Someone please help me understand last line

The effective value of R3 : Don't understand


Thank you

 

According to Ohm`s law, the current through R3 is i=(vin-Avin)/R3.
Therefore, the coresponding conductance is
i/vin=(1-A)/R3 and the corresponding resistance is
vin/i=R3/(1-A) which is much larger than R3 because A<1 (but very close to unity).

This calculation assumes that the capacitor C is rather large (capacitive impedance negligible).
This "bootstrap" method is a kind of positive feedback which increass the input impedance drastically.

 

According to Ohm`s law, the current through R3 is i=(vin-Avin)/R3.
Therefore, the coresponding input conductance is
i/vin=(1-A)/R3 and the corresponding resistance is
vin/i=R3/(1-A) which is much larger than R3 because A<1 (but very close to unity).

This calculation assumes that the capacitor C is rather large (capacitive impedance negligible).
This "bootstrap" method is a kind of positive feedback which increass the input impedance drastically.

Thanks...Trying to understand

"i/vin=(1-A)/R3 and the corresponding resistance is" is i = Vbe/R3?

so current that goes into the base is Vbe/re ?

How has increasing a.c value of R3 done any good? R3 will be in parallel with re at a.c.. no?

Also, Rin = Vin/Iin

Iin is Vbe/(re||R3) ?

thanks

 

First, notice that if the R3 resistor was connected between Vin and GND any change in Vin will cause corresponding change R3 resistor current i = vin/R3 hence rin = R3.

But in this circuit R3 resistor is connected between Vin and emitter. As you shoud know the AC voltage at the emitter follows the input voltage vin ≈ v_emitter. This means that the voltage across R3 is constant, hence for the AC signals, the R3 looks like a constant current source (a bigger resistor).

Examine this


When we have an ideal voltage amplifier with the voltage gain equal to A.

Now let us try to find the input resistance.

Rin = Vin/Iin

In = (Vin - Vout)/R = (Vin - A*Vin)/R = Vin * (1 - A)/R

Rin = Vin/Iin = R/(1 - A)

And if we use as our amplifier the emitter follower (voltage gain close to one)

We will have this situation




This is the bootstrap. The closer the voltage gain to 1 is the less current is drawn from the input source (increase in the input resistance)



And for your circuit Rin becomes :

Rin ≈ (β+1)*(re + RE)

 

First, notice that if the R3 resistor was connected between Vin and GND any change in Vin will cause corresponding change R3 resistor current i = vin/R3 hence rin = R3.

But in this circuit R3 resistor is connected between Vin and emitter. As you shoud know the AC voltage at the emitter follows the input voltage vin ≈ v_emitter. This means that the voltage across R3 is constant, hence for the AC signals, the R3 looks like a constant current source (a bigger resistor).

Examine this
View attachment 177534

When we have an ideal voltage amplifier with the voltage gain equal to A.

Now let us try to find the input resistance.

Rin = Vin/Iin

In = (Vin - Vout)/R = (Vin - A*Vin)/R = Vin * (1 - A)/R

Rin = Vin/Iin = R/(1 - A)

And if we use as our amplifier the emitter follower (voltage gain close to one)

We will have this situation

View attachment 177536


This is the bootstrap. The closer the voltage gain to 1 is the less current is drawn from the input source (increase in the input resistance)



And for your circuit Rin becomes :

Rin ≈ (β+1)*(re + RE)

Thanks Jony...Please post other links as usual

DOn't understand this : Rin ≈ (β+1)*(re + RE)

RE is smaller now

 

DOn't understand this : Rin ≈ (β+1)*(re + RE)

What part you don't understand?

RE is smaller now

If you want to be pedantic

\(R_{IN} \approx \frac{R_3}{1 - A_V}||(r_e+R_X)(\beta+1)\)

Where:

RX = R1||R2||RE

But normally R1 and R2 will be much larger than RE, hence Rx ≈ RE


\(A_V = \frac{R_X}{R_3||r_e + R_X} \approx \frac{R_X}{r_e+R_X}\)

 

What part you don't understand?



If you want to be pedantic

\(R_{IN} \approx \frac{R_3}{1 - A_V}||(r_e+R_X)(\beta+1)\)

Where:

RX = R1||R2||RE

But normally R1 and R2 will be much larger than RE, hence Rx ≈ RE


\(A_V = \frac{R_X}{R_3||r_e + R_X} \approx \frac{R_X}{r_e+R_X}\)

R3/1-A is not in parallel with re???

Anyways, thanks...Guess i should skip the rigor since just starting electronics