enter image description here In many tutorials when explaining hysteresis what I read that when the output is high R3 becomes in parallel with R1; and when the output is low R3 becomes in parallel with R2. Above is a depiction of the situation.

Why is R3 in parallel with R1 when the output is high? And why is R3 in parallel with R2 when the output is low?

I could not figure out the reason.

  • \$\begingroup\$ When the opamp is driving in direction of vcc or gnd, redraw the diagram yourself with the opamp replaced by vcc or gnd respectively. \$\endgroup\$ – PlasmaHH Mar 1 '16 at 12:30
  • \$\begingroup\$ "When the opamp is driving in direction of vcc or gnd" Driving current? \$\endgroup\$ – user16307 Mar 1 '16 at 12:34
  • \$\begingroup\$ I cannot visualize it \$\endgroup\$ – user16307 Mar 1 '16 at 12:34
  • \$\begingroup\$ In this circuit the opamp behaves as a comparator. This means the output is "high" (connected to Vcc) or "low" (connected to ground). Now if the opamp's output is high (Vcc) the right side of R3 is connected to Vcc. What other resistor is also connected to Vcc ? \$\endgroup\$ – Bimpelrekkie Mar 1 '16 at 12:40
  • \$\begingroup\$ sorry i forgot to connect output to Vcc. i edited the picture \$\endgroup\$ – user16307 Mar 1 '16 at 12:49

You guys failed to recognize OP issue.
If those explanation works. He won't ask about this in the first place. (methinks)
Let's see the most simplest form of explanation

How R3 parallel with R1? R1 Anything wrong with that?

Ditto for R2 R2

Now you can read other answers which have good explanation why op-amp work this way.

Some note on R4: Why it is there?
In actual cheap op-amp component the op-amp cannot drive voltage to Vcc. (max at Vcc - 1.5V for instance) There are rail-to-rail op-amp which have higher cost in general. The op-amp in question is likely to be Open-drain op-amp hence the pull-up resistor. It will work perfectly fine without much calculation because it will connect to digital input pin that likely to be a height impedance pin.


Since there's no negative feedback, and the op-amp has ideally infinite gain, only two things can happen:

  1. the op-amp will try to drive the output to Vcc
  2. the op-amp will try to drive the output to ground

Or if you are using a comparitor with an open-collector output, almost the same thing happens:

  1. the op-amp will provide no base current to the output transistor, allowing R4 to pull the output up to Vcc
  2. the op-amp will provide enough base current to the output transistor to saturate it, pulling the output down to ground

If you then redraw the schematic without the op-amp's output pulled up through R4, or actually connected to ground, you get these:


simulate this circuit – Schematic created using CircuitLab

If it's still not clear, it might help to think of the op-amp as a switch, operated by a fairy which compares the inputs and flips the switch accordingly:


simulate this circuit

  • \$\begingroup\$ this is also very good explanation thanx! \$\endgroup\$ – user16307 Mar 1 '16 at 17:15
  • \$\begingroup\$ i was reviewing the inners of schmitt trigger recently. just a question.. isnt R4 shorted in your last figure? how is that parallel it looks like baypassed? \$\endgroup\$ – user16307 Oct 5 '16 at 20:53
  • \$\begingroup\$ @user16307 Indeed, it doesn't really do anything in that figure as drawn, so I edited it a bit. R4 is a pull-up resistor which is really only necessary when using a comparitor with an open-collector output. \$\endgroup\$ – Phil Frost Oct 6 '16 at 12:33
  • \$\begingroup\$ Alright now it makes very sense thanx \$\endgroup\$ – user16307 Oct 6 '16 at 12:37

Think about the operation of a schmitt trigger. It's supposed to clean up a digital signal so it's always held at a 1 or a 0. For example, a digital signal might have a very slow rise/fall time, so it spends a significant amount of time in "no-man's" land as it changes state. A schmitt trigger cleans this up by slamming the output to one of the IO rails once the input crosses a certain voltage threshold.

One of the characteristics of an op amp is it has very low output impedance, which, in the case a schmitt trigger, you can think of the output as always being connected to one of the supply rails. This isn't the case in actuality, but the concept helps clear up some details that aren't really relevant.

So, since the output of the op amp is always connected to one of the supply rails (but never both at the same time!), R3 will always be in parallel with R1 or R2 (again, not exactly true due to non-ideal opamp characteristics, but close enough for this case).

EDIT Here's a little more detail about opamps:

Take a look at the internal circuitry of an op amp:

enter image description here

The output is between Tr5 and Tr6. Assume R8 and R7 are 0 ohm. We know from the operation of a schmitt trigger that the output will always be either V+ or V- (aka, the power supply rails). This means that Tr5 or Tr6 (but not both) will be on full bore. Ignoring the internal resistance of those BJTs (it's low enough not to matter), they will connect the output to either Vcc or ground.

An op amp amplifies the difference between the two inputs. The output is always the difference between the two inputs scaled by a gain factor. For example, if the + input is at 2.3v, and the - input is at 2.2v, and the gain factor is 10 (made up numbers for easy math), then the output will be at 1.0v. 2.3-2.2 = 0.1. 0.1 * 10 = 1.0.

Now, in real life op-amps, the gain is big. Really big. As in, so big that an itsy bitsy little difference in inputs will make the output drive to one side of the power to the op amp. The minimum voltage an op amp can output is the negative supply to the op amp. The max voltage an op amp can output is the positive supply to the op amp. So, if an op amp can 3.3v supply, is can only swing from 0.0v to 3.3v.

  • \$\begingroup\$ i updated my picture.. but my confusion remains. why when the output goes low R3 is in parallel with R2 and when the output goes high R3 is in parallel with R1. i still dont get it. can u explain it with current directions? \$\endgroup\$ – user16307 Mar 1 '16 at 12:51
  • \$\begingroup\$ does the direction of output current change when the output voltage is ON or zero? \$\endgroup\$ – user16307 Mar 1 '16 at 12:54
  • \$\begingroup\$ @user16307 Ignore current. I'm updating my answer with more details. \$\endgroup\$ – CHendrix Mar 1 '16 at 12:56
  • \$\begingroup\$ a simple drawing would help a lot if possible to visualize \$\endgroup\$ – user16307 Mar 1 '16 at 12:59

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