# Common emitter, o.p. set by feedback resistor

I am analyzing common emitter amplifier where operating point is set by feedback resistor between b-c.

Voltage gain showed by symulator is much lower than (-)R2/R1 and input resistance is lower than B*R1 (B is beta, hfe).

What is R3 impact? I am looking for exact formulas (for "by hand" calculating) for k and R_inp. Knowledge about R_out would be also educating.

My goal is understanding voltage and input impedance drop in circuit like below:

Predicted: k=U_Rc/25mV=6V/25mV=240 V/V, R_input=(25mV/Ib || 1k)*B || 82k=109k || 82k=47k

Simulated: k=113 V/V, R_input=13k

• Do you mean R2/R1? Also, show your power rails - don't skimp on this detail. – Andy aka Apr 6 '18 at 14:30
• Yes. I mean k=R2/R1, I misspelled there – kamillabaarnowak Apr 6 '18 at 14:32

For this circuit

simulate this circuit – Schematic created using CircuitLab

The input impedance is equal to $$R_{in} = \frac{(R_B+R_C)\cdot(r_{\pi} (\beta+1)R_E )}{r_{\pi}+ (\beta+1)(R_C+R_E) + R_B}$$

Or google the Miller effect How does a Miller cap physically create a pole in circuits? or this Simple op amp question, finding gain and input resistance

simulate this circuit

$Q_2$ emitter current will be around $700µA$ (If I ignore the base current)

And $I_{C1} \approx \frac{(V1+V2)- 2V_{BE}(1+\frac{R_5}{R_6})}{R_2} \approx 52mA$

Which means that the in a real circuit the $Q_2$ emitter current will be around $800µA$

and $Q_2$ base current will be around $I_{B2} = 2µA$ hence $I_{C1} \approx\frac{(V1+V2)- (2V_{BE}(1+\frac{R_5}{R_6})+I_{B2}R_5) }{R_2}\approx\ 32mA$

So the AC small-signal parameters are:

$r_{e2} = \frac{26mV}{I_E2} = 32\Omega$ and $r_{e1} = \frac{26mV}{I_E1} = 0.8\Omega$

The voltage gain will be around

$$A_V \approx \frac{R7||(\beta+1)r_{e1}}{R7||(\beta+1)r_{e1} + r_{e2}}\cdot\frac{R_2}{r_{e1}} \approx 221 V/V$$

We get such big difference because of the BC547C model you used in the simulation.
In your model we see $R_E = 0.6\Omega$

Which means, that $Q_1$ voltage gain stage is

$$\frac{R_2}{r{e1} +R_E} = \frac{200\Omega}{0.8\Omega + 0.6\Omega} = 143$$

Therefore the overall voltage gain is around $AV = 143*0.9 = 129 V/V$

And the input resistance is equal around:

$$R_{IN} \approx R_6||\frac{R_5}{AV+1}||[(\beta+1)* (r_{e2}+R_7||(\beta+1)*r_{e1})] \approx 12k\Omega$$

• You ve done enormous amount of work. This is it what I looked for. Thank you – kamillabaarnowak Apr 7 '18 at 11:39

simulate this circuit – Schematic created using CircuitLab

This is just off the cuff.. you can work it out.

The assumption is R3 >> R2 >> R1 , if Vin has Rs then AC gain is R3/Rs after HPF breakpoint

1) You are (blindly) using an approximation for the gain (-R2/R1) which may NOT be applied here. Before using any "rule-of-thumb" formula you must know the corresponding conditions/restrictions which do exist.

2.) Did you ever hear about the role and the consequences of negative feedback? In your first circuit, the resistor R3 provides negative feedback (in the second circuit R5-R6) - with consequences on the gain value as well as the overall input resistance (keyword: Miller effect).

3.) My recommendation: Try to understand the working principle of transistor stages and the effects of negative feedback - and do NOT use existing formulas without knowing anything about their region of applicability.

The apparent impedance looking into R3 from the base of Q1 requires some thought.

Since we don't care about the DC currents and voltages at the bias point, we can think of Ohm's Law in this case as:

Ω = dV / dA

Where dV is the change in voltage and dA the accompanying change in current measured in amps.

To find the apparent resistance of R3, start by analyzing the circuit at its quiescent point. Then consider what happens when the base voltage changes a bit. Figure out how the voltage at the other end of R3 changes as a result. Now you have the change in voltage across R3, from which you can compute the change in current flowing through R3 onto the base node.

From the change in voltage of the base node, and the resulting change in current through R3 onto the base node, you can use the equation above to calculate the apparent resistance of R3 as seen from the base node.

I deliberately didn't work through this because it would be a good exercise for you to do.