This is my homework problem. I tried to solve this problem but i got it wrong. Here is the problem:

enter image description here

As it is clear from the circuit that we are taking voltage at the node V0 and some fraction of output voltage is feeding back into input. So according to this observation it should be Voltage-Series OR Shunt- Series topology.

Now my question is that is there any particular convention that i need follow? As here while writing the topology i followed Output-Input i mean first i wrote the Output (Voltage) then Input (Voltage).

However this answer suggest that there is no such convention. And if you see this lecture By Razavi Sir You will find the he is first writing the Output and then Input.

According to answer for this question they are first writing the input and then output.

If anyone can make it clear it would be really helpful.Thank you

  • 1
    \$\begingroup\$ +1 for the research \$\endgroup\$
    – Huisman
    Dec 14, 2019 at 21:55
  • \$\begingroup\$ @Huisman Thanks for Edit. \$\endgroup\$
    – Anurag Rag
    Dec 14, 2019 at 21:56
  • \$\begingroup\$ upvote for identifying the post as a being homework related \$\endgroup\$
    – jsotola
    Dec 14, 2019 at 22:48
  • \$\begingroup\$ I wonder if the answer actually implies a positive and a negative feedback, like, for C you have "series - shunt", does that means that the series one is positive and the shunt one negative? \$\endgroup\$
    – jDAQ
    Dec 14, 2019 at 23:22
  • 1
    \$\begingroup\$ Their argument is that the 1st amplifier does not have its feedback signal connected to AC ground and so is "shunt" while the second does have so is series shunt. Whatever. || Voltage - voltage = series shunt. . | Whatever. again. || VV series shunt, CC shunt series, VC series series, CC shunt series. | And again. || JDAQ's reference is useful. \$\endgroup\$
    – Russell McMahon
    Dec 15, 2019 at 12:41

6 Answers 6


Firstly, let us look at some equivalent terms

series-shunt:voltage series or voltage controlled voltage source (VCVS)

series-series:current series or voltage controlled current source (VCCS)

shunt-series: current shunt or current controlled current source (CCCS)

shunt-shunt: voltage shunt or current controlled voltage source (CCVS)

To identify the type of amplifier topology, identify the feedback resistor first. If the feedback resistor is directly connected to the output, the type of sampling is series or voltage sampling. Next, the type of mixing is to be identified. If the feedback element is connected in series with the input, it is series mixing. If the feedback element is connected directly to the input, the type of mixing is shunt. Here, the topology is series-shunt. topology

  • 1
    \$\begingroup\$ Good explanation... Only, there is a small complication in this topology. R1 introduces a local negative feedback (emitter degeneration) in the first stage that controls the current; Rf introduces a feedback that controls the output voltage. Actually, two currents are summed and flow through R1: the first is the emitter current of the input transistor; the second is proportional to the output voltage that is converted by RF to current. It looks like some kind of a mixed negative feedback... \$\endgroup\$ Dec 17, 2020 at 12:37

C) is correct

These Input-Output nodes are Series-Shunt when you consider the definition for a shunt means in parallel or "Sharing the Node". In this case, the sharing is Feedback Current.

The Input is in series with Vbe1. The Output Vce2 is in shunt

SERIES means there is some voltage OFFSET. Vo= (offset+Vin) * gain

Design Rule: All linear designs must be defined by GAIN and OFFSET (with some tolerances)

I never use this series-shunt terminology due to the confusion or lack of common wisdom and the order of terms ought to be In:Out & not Out:In, only because it more intuitive, not because of any Law of Applied Physics ;) or commercial-common practise. AFAIK.

