Is there anybody built a non feedback amplifier??

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Jan, I'm not following your point. If you reverse the amp, you reverse the fundamental and the distortion components. But the speaker will still distort in phase with how it did before you reversed the amp (i.e., out of phase with the amp distortion component), given that the speaker's distortion mechanism is unchanged.

I agree that the hypothesis is questionable, and the experimental support I've seen for this has been of rather low quality. But it IS plausible and it is testable. I just sort of wish it were tested before going to market...
 
SY said:
I agree that the hypothesis is questionable, and the experimental support I've seen for this has been of rather low quality. But it IS plausible and it is testable. I just sort of wish it were tested before going to market...
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Yeah. The basic premise is sound (no pun intended) but it seems that in terms of practical application it will be a mishmash of hit and miss and everything in between.

se
 
Charles Hansen said:


You can significantly reduce both the distortion and the output impedance of the output stage simply by increasing the number of output devices (as pointed out by another poster).
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So is distortion in output devices random and cancels the way as noise cancels when paralleling transistors in pre-pre-amps ?
 
Bernhard said:


So is distortion in output devices random and cancels the way as noise cancels when paralleling transistors in pre-pre-amps ?
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Hello -

Not at all. In this case I would suggest using a circuit simulator and making a simulation of an output stage.

My favorite simulator is MicroCap. You can download a limited version (100 components) for free. Then run the DC simulations like Douglas Self does in his book on amplifiers. This will show the linearity of the output stage very nicely. MicroCap will let you run a family of curves that change one variable at a time (two variables if you have the full version). Then try changing the number of output devices and see what happens.

If you don't want to go to all that trouble (not much, really) much of this is covered in the latest version of Self's book on amplifier design.

Good luck,
Charles Hansen
 
andy_c said:
Vo/Vi = gm * RL / (1 + gm * RL)

Let gm approach infinity. So it's an increase in local feedback. Or that's one way of thinking about it.
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that works only for fets (voltage controlled current source), tho.

for BJTs (current controlled current source), they usually don't distort as much at low-current regions (not too low tho.).
 
Why not use the Null test to find out if no feedback makes a difference ?

To my understanding the main problem of feedback is the limited bandwith, the performance of the feedback is going down with increased frequency, if the loop is shorter - local feedback - it might work better due to less delay

Also feedback is mostly a closed loop, the input watches the output and as it finds that the output is mistaken, it tries to correct and "waits" for the results and tries to correct again, but then if the bandwith is too small, the input signal has already changed before the output-watch-circuit has been successful.

Plus the feedback = output-watch-circuit can not foresee the results because it passes the nonlinear circuit again, higher orders of higher orders are created in the closed loop.

This is like the use of a cannon in a 18 century battle.
Shoot once and see that it goes too far, correct and see it goes too short, correct again and before you shoot, the other side hits you. :headshot:

Sorry for my naive way of explanation 🙄
 
andy_c said:
Not so. Just substitute

gm = 1/re for BJTs and you have the same result.
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if you parrallel 10 BJT output devices, wouldn't each BJT get 1/10th of the base current had you used just one BJT? so each BJT is outputing just 1/10th of the current. thus no increase in feedback.
 
If you have 10 BJTs all biased at the same DC current as the original single one, it's like having one BJT with 10 times the gm of the original. You can think of the loop gain as gm * RL, so that's like increasing the loop gain by 20 dB.
 
andy, I am more thinking about the incremental current. For the 10 BJTS, if you inject 10ma into their combined base, wouldn't each get 1ma base current? and if they have hfe of 10x, that would translate into 10ma incremental emitter current from each BJT. or 100ma from the combined current.

Now, that's the same as injecting 10ma base current into 1 BJT with hfe of 10x.

Is that right?

in essence, a 10BJT combo has the same current gain as 1 BJT.
 
I think that the THD reduction obtained by using multiple output devices is simply due to :

- Reduced current swing : Current gain tends to be constant for low Ic and decreases progressively/sharply for medium/high Ic so using more devices means using each device in a more linear zone producing smaller distortion components [assuming the output stage is driven from a non zero impedance]

- Increased current gain : Since current gain decreases with increasing Ic, working within a lower Ic range means higher average gain and thus higher average global feedback and lowered average THD [or higher load independence when operating open-loop] [assuming the output stage is driven from a non zero impedance]

- Reduced Vbe swing : Vbe increases in a non-linear fashion with increasing Ic. Using more devices means lower maximum Ic for each device and thus less Vbe distortion component [only applicable when voltage driving of the output stage is dominant]
 
Yes, sounds right. So if you hook your output stage straight into your VAS, then I guess you'd see no improvement. But if you put an EF or two in between, you'll approximate voltage drive (which is what I was assuming, sorry if I didn't make that clear) and then you'll see the distortion improvement with multiple parallel devices.

