choose R6
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Reply to
John Fields

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