Furthermore, a SERIES INPUT here is problematic because the offset Vbe is unstable and depends on the IC1 collect or current and temperature shift. So if you have any high gain the output is unlikely here to be useful as the output voltage never will end up centred @ Vcc/2 . This is corrected in Op Amp series input (non-inverting) circuits and may use precision-fixed Vref or resistor ratios. Here the DC input bias must be precisely tuned for Vbe, hence this design as shown is useless. Proof: If you manage to tweak the input DC within 1 mV to get a symmetrical full swing with a gain of say 50 and the Vbe changes say -2mV/'C then this Voltage input offset (Vio/'C) Sensitivity is amplified with your high V-gain.

p.s. @LvW is correct to prefer V-controlled I-feedback as the config. type.


Vo is in parallel with the input the feedback network (imagine a 2 port network with RF and R1). So the feedback must be either series-shunt or shunt-shunt.

At the input, the amplifier isn't amplifying Vs directly. It is amplifying Vbe of the first transistor. Vs = Vbe - V(R1). Therefor Vbe = Vs + V(R1). Since V(R1) is affected by the voltage from the output Vo via RF, the feedback appears in series to the source, Vs. Therefor the feedback is series-shunt.


I must admit that I do not like terms like "series-shunt" ....etc. Some authors use this kind of feedback classification in the sequence output-input and some others vice versa. Really confusing!!

Therefore, I prefer to describe the situation unique and clearly:

Outdated: [In the actual case we have Voltage-controlled Current feedback" (because two currents meet each other at the emitter of the 1st transistor).]

Corrected: In the actual case we have "Voltage-controlled Voltage feedback" (series-shunt).

Verification: The feedback action influences the emitter voltage. As a consequence, the input resistance at the base of the 1st transistor increases. According to system theory, such an increase is an indication for voltage feedback.

This can be also verified if we compare the two-stage amplifier with an opamp (base=non-inv. input, emitter=inv. input). The principle function of the two-stage amplifier can be compared with a fixed-gain non-inv. opamp with negative feedback (which has voltage-controlled voltage feedback). I think, the relatively low input resistance at the emitter node does not change the principle of feedback as voltage.

Comment These are pure formal considerations only because, in practice, the circuit will not work satifactory. The base of T2 (identical to the collector of T1) will need a DC voltage 0f 0.6..07 volts. Hence, the collector-emitter voltage of the first transistor will be too small.



simulate this circuit – Schematic created using CircuitLab The equivalent circuit is as above. The voltage Vo is sampled here. The Rf and R1 form a voltage divider to give you a fraction of the output voltage (feedback_fraction * Vo)

feedback fraction = R1/(R1+Rf)

The sampled voltage is in series with the input voltage. Hence this feedback is of the type Series-Shunt. Series since the sampled voltage is in series and shunt because the output voltage is sampled. It is voltage controlled voltage source

If it was a shunt-series the ouput current would be sampled and the feedback would be parallel to the input voltage. To clarify if the current were to be sampled there would be a resistor after the second current source(as current needs to be converted to voltage first by passing it through a resistor). The voltage across this resistor is a measure of the ouput current and there will be another resistor from this voltage connecting to the input in parallel (feedback being current). It would be a current controlled current source. An example of CCCS. enter image description here

  • \$\begingroup\$ For the first circuit - don`t you think that the (low) input resistance at the E-node of T1 would be part of the feedback factor? \$\endgroup\$
    – LvW
    Dec 17, 2020 at 18:46

Obviously, it is clear to everyone that the emitter voltage of the first transistor is subtracted from the input voltage in a series manner. Also, a part of the collector voltage of the second transistor (the circuit output voltage) is applied in parallel to the emitter of the first transistor. This part is seemingly R1/(R1 + Rf) if there was not the significant emitter current of this common-base stage as it is seen from the side of the output.

Thus, two currents are summed and flow through R1: the first is the big emitter current of the input transistor; the second is proportional to the circuit output voltage that is converted by RF to current.

So R1 introduces a local current-type negative feedback (emitter degeneration) in the first stage that controls the collector current of the first transistor; Rf introduces a voltage-type negative feedback that controls the output voltage.

It looks like some kind of a mixed negative feedback...


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