LM339 Help

Thread Starter

crash563

Joined Feb 25, 2013
47
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.
I am still lost on this operation though.
 

Thread Starter

crash563

Joined Feb 25, 2013
47
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.
 

Ron H

Joined Apr 14, 2005
7,063
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.
 

WBahn

Joined Mar 31, 2012
29,976
In theory, yes. In practice, it won't split perfectly in two. If nothing else, the current in T2 has to supply the base current for both T5 and T6. Then, the collector-emitter voltage of T6 will vary while Vce for T5 will be pretty well fixed at one diode-drop (which is why it is often referred to as a diode-connected transistor). These result in an input offset voltage which is the value of vd that would have to appear at the inputs in order to get the circuit to behave like it should in the ideal case with vd=0. Because of other sources of offset error, the input offset voltage can be either positive or negative, so the datasheet just specifies a maximum value it can be over range of operating conditions.
 

Thread Starter

crash563

Joined Feb 25, 2013
47
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.

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).
 

Ron H

Joined Apr 14, 2005
7,063
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).
Reread my post. When Ic(T3)>Ic(T6), T7 turns on, Turning off T8. The pullup resistor drives the load in this case.
Just so we're on the same page: ">" means "greater than".
 

Ron H

Joined Apr 14, 2005
7,063
So when - is higher than + t8 will be on
Of course. My gosh, what else could it be? Off? We covered that already.

This is the essence of the ideal LM339:
When (+in)>(-in), Q8 is off.
When (+in)<(-in), Q8 is on.
When (+in)=(-in), Q8 state is indeterminate.

These conditions are subject to the input offset voltage.
 

#12

Joined Nov 30, 2010
18,224
Yes. Theoretically, you could use a comparator like an op amp, with a closed loop feedback path and a known gain. However, comparators are designed to do fast switching better than closed loop amplifying. Use it for what it was designed for, non-linear, 2 states, fast switching.
 
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