hi can anyone try to help me understand what is going on here in this internal circuit diagram.
http://www.unisonic.com.tw/datasheet/LM339.pdf
http://www.unisonic.com.tw/datasheet/LM339.pdf
Tying the collectors of T1 and T4 to ground allows the input common mode range to include the negative rail. If they were connected to the T2 and T3 collectors, T1 and T4 would be saturated when the inputs are at the negative rail.Is it just me, or does it seem like the base of T7 is hard tied to ground?
Also, and I'm a lot less sure on this one, but it seems like the collectors of T1 and T4 be tied to the collectors of T2 and T3, respectively instead of tied to ground?
I believe you've spotted an error in the datasheet.TI's datasheet shows T7's base going to T3's collector:
http://www.ti.com/lit/ds/symlink/lm239-n.pdf
Makes sense. It probably also improves the matching of the two channels since one of the sources for mismatch is no longer in the channel.Tying the collectors of T1 and T4 to ground allows the input common mode range to include the negative rail. If they were connected to the T2 and T3 collectors, T1 and T4 would be saturated when the inputs are at the negative rail.
It works the same as any diff pair. The two NPNs at the bottom are a current mirror.So how does that diff pair work
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.So how does that diff pair work
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