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Old 1st October 2006, 07:35 PM   #1
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Default Feedback basics

Since when I touched a little bit some feedbacks I got an attack on my personality let me touch it deeper snd explain some feedback basics.

Feedback means some part of a signal from output of a device is applied to the input of the same device. Negative feedback means to subtract some part of output signal from the input signal, positive feedback means some part of output signal is added to the input signal.

Negative feedback decreases a total gain of a device, while positive feedback increases it.

Why to decrease a gain?

A first, all known active elements such as vacuum tubes and transistors are not absolutely linear, they distort signals they amplify. A negative feedback makes an amplifier more linear; subtracting distortions made by amplifier itself. The overall linearity of such amplifier is better, and its frequency response is better, which means its gain is more equal on a broader band of frequencies.

We may take for a feedback both output current and output voltage, and apply to both in parallel and in series with input signal. It gives us a great flexibility to increase or decrease output resistance (impedance), to increase and decrease input resistance.

The picture shows several types of a negative feedback (for simplicity purposes I used transistors):

Click the image to open in full size.
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Old 2nd October 2006, 03:38 PM   #2
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Just curious, by what mechanism is the input resistance changed?

By how much?

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Old 2nd October 2006, 03:48 PM   #3
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Quote:
Originally posted by bear
Just curious, by what mechanism is the input resistance changed?

By how much?

_-_-bear

Easy. In the last pic, you will see that any rise in input voltage (at the base) will, through the feedback, give an almost equal rise in emitter voltage.

Because the emitter voltage follows the base voltage, there is no change in input current as a result of the voltage rise at the base (the Vb-e remains the same, almost).

According to ohms law, input resistance is (delta) input voltage divided by (delta) input current.
No change in current with change in voltage means infinite resistance.

Jan Didden
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Old 2nd October 2006, 04:59 PM   #4
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Default re: feedback basics

The input resistance presented to the input source driving the base is:

Rb + ((beta + 1) * Re).

If no base resistor is present, then Rb becomes rbb', the internal base spreading resistance. As an example, if Rb = 0, rbb' = 10 ohms, beta = 100, and emitter resistance Re = 50 ohms, then Rin = 5,060 ohms.

In order for the circuit to present infinite input resistance, the transistor forward current gain, beta, would have to be infinite, OR, Re would have to be infinite. With real world finite values, the input resistance is always finite.

Have I answered the question(s)? BR.
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Old 2nd October 2006, 05:41 PM   #5
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Default Re: Feedback basics

Quote:
Originally posted by Wavebourn
[BThe picture shows several types of a negative feedback (for simplicity purposes I used transistors):
[/B]
You shouldn't, we're in the "Tube" forum
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Old 2nd October 2006, 05:48 PM   #6
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If the base resistance is 5060ohm (this range), where does the "negative base impedance" coming from?
It is usually cured by base stoppers about 100-220ohm. 220ohm is small compared to 5060ohm?
What is really happening in the case of "negative base impedance"? Why is base/grid stoppers are needed?
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Old 2nd October 2006, 06:11 PM   #7
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Quote:
Originally posted by lumanauw
If the base resistance is 5060ohm (this range), where does the "negative base impedance" coming from?
It is usually cured by base stoppers about 100-220ohm. 220ohm is small compared to 5060ohm?
What is really happening in the case of "negative base impedance"? Why is base/grid stoppers are needed?
http://www.diyaudio.com/forums/showt...311#post424311

The above thread discussed this issue, and my answer is included. Basically, the negative resistance appears due to the emitter impedance, Ze, being reactive as well as resistive, and the forward current gain, beta or "hfe", being complex, not pure real. The product of two complex numbers, can contain a negative real part (resistance) even if both resistances are positive. I hope this helps.
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Old 2nd October 2006, 06:25 PM   #8
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Default Re: re: feedback basics

Quote:
Originally posted by Claude Abraham
The input resistance presented to the input source driving the base is:

Rb + ((beta + 1) * Re).

If no base resistor is present, then Rb becomes rbb', the internal base spreading resistance. As an example, if Rb = 0, rbb' = 10 ohms, beta = 100, and emitter resistance Re = 50 ohms, then Rin = 5,060 ohms.

In order for the circuit to present infinite input resistance, the transistor forward current gain, beta, would have to be infinite, OR, Re would have to be infinite. With real world finite values, the input resistance is always finite.

Have I answered the question(s)? BR.

I think he wanted to know what the feedback did to Zin. That is what I tried to explain in understandable terms. I may or may not have succeeded...

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Old 2nd October 2006, 06:39 PM   #9
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Hi, Claude

Thanks for the explenation Make things alot clearer for me.
I got 3 questions :

1.Is there any other way to deal with this besides base stoppers? You said Re cannot help.
2.Is this only happens with EF? Why is that? Common emittor doesn't experience this?
3.Why I never see C in the attachment below? Inductor in place of base stoppers resistor?never see C in the attachment below? Inductor in place of base stoppers resistor?
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Old 2nd October 2006, 07:19 PM   #10
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Quote:
Originally posted by lumanauw
Hi, Claude

Thanks for the explenation Make things alot clearer for me.
I got 3 questions :

1.Is there any other way to deal with this besides base stoppers? You said Re cannot help.
2.Is this only happens with EF? Why is that? Common emittor doesn't experience this?
3.Why I never see C in the attachment below? Inductor in place of base stoppers resistor?never see C in the attachment below? Inductor in place of base stoppers resistor?

But, David, you see the C often, its the miller comp cap!
And sometimes you see the inductor in the form of a bead.

Jan Didden
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