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I'm building a simple circuit which has an LM211 comparator at its heart. The positive input is connected to a 10k-10k resistor divider across the supplies, stabilized by a 1uF capacitor. Hysteresis is provided by 1.8MOhm from the output to the positive input.

The signal on the SENS line is a slowly changing resistance hooked up to +5V on the other side.

Resistance threshold comparator

Everything works perfectly on a breadboard, but I wanted to check a few things before finalizing the board layout, and came across a TI application note which says in paragraph 6:

It is a standard procedure to use hysteresis (positive feedback) around a comparator, to prevent oscillation, and to avoid excessive noise on the output because the comparator is a good amplifier for its own noise. In the circuit of Figure 2, the feedback from the output to the positive input will cause about 3 mV of hysteresis. However, if the value of Rs is larger than 100Ω, such as 50 kΩ, it would not be reasonable to simply increase the value of the positive feedback resistor above 510 kΩ.

However, it doesn't say why a high value feedback resistor is "not reasonable". I built up my circuit on a breadboard, and it seems to work fine, I can definitely see good hysteresis behaviour with the 1M8 resistor as compared to without it.

So, what is the drawback?

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Olin is right about the problem with C23: it lowers the bandwidth of the positive feedback rendering it useless. Just decoupling the supply close to the divider will do the job.

In fact, that is probably also the reason why TI wants to keep the feedback resistance low in value. If the value is too high, the input capacitance of the op-amp alone will already lower the bandwidth of the positive feedback, so on a noisy signal the hysteresis will not work. Normally you would like the hysteresis to be faster than the response of the op-amp; this will be probably be the 510K TI talks about.

If you make sure the highest (noise) frequency that can reach the negative input will be lower than the feedback response, you should be in the clear; however, making the feedback faster then the op-amp can react will be the better option.

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  • \$\begingroup\$ So many good replies, but this one actually answers the question. Thanks! \$\endgroup\$ – Medo42 Oct 27 '14 at 15:19
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1 MΩ for the positive feedback should be fine in your case from a impedance point of view. Whether it always provides enough positive feedback under all conditions is something you have to decide.

What bothers me about your circuit is that the positive feedback is fighting against C23. That is going to slow it down, which probably in part defeats its purpose. To fix this, you can add some resistance between C23 and the + input. The immediate effect of the positive feedback will be based on the feedback resistance and this deliberate resistance. In the long term, the 5 kΩ impedance of the power supply divider will be added to the deliberate series resistance.

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You can parallel the 1M8 resistor with 2.7nF and avoid the objections while still maintaining a 1uF filter cap.

That's maintaining the 1800K/5K ratio of feedback resistance.

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  • \$\begingroup\$ I entered this into LTSpice and found it works nicely to speed up the edges. Would you recommend adding ~1 Ohm in series with the 2.7nF to avoid peaking due to parasitic inductivity? I guess it's total overkill, but I can't help myself when playing with the frequency analyzer in LTSpice... :) \$\endgroup\$ – Medo42 Oct 27 '14 at 16:29
  • \$\begingroup\$ I wouldn't bother. \$\endgroup\$ – Spehro Pefhany Oct 27 '14 at 16:51

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