I am still lost on this operation though.The current source at the top makes it so that the total current going through both transistors in the pair is a constant. The two input signals determine how that constant current is split between the two paths. Imagine relacing the current mirror at the bottom with two resistors, one in each path. In the output branch (the one that the base of the next stage input is connected to) as the input signals change the current that goes down that branch, the voltage at the input to the next stage changes. The bigger the resistor, the more the voltage changes by for the same change in current. The current mirror is an "active load" and is basically a current source with a high output impedance. It's output impedance acts as the load resistor that the output voltage (the input voltage to the next stage) is developed across.
Do you know how a differential pair works? Scroll down and you will find a discussion of a differential pair with a current mirror load. The circuit is the complement (NPN diff pair) of the LM339's diff pair,but the principle is the same.I am still lost on this operation though.
When the differential input is zero, ideally the T2 and T3 collector currents are equal.Yes, I am lost about the current divided and active load. Ron the differental gain (AD) is usally very high the difference between the inputs will cause the output to follow the supply voltages or saturation voltages. Thus making the comparator a nonlinear circuit.
When the differential input is zero, ideally the T2 and T3 collector currents are equal.
Ignoring the base currents of T5 and T6, T5 collector current=T2 Collector current.
Since T5 and T6 are identical, and their B-E voltages are equal, then the collector current of T6=collector current of T5. This is called a current mirror. Do you see why?
Note that this means that the collector current of T6=collector current of T3.
Therefore, no current flows into the base of Q7, so it will be off.
NOW - Make in(+) slightly lower than in(-).
The T3 collector current will increase.
The T2 collector current will decrease.
The T5 collector current will decrease. The collector current of T6 will decrease.
Now, Ic(T3)>Ic(T6). The difference flows into the base of T7, and it turns on.
Reread my post. When Ic(T3)>Ic(T6), T7 turns on, Turning off T8. The pullup resistor drives the load in this case.If this was the case T8 would turn off and +Vsat voltage from the pull-up resistor will flow to the load correct. Is T7 turning off and T8 is turning on when Ic(T3)>Ic(T6).
In that case, T8 turns off.But isnt it the case when a the + is higher than -
Of course. My gosh, what else could it be? Off? We covered that already.So when - is higher than + t8 will be on
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