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Old 14th August 2003, 04:40 AM   #11
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Quote:
Originally posted by SY
Not even that, since it's degenerated by (at minimum) Rbe.
Mmmm. I don't see the negative feedback there. Even though you have a resistor from base to emitter, the emitter's going straight to ground so no emitter degeneration.

se
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Old 14th August 2003, 06:11 AM   #12
sajti is offline sajti  Hungary
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Eddy,

I think the keyword is the LOOP. You can't creat loop with one stage only. The feedback You mentioned is LOCAL feedback.
The biggest problem with the loop feedback is the delay between the input and the output of the amplifier. This means that the feedback signal allways delayed compare to the input.
In case of local feedback this problem not exist. The feedback caused the input signal itself, so there is no delay.

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Old 14th August 2003, 07:26 AM   #13
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Quote:
Originally posted by sajti
I think the keyword is the LOOP. You can't creat loop with one stage only. The feedback You mentioned is LOCAL feedback.
You seem to be contradicting yourself here. You say the feedback I'm talking about is local feedback and that you can't create a loop with one stage only. But you can't have feedback without a loop. So clearly there is a loop in even this single stage. Yes, it's local feedback. But then I didn't say it wasn't.

Quote:
The biggest problem with the loop feedback is the delay between the input and the output of the amplifier. This means that the feedback signal allways delayed compare to the input.
Delay's only a problem in that it effectively reduces the amount of feedback as frequency increases. Some would argue that it's the feedback itself that's the problem, reducing low order distortion at the expense of increasing higher order distortion.

Quote:
In case of local feedback this problem not exist. The feedback caused the input signal itself, so there is no delay.
As long as there's a loop of any non-zero size there will be some delay. And any reactance within that loop will also add to the delay. Whether the feedback is global or local doesn't change that.

se
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Old 14th August 2003, 07:45 AM   #14
sajti is offline sajti  Hungary
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"You seem to be contradicting yourself here. You say the feedback I'm talking about is local feedback and that you can't create a loop with one stage only. But you can't have feedback without a loop"

There is feedback without loop. In this stage the feedback caused by the emitter current itself. As the emitter current increase, the emitter voltage increase too. The ouput point and the feedback point is same. I can inmagine loop between two points. But in this case we have only one.

"Delay's only a problem in that it effectively reduces the amount of feedback as frequency increases"

Delay can be problem at any frequency. You can measure this problem even with 1kHz signal.
In well designed amplifier the results is very-very small, of course.

"And any reactance within that loop will also add to the delay. Whether the feedback is global or local doesn't change that."

Would You explain the parts of the loop, with this single transistor emitter follower?

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Old 14th August 2003, 09:33 AM   #15
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Quote:
Originally posted by sajti
"You seem to be contradicting yourself here. You say the feedback I'm talking about is local feedback and that you can't create a loop with one stage only. But you can't have feedback without a loop"

There is feedback without loop. In this stage the feedback caused by the emitter current itself. As the emitter current increase, the emitter voltage increase too. The ouput point and the feedback point is same. I can inmagine loop between two points. But in this case we have only one.
I'm sorry but I've never heard of feedback without a loop.

In this case, the loop is the base/emitter loop with the emitter being fed back to the base via the load.

Quote:
"Delay's only a problem in that it effectively reduces the amount of feedback as frequency increases"

Delay can be problem at any frequency. You can measure this problem even with 1kHz signal.
In well designed amplifier the results is very-very small, of course.
What exactly is it that you're measuring? As I said previously, the delay effectively reduces the amount of feedback due to the phase difference between the input and output signal and when you reduce the feedback, you increase nonlinearity. You'd get the same result by simply reducing the feedback though the reduction in feedback wouldn't be frequency dependent the way it is when there's a delay involved.

Quote:
"And any reactance within that loop will also add to the delay. Whether the feedback is global or local doesn't change that."

Would You explain the parts of the loop, with this single transistor emitter follower?
Sure. The loop is formed by the base, the emitter, the load, and then back to base.

se
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Old 14th August 2003, 09:53 AM   #16
sajti is offline sajti  Hungary
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"emitter being fed back to the base via the load"

Sorry, but I don't understand it. How the emitter feed back the base, via the load?
The load is not connected to the base.

"What exactly is it that you're measuring? "

Apply 1kHz square wave on the input, and measure the collector of the first stage. You will see overshot on the squarewave. That is the delay.

Sajti
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Old 14th August 2003, 12:31 PM   #17
SY is offline SY  United States
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Quote:
I don't see the negative feedback there. Even though you have a resistor from base to emitter, the emitter's going straight to ground so no emitter degeneration.
The emitter terminal goes straight to ground, but the emitter resistance, it seems to me, should be degenerating things, otherwise you'd have infinite gain. If I think of a T model for a bipolar, there's an equivalent resistance re in series with the emitter lead.
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Old 14th August 2003, 01:46 PM   #18
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Quote:
Originally posted by sajti
Eddy,

You can't creat loop with one stage only.
What about a common emitter gain stage with little or no degeneration, the gain of which is set by a resistor or capacitor from collector to base?

In this you can conceive an open loop configuration (i.e. to impedance between collector and base), hence there is an open loop voltage gain. The feedback impedance then takes the voltage signal and takes it back to the input where it is current-added with the input signal.

Speed and distortion wise, this is probabyl similar to emitter degeneration, which is in fact a current derived feedback. But I have more trouble defining what the open loop case is here...
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Old 14th August 2003, 02:14 PM   #19
sajti is offline sajti  Hungary
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capslock,

You have right. With common emitter connection You can create feedback loop.

But as I know the emitter degeneration resistor is not feedback loop.

Sajti
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Old 14th August 2003, 05:09 PM   #20
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Default Re: Re: Re: Re: Re: Re: Re: I agree with Charlie........

Quote:
Originally posted by mikek



Correct....another 'follower' of interest perhaps is the common-base stage.....which possesses 100% current (series) derived, current (shunt) applied, negative feedback....ergo...a current buffer.
mikek,

sorry, but common-base stage does not amplify the current .......

Pavel
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