lorki

modified 10 years ago

AB Amplifier without negative voltage NEW

961

16:10:09

The Input ground level is set to Vcc/2 by the 1:1 voltage divider. With this you haven't got any negative voltage in the Input thus you don't need a negative Vcc. 

Now with Cap on output to get rid of DC offset. And also with 9V so you see how it is useable with a 9V block battery.

published 10 years ago

rbrtkurtz

10 years ago

Put another 470uF on the output, and you'll get rid of that DC offset.

lorki

10 years ago

Okay I'll do it

thebugger

10 years ago

This is not class AB but just class B and what you call no negative voltage is actually called biasing which you dont really need because thats how class B amps operate. The NPN transistor amplifies the positive cycle while the pnp amplifies the negative one. The two diodes are there to minimize the crossover distortion. Remove the 100k resistors and the amp will still work the same way. And also you wont need negative VCC because you have a decoupling capacitor at the output. Once you remove it only then you will need a split rail power supply to operate this puah pull amp.

hurz

10 years ago

The 100k are indeed not needed. The diodes and 80Ohm are voltage divider enough. But this thing is in AB opreration and this is what the headline says. This is not a pure class B amp. Its idle current is about 90mA. You can make a pure class B out of it but for the price of higher crossover distortion. and this one as is has 0.23% THD at this given operating conditions.

lorki

10 years ago

As stated in my teachers book its an AB hybrid. A because it also amplifies lower voltages than 0.7V. B is clear to see

lorki

10 years ago

I always wonderd what THD is standing for. Is it like the noise of the amp? (never learned this in school)

hurz

10 years ago

Total Harmonic Distortion, the ouput of an amp does not only have the wanted output, in your case 1kHz spectral signal. There are also harmonics caused by nonliniarities of your amp. The ratio in percent of unwanted and wanted output is expressed in percent. e.g. 1% THD means 99% 1kHz sinwave and the rest 1% are unwanted harmonics.

Sine_eyed

10 years ago

All this techno-talk and electro-rhetoric, but nobody noticed the input signal isn't amplified at all..

rbrtkurtz

10 years ago

That's because you're not going to get voltage gain out of it.

hurz

10 years ago

More techno-electro-shit for Sine, gain is the ratio of output power to input power! 18.1mW in 48.4mW out. Not a big gain, but its gain. Hope that helps

lorki

10 years ago

Gain increases with increasing Vcc. It may not be a huge voltage difference (In vs Out) but the amplifier drives much more current -> more power (V*I or I*I*R)

rbrtkurtz

10 years ago

Exactly. You can almost think of it as buffer. Let's say you've got an op amp that's output swings from 0 to 9V, but it's only capable of sourcing 40mA. That's not going to drive an 8 ohm load by itself. Put a push-pull on the output, and it's a different story. Now the output still swings from 0 to 9V (or -4.5 to 4.5 after you AC couple it), but the push-pull can easily drive that 8 ohm load. So, while there isn't voltage gain, there is power gain.

hurz

10 years ago

Increase of Vcc wont help. (It would just cause a higher power disipation at transistors but not given to the output). The gain of your circuit is fixed! The only way to increase the ratio of output to input is to change the impedance of output. E.g. derease 8Ohm to 4Ohm will double the output current and according to I*I*R it will double the output power to 88mW.

thebugger

10 years ago

How is this class AB amp when it doesnt have the class A part.

hurz

10 years ago

Reduce the input voltage to 0V and you can see the "class A part" which is physicaly not present, its 90mA idle current. Both transistors are working against each other. One is using the other as working resistor. And the decision when its class A, AB or B is fuzzy. I dont think there is a clear definition. But in this case i would put it into class AB. And if you increase both 80Ohm to lets say 500Ohm i would put it into class B.

thebugger

10 years ago

Yeah but it lacks the class A part. In class AB mode a class A stage should amplify the signals in the lower end of the output power so that it is more linear at lower powers where inefficiency does not really matter. Once it gets past a certain threshold only then the class B stage kicks in and amplifies. Well here the class B part amplifies all of the signal so no class A is involved therefore its not a Class AB but a Class B only. Correct?

hurz

10 years ago

You dont make a class B amp to a class AB by adding a class A stage! This is not what the class definitions are all about. Its not adding letters A + B to get AB. Its e.g. a class B stage which turns into AB by changing the operating point.

zorgrian

9 years ago

Does he actually listen to your advice? Read this stuff from a thousand websites, actually read books?

hurz

9 years ago

I don't know, maybe. So many times of discussions about his bullshit theories. Anyway, if he is still a stubborn, others might take it as useful information.

Ernar

9 years ago

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