As Charles mentioned, you can see the effect of this with your simulator.
 
andy_c said:
You can think of the loop gain as gm * RL, so that's like increasing the loop gain by 20 dB.
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Thanks for bringing that up.

I'd been mulling that over last night and today and think it might provide the answer to feedback vis a vis Darlington vs. Sziklai.

So for a simple EF, we have REgm/(1 + REgm) (the latter portion reflecting the 100% feedback of the EF).

Charles argued that the Sziklai was two common-emitter sections tied together and because of that it must have a greater amount of feedback than the Darlington.

But the common-emitter's gain is also RCgm. The difference being that in the EF, that gain is brought back down to essentially 1 due to it having 100% negative feedback.

So if we look at the first device in a Darlington pair, it's essentially an EF whose RE is the input impedance of the second device, which will be (hfe + 1)RL

So we can rewrite the gain of the first device as gm((hfe + 1)RL).

The second device will be RLgm.

If we assume two identical devices each with a gm of 1S, hfe of 100, and a load resistor RL of 100 ohms, then we have a gain for the first device of 10,100 and for the the second device, 100.

For the Sziklai, the RC of the first common-emitter device will be the input impedance of the second common-emitter device which will be hfere. Since gm for this example is 1, and gm = 1/re, re = 1.

So gain for the first common-emitter device is 100. And the gain for the second device is 100.

Am I overlooking something here?

se
 
Steve Eddy said:
(...)So for a simple EF, we have REgm/(1 + REgm) (the latter portion reflecting the 100% feedback of the EF).

(...)So if we look at the first device in a Darlington pair, it's essentially an EF whose RE is the input impedance of the second device, which will be (hfe + 1)RL
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I'm fine up to this point. It's when you get to this:
So we can rewrite the gain of the first device as gm((hfe + 1)RL).
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that I think something's amiss. I have two problems with what you're saying. The first is just simple algebra, namely that the (hfe + 1)RL term should replace RE everywhere it appears in your first expression for the gain of the first EF. But maybe you're trying to call this the open-loop gain somehow? The second concern I have is more fundamental. It appears to me that you're trying to make the cascade of two two-ports into something it isn't - a loop, through some kind of mathematical manipulations. To me, this seems a stretch. But it could also be said that my unsubstantiated claim, namely that you could consider gm * RL to be the "loop gain", to be a stretch as well. So I'll attempt to justify that now.

For a classical feedback arrangement, you can write:

Vo = A * (Vi - B * Vo)

where B = feedback factor and A = open-loop gain, AB = loop gain

For the EF or source follower, you can write:

IL = gm * (Vi - Vo)

Setting IL = Vo / RL, you get

Vo = gm * RL * (Vi - Vo)

Comparing the last equation with the first, one could say B = 1 and A = gm * RL. Or equivalently AB = loop gain = gm * RL

In this case the "loop" is just the input voltage loop. Vi - Vbe (or Vgs) - Vo = 0. But I fail to see how you can make a loop out of a cascade, as in the Darlington case. Computing the loop gain of the CFP would involve lots of algebra. I'd prefer to use the Middlebrook technique with SPICE for that.
 
andy_c said:
Computing the loop gain of the CFP would involve lots of algebra. I'd prefer to use the Middlebrook technique with SPICE for that.
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Would you care to do it?

I'm just flying by the seat of my pants here. I'd hoped someone who really knew what they were talking about could definitvely answer the question. I guess I was hoping for too much. 🙂

se
 
Steve Eddy said:


Would you care to do it?

I'm just flying by the seat of my pants here. I'd hoped someone who really knew what they were talking about could definitvely answer the question. I guess I was hoping for too much. 🙂

se
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Okay, but it will have to wait until tomorrow night. I'm about ready to crash, and I don't get back from work until about 7:30 PM PST usually.
 
andy_c said:
Okay, but it will have to wait until tomorrow night. I'm about ready to crash, and I don't get back from work until about 7:30 PM PST usually.
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Thanks, Andy!

No rush. I'll be packing it in myself here in a couple hours and tomorrow I'll be busy stuffing and soldering circuit boards.

Oh, by the way, the question isn't simply how much feedback is employed in the CFP, but relative to the Darlington.

G'night!

se
 
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