Sioux12
Newbie level 4
Hello!
Consider page 2 of the datasheet for current differential amplifier "LM2900/LM3900/LM3301 Quad Amplifiers", "Schematic and Connection Diagrams".
They are 4 identical differential amplifiers. The first stage gets two currents as its inputs: I_IN+ and I_IN-. Their difference (I_IN+) - (I_IN-) is the current input of the second stage.
As stated in the "General Description",
Assuming that a DC operating point is defined, consider the small-signal equivalent circuit of the first stage. The small signal impedances from the two input pins, obtained considering just a small AC voltage and current variations around the DC operating point, look different to me.
The +INPUT pin faces two small-signal impedances in parallel:
The -INPUT pin faces something very different. The two parallel small-signal impedances seen by this pin are:
If that is true, the +INPUT and -INPUT pins are in a different condition, so the current entering into the +INPUT pin is not treated by the circuit the same way as the current entering the -INPUT pin.
Is this correct?
And, if yes, shouldn't a differential amplifier treat instead its two inputs the same way?
Consider page 2 of the datasheet for current differential amplifier "LM2900/LM3900/LM3301 Quad Amplifiers", "Schematic and Connection Diagrams".
They are 4 identical differential amplifiers. The first stage gets two currents as its inputs: I_IN+ and I_IN-. Their difference (I_IN+) - (I_IN-) is the current input of the second stage.
As stated in the "General Description",
So, the "diode" connected to the +INPUT pin actually represents a diode-connected transistor, which is the master branch of a current mirror; the transistor of the slave branch is on his right hand side and it is explicitly shown.These amplifiers make use of a current mirror to achieve the non-inverting input function.
Assuming that a DC operating point is defined, consider the small-signal equivalent circuit of the first stage. The small signal impedances from the two input pins, obtained considering just a small AC voltage and current variations around the DC operating point, look different to me.
The +INPUT pin faces two small-signal impedances in parallel:
- the one of a diode-connected transistor, which is very low (being the input of a current mirror);
- the one looking into the base of an ordinary BJT, which is higher.
The -INPUT pin faces something very different. The two parallel small-signal impedances seen by this pin are:
- the small-signal impedance seen from the collector of an ordinary BJT, which is very high;
- the small-signal impedance seen into the base of an ordinary BJT, which, as before, is high.
If that is true, the +INPUT and -INPUT pins are in a different condition, so the current entering into the +INPUT pin is not treated by the circuit the same way as the current entering the -INPUT pin.
Is this correct?
And, if yes, shouldn't a differential amplifier treat instead its two inputs the same way?