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This is a BJT amplifier we have learned

I read from the book that the feedback type of it is voltage series negative feedback. I have 2 questions :

  1. If this is a negative feedback circuit how can we compute its open loop voltage gain?
  2. Is this thing(learn how to analyse bjt/mos negative feedback)important?

enter image description here

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  • \$\begingroup\$ The concept and application of negative and positive feedback in circuits is very important. Can you identify in this circuit why it is negative feedback? \$\endgroup\$ – JIm Dearden Oct 17 '16 at 15:22
  • \$\begingroup\$ Learning BJT/transistor design is important if you plan to design circuits that use transistors, or understand the deeper details of how to use circuits designed with transistors. I use this knowledge almost every day. Negative feedback is important if you plan to work with control systems or analog circuits of any type, including power supply circuits. \$\endgroup\$ – mkeith Oct 17 '16 at 15:36
  • \$\begingroup\$ The circuit you show is an emitter follower. The voltage gain will be 1. I don't think it is useful to think of feedback in this circuit. \$\endgroup\$ – Peter Bennett Oct 17 '16 at 16:11
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  • My answer to 2.: Yes it is very important because the principle of negative feedback is most important for all electronic amplifiers, filters, oscillators,.. More than that it plays a dominant role for all stability analyses (remember: Amplifiers with feedback can become unstable).

  • My answer to 1.: I think, it is not possible to calculate the "open-loop voltage gain" because the transistor is NOT a voltage amplifier. Instead, it acts as a "voltage-controlled current source" [Ic=f(Vbe)]. That means: The open-loop transfer characteristic is the transconductance gm=Ic/Vt.

(As another example, the same situation exists for the integrated operational transconductance amplifier (OTA), having an "open-loop gain" which also is identical to the transconductance gm of the device).

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  • \$\begingroup\$ Actually it's easy to design a single transistor high gain open loop and then close the loop but for lower input and output impedance. \$\endgroup\$ – Sunnyskyguy EE75 Oct 17 '16 at 19:38
  • \$\begingroup\$ Tony Stewart, I must admit that - in the context of the subject to be discussed - I do not understand your comment. Please, can you explain? \$\endgroup\$ – LvW Oct 17 '16 at 21:09
  • \$\begingroup\$ I am speaking in terms of implementation topology not the fundamental bipolar model and the fact that negative feedback "may" be added to any model with high (open loop) gain to improve performance with closed loop gain. But I realize this H Bias topology is neither. ( does not use NF or Hi Gain). \$\endgroup\$ – Sunnyskyguy EE75 Oct 17 '16 at 22:05
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You show a emitter follower. You can think of it as having negative feedback, but that's a little awkward. It's easier to think of the B-E voltage being roughly constant. When the load draws more current, it pulls down on E. That increases the B-E voltage, which causes more C-E current, delivering more to the load.

The net result is that the load sees a low impedance source. At first approximation (B-E voltage is constant), the output impedance is the impedance the base is being driven with divided by (transistor gain + 1).

Yes, understanding the few basic single-transistor amplifier configurations, how they work, and their general characteristics, is important if you ever want to be good at designing circuits.

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  • 1st fig. is a voltage controlled current sink (thus inverting out on collector)
  • H Bias very stable, less sensitive to hFE but low V gain.
  • Shared current produces a gain of -RL/RE enter image description here
  • 2nd fig. has a Cap that shunts Re so gain above ω > 1/RC depends on rBE junction (small Ω) to make large gain ratio controlled by Ib bias current.
  • more V gain but more sensitivity with hFE enter image description here

  • 3rd fig. has very high open loop gain in Q1 and Q2

    • depends on high hFE and critical choice of RL ( which shud be ~1.8k )
    • can amplify 1mV with gain (500~1000x)
    • Zin = 10K
    • skill test, What would you add to improve sensitivity from hFE , keep gain >500 and improve THD using only one negative feedback R? and what value? enter image description here

Question, What is the highest V gain design you can get with 1 transistor with hFE of 200 at 12Vdc?

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  • \$\begingroup\$ Comment/question to 2nd Fig: You wrote "rBE junction (small Ω)". My question: Small resistance? Normally, it is in the kOhm range! It seems you have mixed it up with transconductance, which is the relevant quantity (gain=-gm*RL)! \$\endgroup\$ – LvW Oct 17 '16 at 21:12
  • \$\begingroup\$ Yes for 10 uA currents but 25Ω @ 1mA Ebers-Moll results in rBE= 25 Ω/mA for Ibe @ 20'C \$\endgroup\$ – Sunnyskyguy EE75 Oct 17 '16 at 22:14
  • \$\begingroup\$ Both methods can be used. gm or rBE \$\endgroup\$ – Sunnyskyguy EE75 Oct 17 '16 at 22:18
  • \$\begingroup\$ Does the voltage gain depend on the transconductance gm or on rBE? \$\endgroup\$ – LvW Oct 18 '16 at 7:15
  • \$\begingroup\$ The voltage gain (common emitter) is A=-Rc/(1/gm + RE) . For RE>>1/gm we have app. A~-Rc/RE and for RE=0 (without signal feedback) we have A=-gm*Rc. \$\endgroup\$ – LvW Oct 18 '16 at 9:38

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