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HARD Feedback question !!

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mtkee2003

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15 Point for everyone who give true answer!

hi



1- what is the feedback topology in this circuit?

2- what is the Beta (feedback factor) in this circuit?

AND WHY?

regards
 

_

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Re: Feedback question

Is R_C meant to be connected to voltage source?
 

ptmsl

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Re: Feedback question

I believe it is a voltage-series type of feedback since the output voltage (assuming infinite load resistance) is mirrored at the collector resistor that is in series with the input. As for the feedback factor it is close to R_C/R_L=1 assuming that the transistors are "ideal" (very large DC and AC gain and large output resistance). For a more accurate model you must enter the transistor model that you use and see what you get.
 

mtkee2003

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Feedback question

hi "_"

this is the ac circuit and biasing circuit is neglected.
regards

and this is not voltage-series feedback
 

v_c

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Re: Feedback question

This looks like an emitter follower topology. The output resistance seen by R_L is the total resistance in the base divided by the total \[\beta\]. The total amplification factor \[\beta = \beta_{pnp} \beta_{npn}\]. The output resistance is \[10k\Omega/\beta\]. I agree that it is voltage feedback since the variations in the output voltage get reflected back to the base of the NPN transistor. Why? because in the large signal sense, the base voltage is with 0.7V of the emitter voltage. Note that due to this the phase of the base voltage is the same as the phase of the emitter voltage. In the small signal sense these signals are pretty much identical -- so the voltage feedback factor is nearly 1 (usually a little less since \[\beta\] is finite).
So this is pretty much an emitter follower but the output impedance is inversely proportional to \[1/\beta^2\] instead of \[1/\beta\].

Best regards,
\[v_c\]
 
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mtkee2003

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Re: Feedback question

Added after 2 minutes:

v_c said:
This looks like an emitter follower topology. The output resistance seen by R_L is the total resistance in the base divided by the total \[\beta\]. The total amplification factor \[\beta = \beta_{pnp} \beta_{npn}\]. The output resistance is \[10k\Omega/\beta\]. I agree that it is voltage feedback since the variations in the output voltage get reflected back to the base of the NPN transistor. Why? because in the large signal sense, the base voltage is with 0.7V of the emitter voltage. Note that due to this the phase of the base voltage is the same as the phase of the emitter voltage. In the small signal sense these signals are pretty much identical -- so the voltage feedback factor is nearly 1 (usually a little less since \[\beta\] is finite).
So this is pretty much an emitter follower but the output impedance is inversely proportional to \[1/\beta^2\] instead of \[1/\beta\].

Best regards,
\[v_c\]
hello my friend.
it is a DARLIGTON.
i'm sure that it is not a voltage-series feedback!
and also i'm sure that Beta is equal to -1/R_C.
regards
 
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v_c

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Re: Feedback question

Yes -- I know it is a Darlington pair that is why the beta for the transistors multiplied to give you a total beta.

I guess I got confused by your question -- what I meant to say is that the gain of the circuit is approximately one (not the feedback factor). Gain is beta/(1+beta), where beta is the feedback factor
 

mtkee2003

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Feedback question

hi v_c,
yes gain is (beta*A)/(1+beta*A), where beta is the feedback factor and A is forward gain.

and also i'm sure that Beta is equal to -1/R_C.

thanks
regards
 

mtkee2003

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Re: Feedback question

hi
can anyone solve this hard problem?

ifound that Beta is equal to -1/R_C.
regards
 

ptmsl

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Re: Feedback question

Are you sure that the feedback factor is depenent only on the value of one resistor? Output is voltage driven. The only relation between the output voltage and the input feedback is the voltage between positive supply and emmiter of the input transistor depended on the ratio of the R_E and R_C resistors . The feedback would be 1/R_C only if the output is current driven and the output current is sampled by R_C. Then Vfeedback=Iout*b=Iout*1/R_C so b=-1/R_C
 

rkodaira

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Re: Feedback question

The name of this transistor configuration is "SZIKLAI CONNECTION" or Complementary Feedback Pair (CFP), also named Complementary Darlington.
Used in output stage of power amplifiers. The advantage is that there is only one Vbe and not 2x Vbe as in the common Darlington connection. The gain is the product of the transistors gains. This was invented by George Sziklai.
 

vicky

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Re: Feedback question

if i amnot wrong itys look like some sort of darlington pair is it right
 

mtkee2003

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Re: Feedback question

hi

ptmsl said:
Are you sure that the feedback factor is depenent only on the value of one resistor?
yes, i'm sure.
and also i'm sure that this is a current-voltage feedback topology.
thank you my friend.

rkodaira said:
The name of this transistor configuration is "SZIKLAI CONNECTION" or Complementary Feedback Pair (CFP), also named Complementary Darlington.
Used in output stage of power amplifiers. The advantage is that there is only one Vbe and not 2x Vbe as in the common Darlington connection. The gain is the product of the transistors gains. This was invented by George Sziklai.
thank you for your info. but my question is about the Beta(GAIN) of feedback network.

magnetra said:
That is not a darlington pair, there is no DC biasing for the transistors used.
already i said that i only draw the ac circuit.

again i repeat my question:
1- what is the feedback topology in this circuit?
2- what is the Beta (feedback factor) in this circuit?

AND WHY?

regards
 

teteamigo

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If the previous circuit have topology current-voltage what is the topology of the next circuit?

Think if you are really correct.
 

teteamigo

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mtkee2003 said:
hi

ptmsl said:
Are you sure that the feedback factor is depenent only on the value of one resistor?
yes, i'm sure.
and also i'm sure that this is a current-voltage feedback topology.
thank you my friend.
TOPOLOGY [SUM-SAMPLE] : VOLTAGE-VOLTAGE

feedback factor is a voltage gain between output and voltage in resistor.

Then Beta=1

Added after 5 hours 32 minutes:

I'm sure that my analysis of feedback topology by [sum-sample] is correct.

But this isn't enough, because I have to guarantee that I'm in the conditions to apply the feedback theory:

1- Feedback diport should be unidirectional

To topology voltage-voltage, i can't guarantee that this is correct.
And the aproximation make us fall in error.

I present the explanation using topology [sum-sample] current-voltage, for this you have to see the Norton equivalent of source+ input resistor of amplifier.

I tryed to solve using current-voltage topology, but i fall in same problem that in voltage-voltage, the feedback block isn't sufficiently unidirectional.

the feedback factor is -oo (minus infinite).

for input null:
Vout=-A*Vout <=> Vout=0 when amplifier works in linear region

When We put a source or active circuit at output, the amplifier can go out of the linear region, then saturates.
 

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