The circuit shown below is an positive feedback opamp with positive pulse of 6V at the negative terminal with a 0.7V diode in series!

The opamp being in positive feedback with Vn @ 4V has an output of 0volts and when the input is a pulse of sufficiently large magnitude that the positive terminal goes high above refernce voltage and output remains at Vcc even after the pulse is removed!

The question is : But,when the pulse is not high enough but still appears positive terminal of opamp , i still see a postive pulse at output in millivolt range, Why does that occur ? Also when i vary the resistor values ,the output pulse magnitude varies and at some point i see positive feedback dominate and take output high forever!! Why?

Here is a pic of output : enter image description here

  • \$\begingroup\$ I could use heavy needlenose pliers to drive nails, but things go smoother if I use a hammer. If you need a comparator, use a comparator. They are optimized quite differently from an op amp in positive feedback. Many such issues go away when you use the right tools for the right job. Also, the simulator is not the chip. It can simply be failing in a mode the chip doesn't fail in. \$\endgroup\$ – Scott Seidman Sep 29 '15 at 14:02
  • \$\begingroup\$ If the ouput of the opamp starts at 0V when t = 0 (from the plot: it does) then the input pulse is not high to pull the + input of the opamp above 4 V ! R1, D1 and R2 form a voltage divider, so you would need at least a pulse of 8.6 V or more. \$\endgroup\$ – Bimpelrekkie Sep 29 '15 at 14:06

The op-amp output going high and staying high once it goes should not be a surprise- that's what the circuit does. There's nothing to bring the input below 4V since the diode is reverse biased once the output has gone high (provided only that the supply voltage is high enough).

When the op-amp output is low and you feed a pulse into the input that is high enough to get the diode to conduct substantially but not high enough to cause the non-inverting input to exceed 4V, the output of the op-amp is called upon to sink current. LT have supplied the full schematic and you can try to analyze the output stage, but in any case the performance looks like this:

enter image description here

When the voltage at the non-inverting input reaches 4V the output is sinking more than 3mA, so this is off the above graph, but you can expect the output voltage to typically approach 1V (maybe ~0.8V as shown in the next figure on the datasheet).

In the case of your circuit, you've supplied 6V, so the current is about 2.5mA and you'd expect the output voltage to be in the 300-400mV range, which it is.


Those few millivolts are just due to the design of the output stage of the opamp.

You have selected a bipolar version and bipolar transistors always have a few millivolts when saturated.

If you select one of the "ideal" opamps in LTSpice or a CMOS opamp (e.g. LT2050) the voltage should be zero.

With that circuit it will latch-up at power up under some conditions. I would put a switch between the opamp +in and ground and activate that for a few microseconds after power up (use a Voltage controlled switch driven by a Pulse output from a voltage source in a similar fashion to your other pulse generator).


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