Hello, I often see many differential amplifiers using constant current supplies on one side of their power supply rail. For example in the circuit below (not my design just grabbed it off a web page) the differential amplifier uses no constant current source
Lemme explain in words, as opposed to equations that clarify nothing...
A differential amplifier is ,in essence ,a balance. You apply a voltage differential ( one input sits at a different level than the other) to the inputs. That means, one of the two transistors will conduct more current than its twin.
In the first schematic : lets say the left transistors base sits at a higher level , in respect to ground, than the right one
That means the left one will pass more current. At the point where the three resistors meet the currents are added . The current coming from the left is added to the current from the right and that sum flows through the bottom resistor to the ground.
The problem is that that node is not 'hard' . If i send the left transistor more into conduction i get a larger current there , that causes the voltage across the 75 k resistor to go up. So the right transistor now has less voltage between its base and that 75k resistor.
The total amount of current going through the 75k resistor is not a constant in this case. So the real behavior of the balance is not fixed.
Think of it this way. You have a teeter totter . normally the legs extend a same distance from the pivot point. In the first circuit that center point shifts, depending on which leg is up above the other.
So , to fix that we enter the current source.
now that current is a constant. So the sum of the current coming from the left and the right is always equal to the current in the current source.
Let's say the current source is set for 1 milliamp. If the left transistor lets 0.8 milliamp flow out of its emitter then the right transistor can only let 0.2 milliamp flow. And vice versa.
There is no more impact from the voltage drop across the emitter resistors (the 1k resisotrs in the first schematic) . The centerpoint of the balance no longer shifts depending on which leg is up.
That centerpoint shift causes non linearity. The current source solves that.
Now, both the 75k resisotr and the current source have a voltage drop across them. We call that the common mode voltage.
If you apply input voltages that are lower than Vbe+Vcommon the balance no longer works as you cannot send the transistors in conduction.
With the first schematic that common mode point shifts a bit depending on input voltages. with a current source it doesnt shift anymore. if the common mode shifts , the amplification shifts .. For an amplifier that is not a good idea.
You will find opamps that have the current source sitting between the negative rail and inputs, and others have it sitting between positive rail and inputs.
We call these top driven or bottom driven . A top driven amplifier can work with signals that go all the way to its negative power supply , but stops working if the signals are at vcommon off the positive supply.
A bottom driven on works for signals from its positive supply to vcommon off the negative supply.
Rail-to-rail input amplifiers (dont confuse those with rail to rail output , that is something else, to many times people talk about rail to rail opamps but do not specify if they mean input,output or both ...)
So , rail to rail input means you can go from positive rail to negative rail (rail meaning power supply voltage) . Do these guys have no drop across the current source ? Yes they do, they just employ a trick using a few additional transistors so you do not see the effect of the cinternal common mode voltage. For an applied input range of v+ to v- , you actually, internally in the device stay within v+ to vcommon range. They simp,y scale you up , or down , depending on where the current source sits.
There are even opamps that can scale you well above their positive rail or negative rail. A simple matter of a voltage divide made from two resistors.
Such opamps are found in current sensing applications. They sense a small differential voltage but at a common mode well above , or below, their own power supply voltage.