For the given NPN transistor with \$\beta=99\$ and \$r_e=0.02 k\Omega\$, I need to find the voltage gain using Small signal Analysis.


I drew the equivalent \$r_e\$ model as below.Re model diagram

To find the Voltage Gain \$Vo/Vi\$, I First went from \$Vi\$ to ground which gave:

\$i_b = \frac{V_i}{(\beta + 1)(r_e + R_E)} = \frac{V_i}{(99 + 1)(20\Omega + 1k\Omega)} = \frac{V_i}{103.02k\Omega}\$

I tried to write Kirchoff's current law at Node 1 which gives, KCL at node 1

I don't have more equations to solve this, Can someone please help in relating \$Vo\$ and \$Vi\$?

  • 2
    \$\begingroup\$ I think you must assume the load resistor is infinity, so \$i_o = 0\$, then you can work the voltage gain easily. Voltage gain depends on load resistor here so you will get a messy expression if you include it \$\endgroup\$
    – across
    Mar 5, 2022 at 6:17
  • \$\begingroup\$ Does this answer your question? Negative feedback - equilibrium No yet that's what you accepted. There is no AC NFB \$\endgroup\$ Mar 6, 2022 at 18:48
  • \$\begingroup\$ Is it a problem we allow illogically biased designs to be analyzed that have no AC negative feedback yet allow the OP to accept an answer with NFB from a true voltage source that shunts all AC negative feedback(NFB) and only allows poor DC NFB with an almost saturated collector voltage, poor input impedance , low gain and high offset DC output. I voted to close yet answered by changing the question with Rs=50 and Re=0, which IS an acceptable design. \$\endgroup\$ Mar 6, 2022 at 18:51
  • \$\begingroup\$ This design makes no sense to use regardless of how you analyze it. Normally the goals are to balance the Vc output for large signal or high symmetry current swings or low Vbe modulation or low THD or large swing output or high impedance input , or tolerant to high source impedance or at least an output impedance the same or lower than the source or at least have some design goals incl. BW or tr or ft toggle rate. Which of these are demonstrated here? none? The title should be what's wrong with this NPN CE design? and sent back to author. \$\endgroup\$ Mar 10, 2022 at 14:53

3 Answers 3


If you define the voltage gain as Vo/Vi you must consider Vi as an ideal voltage source. In this case, there is no signal feedback from the collector to the base because each feedback signal will be shorted (assuming a negligible signal voltage across C1).

Therefore, the gain expression is the classical one: Open-loop voltage gain divided by (1-LG) with loop gain LG=-gmRe (gm=transconductance and Re=1kOhm). Using your notation we have: gm=1/re.

Comment: As mentioned by Tony Stewart, the base divider resistors do not allow a suitable bias point.

  • \$\begingroup\$ using your notation, I see, \$Av=\dfrac{Rc/Re}{ 1-LG}=\dfrac{10}{1+Re/re}\$ I fail to see a correct result here. \$\endgroup\$ Mar 6, 2022 at 2:27
  • \$\begingroup\$ If you insist on analyzing this poor academic design with an ideal voltage source then there is no negative AC feedback and the analysis of a closed loop system is false. It is still an open loop system but has an added AC load. The NFB is only for DC.` \$\endgroup\$ Mar 6, 2022 at 18:44
  • \$\begingroup\$ No - the equation for Av in your 1st comment is wrong. The numerator (open-loop voltage gain) is -(gmRc)=-Rc/re. In words: Transconductance x effective collector resistance. What you have written (in the numerator) is a rough approximation for the closed-loop gain. \$\endgroup\$
    – LvW
    Mar 6, 2022 at 20:58
  • \$\begingroup\$ my error for misreading you "Open-loop voltage gain divided by (1-LG) with loop gain LG=-gmRe (gm=transconductance and Re=1kOhm) Aol=-Rc/(re+Re) and you had LG=-gmRe But there is no AC closed loop. with negative feedback. Just DC NFB since you assumed true Voltage source .. meaning your answer and my comment on your answer are both wrong . The whole question ought to be deleted \$\endgroup\$ Mar 6, 2022 at 21:12
  • \$\begingroup\$ Of course, there is an ac closed-loop caused by Re. Do you want me to show it? More than that, Aol=gmReff=Reff/re with Reff=10k||10k=5k (assuming an ideal signal source with no internal source resistance). Why do you recommed to delete the question? \$\endgroup\$
    – LvW
    Mar 9, 2022 at 18:43

\$r_e\$ is irrelevant to gain here with Re=1k.

  • The base resistor values are wrong.

  • Base to ground must be about 40x Re =40k

  • Rcb feedback must be much greater than 40k to bias collector near Vcc/2

  • Source impedance must be 50 ohms and if higher, then input impedance being reduced by NFB will attenuate source.

  • With 0.5 mA = Ie you can expect up to 85% of open loop gain of 10=Rc/Re upto Rc/Re=100

  • this % depends on input feedback attenuation and gains >100 are difficult to optimize , so don't try.

  • for Av=-100, Re=0 Rs= 50, Rcb=10k , Rc= 10k, Vcc=10V, gain increases with Ic and thus Vcc.


The way to see what the actual gain is, is to embrace something LvW absolutely got correct -- that there is no negative feedback to worry about. That's because the base is being directly driven by an ideal source. The next thing to realize is that your effective collector resistance is the parallel of base-collector resistance (I'll call it \$R_{_\text{B}}\$) and the positive supply-collector resistance (I'll call it \$R_{_\text{C}}\$), or \$5\:\text{k}\Omega\$. The reason that this is true is that the base end of \$R_{_\text{B}}\$ is driven by an ideal supply and the high end of \$R_{_\text{C}}\$ is driven by \$V_{_\text{CC}}\$. So the collector "sees" two resistors in parallel.

At this point, it really is \$A_V\approx -\frac{R_{_\text{C}}\,\mid\mid\,R_{_\text{B}}}{R_{_\text{E}}+r_e^{\:'}}\$.

By using a large enough \$V_{_\text{CC}}\$ you can get an acceptable DC bias. It's just that it would need to be higher than often expected. The DC bias could definitely be improved.

Let's look at what they claim. They assert that \$r_e^{\:'}=20\:\Omega\$. This implies that \$I_{_\text{E}}=1.3\:\text{mA}\$ (give or take a little depending on what you take as the device temperature.) This means that the BJT terminal voltages are: \$V_{_\text{E}}=1\:\text{k}\Omega\cdot 1.3\:\text{mA}=1.3\:\text{V}\$, \$V_{_\text{C}}=V_{_\text{CC}}-10\:\text{k}\Omega\cdot \left(1.3\:\text{mA}\frac{\beta}{\beta+1}\right)\$, and \$V_{_\text{B}}=V_{_\text{C}}-10\:\text{k}\Omega\cdot \left(1.3\:\text{mA}\frac{1}{\beta+1}\right)\$. Given that for active operation it must be that \$V_{_\text{C}}-V_{_\text{E}}\ge 1\:\text{V}\$ (the \$1\:\text{V}\$ comes from an estimated base emitter voltage of \$700\:\text{mV}\$ plus another \$130\:\text{mV}\$ needed across \$R_{_\text{B}}\$ to supply the base current plus some allowance for output signal excursions, etc. -- it really should be much more than that, in practice) you can work out that in your case \$V_{_\text{CC}}\ge15.13\:\text{V}\$. So it can be DC biased. But with caveats. If you wanted any room for output signal, you'd need more.


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