Re model for Collector feedback bias with emitter resistance

Question

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.

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,

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

Answer 1

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.

Answer 2

\$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.

Answer 3

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.