I was trying to design a differential amplifier in which the collector resistances of the transistors are 2.2 kΩ and are biased at 1 mA and are supplied with Vcc= +/-5 V . To find the Ro (the resistance of the current source), I tried doing 5/1 mA=5 kΩ.
Is my approach correct or should I just assume any value for that resistance?
Anya - you want to design a two-transistor differential amplifier. This means: It is the purpose of this amplifier to amplify only the voltage difference between both signal inputs.
In the following, you find some background information regarding the selection of a proper Ro value.
In this context, one should know that for all real applications this amplifier will also amplify the so-called "common mode" part of the signals, which is DC voltage active at the same time at both inputs. This common-mode part is nothing else than the mean value of both signals. This unwanted signal is zero only for the special case that we have two equal signals with different polarities.
For a "good" design, it is, therefore, our goal to suppress this unwanted part of the signal as much as possible. In this context, a measure of quality was defined - called "Common Mode Rejection Ratio", which is nothing else than the ratio of two gain values: CMRR=Vd/Vc (Vd: Diff. gain; Vc: Common-mode gain).
For the shown simple circuit, the CMRR value can be calculated as CMRR=gm*Ro (transconductance gm=Ic/Vt; emitter resistor Ro).
Inserting the values Ic=1mA and Ro=2.2kOhms we arrrive at CMRR=84.5 (38.5 dB).
For many applications this value is too low. Therefore, the resistor Ro in the common emitter leg must be increased. Because a larger ohmic resistor needs a larger negative supply voltage, this method for improving CMRR has some limits. This is the only reason for replacing the ohmic resistor Ro with the very large dynamic output resistance of a 3rd transistor in the common emitter leg.
The differential pair (figuratively called "long-tailed pair" in the past) consists of two common-emitter stages with a common current source in the emitters. Depending on the mode of operation, they either cooperate (common mode) or oppose (differential mode). The output voltages reflect the "efforts" of the two transistors in this interaction - in the first case they are insignificant (small gain), and in the second significant (high gain). Usually one of the outputs is used which causes problems with the common-mode voltage.
I was trying to design a differential amplifier in which the collector resistances of the transistors are 2.2 kΩ and are biased at 1 mA and are supplied with Vcc= +/-5 V.
I have illustrated your arrangement (with corrected Ie = 2 mA) by a little unusual CircuitLab simulation consisting only of voltage sources and meters (see the schematics below). My idea was to visualize the most important electrical quantities without cluttering the schematic. For this purpose, I have presented the collector resistors as ammeters with an internal resistance of 2.2 kΩ, and the emitter resistor - through a voltmeter with an internal resistance of 2.2 kΩ. Simply put, think of them as "visualized resistors". The collector and emitter voltages are visualized with ordinary (perfect) voltmeters. Note that their "black probes" are connected to ground (the midpoint of the power supply). The current directions are represented by arrows in green, and the voltage-change directions by arrows in red.
I suggest revealing the ideas behind this famous circuit by doing a number of step-by-step experiments. The first of them (framed in black) is your circuit; the others (framed in pale gray) explore its operation modes.
First we need to set the "initial state" of the circuit, or as it is commonly said, "establish the quiescent operating point". This means, at zero input voltages, adjust the emitter resistance Re so that its output currents and voltages are in the middle of the output range. This basically applies to any amplifier, but here it is a little more complicated because we have two types of signals - common-mode and differential-mode.
Ic1 = Ic2 = 1 mA: Assuming that at the common mode Vout will vary from -1 V to 1 V, we can agree with your current values of 1mA.
simulate this circuit – Schematic created using CircuitLab
Ic1 = Ic2 = 1.5 mA: Now let's experiment a bit by first setting a higher current (less Re)...
Ic1 = Ic2 = 0.5 mA: ... and then a lower current value (higher Re).
Operation: The mechanism of this biasing is as follows. When we change Re with the purpose to change the emitter and accordingly the collector currents, the transistors react to this intervention by changing their collector currents so that to make their sum equal to the emitter current.
If we change (even slightly) the input voltages in opposite directions...
Vin1 = 10 mV, Vin2 = -10 mV: ... one way...
Vin1 = -10 mV, Vin2 = 10 mV: ... or the other...
... the output voltages change significantly in opposite directions because the midpoint between the emitters is "fixed"; it acts as a virtual ground.
But if we change (even significantly) the input voltages in the same direction...
Vin1 = Vin2 = 1 V: ... e.g. by "lifting"...
Vin1 = Vin2 = -1 V: ... or "dropping" both...
... the output voltages change slightly because the emitter midpoint "moves" in the same direction with the input voltages.
Normally, the input voltages are varied differentially against some common-mode voltage. Let's try it by first "lifting" the input voltages by 1 V, and then differentially change them by only 10 mV...
Vin1 = 1.01 V, Vin2 = 0.99 V: ... in one...
Vin1 = 0.99 V, Vin2 = 1.01 V: ... or the other direction.
The result is the same as in the pure differential mode above - the output voltages change significantly because the midpoint between the emitters is moved but still "fixed"; it acts as a "shifted virtual ground".
As can be seen from Schematics 3.1 and 3.2, the output voltages, although not much, still change when we change the two input voltages simultaneously (common mode). This is a problem because we usually use one of them as a single-ended output.
The straightforward solution invented a century ago was to simultaneously increase Re e.g. to 500 kΩ, and V- to -1000 V (although unrealistic); hence the name "long-tailed pair".
Vin1 = Vin2 = 0 V: Then the emitter and collector currents, and the collector voltages will be the same as in the initial Schematic 1.1.
Vin1 = Vin2 = 1 V: But when we simultaneously increase both Vin1 and Vin2...
Vin1 = Vin2 = -1 V: ... or simultaneously decrease them...
... nothing will change. In other words, the common-mode input voltage gain is 0.
But later they thought that it was not reasonable to increase the voltage V- so much and came up with a smarter solution - to change Re simultaneously and in the same direction with both Vin... to make it dynamic. Let's simulate it by replacing the constant resistor Re with a variable one.
Vin1 = Vin2 = 1 V: Now, when the input voltages rise to 1 V, we increase Re to 2.7 kΩ...
Vin1 = Vin2 = -1 V: ... and when the input voltages drop to -1 V, we decrease Re to 1.7 kΩ.
The result is even better than above - absolutely nothing changes.
In fact what we have done above is what they call a "current source"; only it is manually controlled. Let's replace it with a standalone current source from the CircuitLab library and repeat the above experiments.
Vin1 = Vin2 = 0 V: Whether the input voltages are 0 V...
Vin1 = Vin2 = 1 V: ... or 1 V...
Vin1 = Vin2 = -1 V: ... or -1 V...
... the result is the same - no change.
Let's finish this story about the famous long-tailed pair with a practical implementation of the emitter current source. We use the property of the transistor at a constant input (base-emitter) voltage to behave as a dynamic "resistor" that keeps its output (collector) current constant.
Vin1 = Vin2 = 0 V: Initially it has "resistance" 2.2 kΩ...
Vin1 = Vin2 = 1 V: ... then 2.7 kΩ...
... and finally 1.7 kΩ.
Vin1 = Vin2 = -1 V:
The result is almost the same - there is a small change (the transistor is real, though).
See also my Codidact paper about the same circuit accompanied by real experiments.
