When the negative input of the TI LM2901 comparator is negative, the positive voltage drops. Why?

陈大侠

When the negative input of the TI LM2901 comparator is negative, the positive voltage drops. Why? [Copy link]

T1 and T2 are PNP transistors, and the collector is grounded. So when the input IN+ and IN- are both greater than zero, the collector junctions of T1 and T2 are reverse biased, and they either work in the amplification state or in the cutoff state. Since the emitter current is 10uA, T1 and T2 must work in the amplification state. Similarly, T3 and T4 also work in the amplification state. The emitter potential of T1 is IN++Ueb, and the emitter of T2 is IN-+Ueb. 1 IN+=IN- The base potentials of T3 and T4 are equal. Since T3 and T4 are the same tubes, Ic3=Ic4. The collector voltage of T5 is equal to the base voltage. T5 works in the amplification state (critical saturation). T5 and T6 form a mirror current source, Ic5=Ic6, and Ic3=Ic4. The base current Ib7 of T7 is zero. T7 is cut off, and the 80uA current flows from the base of T8, T8 is turned on, and the OUT output is low level. 2 IN+>IN- Since the base voltages of T3 and T4 are not equal, Vb3>Vb4, and since Ve3=Ve4, Vbe3Ic3, and according to the characteristics of T5 and T6 forming a current source, at this time, the current transmitted from the collector of T6 to the base of T7 is >0, and T7 is turned on. In this case, the OUT output is divided into two cases: 1) In the first case, the output of T8 is in the linear amplification area. At this time, T7 is in the amplification state and Ib8=80uA-Ic7. If T8 is to be in the linear area, the current of Ic7 must be slightly less than 80uA, then converted to Ib7=(several uA)/beta. Since this Ib7 is very small, about tens of nA level, it can be inferred that Vin+ only needs to produce a very small voltage compared to Vin- to make T8 leave the amplification area and enter the cut-off area. 2) In the second case, the output of T8 is in the cut-off area. At this time, T7 can be in the amplification state as long as the collector current of Ic7 is guaranteed to be 80uA, or T7 can be in the saturation state. At this time, OUT is in a high impedance state. 3 IN+Vbe4, so Ic4

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"In the experiment, when the input to IN- is -0.9V, the voltage at IN+ drops from 2.5V to about 1.25V. How does the voltage drop at IN+ occur?" The original voltage at IN+ is 2.5V, and the voltage at IN- is also 2.5V. After the voltage at IN- drops to -0.9V, the voltage at IN+ drops to about 1.25V. How do you apply the voltage at IN+ to the pin? It is probably applied by a power source with a very high internal resistance.

The voltage at the IN+ terminal is 2.5V after the 5V power supply on the PCB is divided by the resistors R50 and R52.

When the negative input of the TI LM2901 comparator is negative, the positive voltage drops. Why?

陈大侠

maychang

"In the experiment, when the input to IN- is -0.9V, the voltage at IN+ drops from 2.5V to about 1.25V. How does the voltage drop at IN+ occur?" The original voltage at IN+ is 2.5V, and the voltage at IN- is also 2.5V. After the voltage at IN- drops to -0.9V, the voltage at IN+ drops to about 1.25V. How do you apply the voltage at IN+ to the pin? It is probably applied by a power source with a very high internal resistance.

陈大侠

The voltage at the IN+ terminal is 2.5V after the 5V power supply on the PCB is divided by the resistors R50 and R52.