I'm new to PSpice and here I was trying to simulate a square wave oscillator using a LM193 on PSpice. Here is the circuit that I've created there:

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Edit: The capacitor C1 has an IC set to 0

However, when I plot the waveform of Vout, this is the obtained result:

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Even though this form "resembles" a square wave, it's not even close to what I wanted. I'd be really thankful if someone could explain a couple of things to me:

  1. Why is the signal oscillating only between 0 and -5V? Shouldn't it be between -Vsat and + Vsat, where Vsat is used to designate the saturation voltage of the Op Amp? What have I done wrong in this circuit?

  2. Moreover, isn't the output voltage supposed to be more symmetrical (therefore resembling more a square wave)? If someone could give me some advice on how to make it more like it I'd appreciate it.

Thanks a lot for your attention!


2 Answers 2


The LM193 has an open collector output therefore, it needs a pull up resistor: -

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Examples of a relaxation oscillator using a generic op-amp with push-pull output on the left and a single-supply connected LM193 (that requires a pull-up) on the right: -

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Without a pull up resistor, the 100 ohm positive feedback resistor and 1 kohm resistor power the open collector output but not from the positive supply but from 0 volts. Hence that is why you see such an asymmetrical output voltage. Use a 1 kohm pull up resistor to V+ and increase R1 from 100 ohms to something like 10 kohm. 100 ohm is much too low for decent operation.

  • \$\begingroup\$ Thanks a lot! Managed to solve it! By the way, could you enlighten on the role of the negative feedback resistor R on the waveform and also what would be a reasonable range of values for it? Thank you! \$\endgroup\$ Jun 22, 2020 at 13:37
  • \$\begingroup\$ I believe you mean the 50 kohm in your question yes? A range of a few hundred ohms to many tens of kohm is usually fine without any small_print. It controls how fast or how slow the capacitor is charged in the opposite direction. \$\endgroup\$
    – Andy aka
    Jun 22, 2020 at 13:48

A few problems here.

LM193, LM293, and LM393 are open collector 0V to Rout to V+ outputs. Your design is missing the pullup R.

Then you want to minimize the swing to Vin+,- so R1 should be much larger than R2. e.g. 10x

Also feedback R's should never be 100 Ohms. This is too low and will be current limited.

See page 3 https://www.st.com/resource/en/datasheet/lm193.pdf

Now try again using R2=100K split into 2 R's to Vcc and 0V. Then make R1= same or bigger than R2.

NOW you can use a single supply.

What happens when you reduce the feedback R on Vin-?

How is that different from increasing R1 for lower hysteresis?

See here


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