How Does a Constant Current Circuit Boost Resistance in Differential Amplifiers?

[chaoseverlasting]

In a differential amplifier in common mode, a constant current circuit is appended to emmitter part to increase the resistance R_e. This results in a very low common signal amplification (which reduces noise in the ckt).

Why does a constant current circuit increase the resistance?

 

[chaoseverlasting]

http://www.ecircuitcenter.com/Circuits/BJT_Diffamp1/BJT_Diffamp1.htm

The link shows a DA circuit, here the resistor R_e is replaced by a constant current circuit.

 

[berkeman]

If you use a real resistor of high resistance, you get way too much voltage drop. Instead, you can use a current sink (or source) circuit to generate the bias current, but still present a high impedance.

 

[chaoseverlasting]

Then what we're trying to do is control the resultant current, and its not the resistance but the current that causes the common mode gain to be small?

 

[berkeman]

Then what we're trying to do is control the resultant current, and its not the resistance but the current that causes the common mode gain to be small?

Not sure I understand the question. The current is the total bias current for the diff pair. The splitting of the current by the diff pair is what gives differential gain, as the current is split unevenly by the difference in input voltages. If the input voltages are the same but varying, the current should not be different in the two sides of the diff pair.

 

[Phrak]

In a differential amplifier in common mode, a constant current circuit is appended to emmitter part to increase the resistance R_e. This results in a very low common signal amplification (which reduces noise in the ckt).

Why does a constant current circuit increase the resistance?

The dynamic resistance. The dynamic resistance of a current source is greater than a resistor.

R_{dyn} = \frac{dE}{dI}

For an ideal current souce, the voltage changes nothing for any change in current.

 

[chaoseverlasting]

The common mode voltage gain is given by:

A_c=\frac{\beta R_c}{r_i +2(\beta +1)R_E} (Boylestad, p.600)

To minimize this gain, its the resistor R_E that we maximize.

The bias current also depends on this resistor:

I_c=\frac{1}{r_i+2(\beta +1)R_E}

If we use a constant current circuit to control this current and minimize it, we effectively increase the magnitude of the resistor R_E and that causes the common mode gain to be small. Is that right?

That is what I meant to say when I said that we use the current to control the gain.

 

[berkeman]

Looks good to me, chaos.

 

[chaoseverlasting]

Finally!