Queries regarding this circuit:

  1. I am able to say that this circuit has negative feedback but my procedure is disorganized. I want a step wise procedure to identify the polarity of feedback.
  2. In this circuit the 10K resistor is acting as feedback (correct me if I am wrong) but the resistor I am unable to fit the whole thing in format of basic amplifier and feedback amplifier for shunt shunt feedback (again correct if I have wrongly identified the type of feedback topology) I mean the ports of feed back resistor are not lying at end like in below diagram but in some what middle thus how will I do feedback analysis of the circuit? Consider the circuit shown is incremental model of actual circuit where BJT is biased in active region

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  • \$\begingroup\$ Does anything I said make sense to you? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 20 '19 at 21:58
  • \$\begingroup\$ @Sunita - The circuit is PROBABLY wrong as drawn (or incomplete and unable to be dealt with without more information). Components should have names / labels so they can be referred to unambiguously, Call the right hand 10k with V2 across it = R4. This is connected from output to ground. If + and - are intended to be the power supply = Vcc then R4 cannot be involved in any way. If i2 does NOT flow from Vcc then there is no obvious source. If this is a fragment of a circuit then you need to explain all relevant conditions in your question. . \$\endgroup\$ – Russell McMahon Jun 21 '19 at 5:23
  • \$\begingroup\$ The circuit is incompletely/poorly labeled. What's the input and what's the output? I'm going to make the assumption that V2 is the supply voltage, and V1 is the output voltage. In that case there is no negative feedback. V1 has no path to the input at the base of the BJT. \$\endgroup\$ – Shredder Apr 16 at 4:38

Since Vbe =0 it cannot operate.

Since Vce= 0 it cannot operate.

To invert input voltage with respect to output collector voltage (which is always the case for relative changes in voltage being both positive with respect to emitter or 0V, when DC biased properly.

Furthermore, the NPN collect Vce must be >= Vce(sat)for some linear operation. Thus the 10k on the right must be not to ground but to Vcc.

In that case any resistor from the inverting collector output to the base forms a negative feedback. The resistance ratio determines how much negative feedback is done which affects voltage gain. But transistors are not as ideal as Op Amps so they ALSO need some input bias current, which I suggested and you ignored in previous questions with a pullup from Vbe to Vcc.

The same analysis works for FETS using the inverting Drain output.

SO let me ask you what Vcc is needed to replace the ground symbol on the left side in order to begin to operate as a negative feedback amplifier assuming Vbe=0.6V @ Ic=1mA.

Let’s assume The Rb and Rload are both tied to Vcc. What is that Vcc for Vce=1V. THen you have the minimum voltage for these R ratios to get negative feedback for DC and an AC coupled input Vac signal.

To clarify, what you have shown is some textbook AC equivalent circuit without any bias voltages. You did not mention this, so I thought I would so you understand the differences in your thought process. Either follow the book or my previous answers and try not to confuse AC circuit analysis with real DC operating conditions what work better.


For a simple thought experiment with rising input voltage causes rising input current and sinking downward collector current and voltage thus inverted collector voltage changes relative to the base. Thus negative feedback via Rcb.

To be more emperical

Let the Thevenin equivalent circuit of 1k+10k = Rfb +Vfb

If I1 and I2 are constant and Ic=hFE*Ib

Vfb = (I2 -Ic)Rfb as I2 is shared. ....(1)........
Since Ib=I1 =Ic/hFE. And Vc=Ic
load ( load = 10k//1k+10k).
You can conclude from the 1st equation Vfb is proportional to negative Ic and thus inverting or negative feedback.

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  • \$\begingroup\$ "Since Vbe =0 it cannot it cannot operate", but in the problem statement "Consider the circuit shown is incremental model of actual circuit where BJT is biased in active region" Be nice. You are disagreeing with the OP's professor's approach to the problem. \$\endgroup\$ – TimWescott Jun 20 '19 at 20:13
  • \$\begingroup\$ Yes that is safe, but perhaps not stated , so I did \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 20 '19 at 20:13

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