I am searching for a way to use a comparator with possible negative voltage on its inputs.

My thoughts and attempts so far:

  • Op amps and comparators with V+, V- and GND: the circuit utilizes a single LiPo battery.
  • LM339/393: works correctly only when Vin >= V- - 0.3V, which is not enough.
  • Diodes: too low voltage, I suppose

Requirements are:

  • Vin range -2.0V .. + 2.0V (and around 0.1V on the reference)
  • Max. current provided for comparator input: around 500 uA

Recently I have found an interesting comparator - LT1716 (datasheet/website). It provides a "Valid Output with Either Input 5V Below V– ". If there is one, then the idea is not new and there should be more, right?

(I am new to electronics, and I am sorry if I did not provide some related info - please let me know what I forgot)

  • \$\begingroup\$ Hi, welcome to EE.SE, there is one problem with your question, its a shopping question. These are not good questions to ask because you could easily answer them yourself if you would learn to use google. If you want to ask a design question then go right ahead. The question is equivalent to, "I don't like this kind of orange juice please find me a better brand" \$\endgroup\$
    – Voltage Spike
    Apr 8, 2016 at 22:21
  • \$\begingroup\$ I got the point. However, I attempt to respect time of others, and quite a lot of googling did not bring any results. I am not very experienced in the topic, which might make searching difficult without some hints. Hints would be great already, not necessary to give me some specific model. \$\endgroup\$
    – Art
    Apr 8, 2016 at 22:43

1 Answer 1


This may give you some ideas:


simulate this circuit – Schematic created using CircuitLab

Figure 1. Comparitor to handle negative voltages.

  • R3 sets a threshold between 0 and V+. If this is too coarse add series resistors at the top and/or bottom of the pot to narrow the range of adjustment for easier adjustment.
  • The trick of the circuit is to feed the non-inverting input from a potential divider, the top of which is connected to V+. In this example \$ V_A = V_+ - \frac {V_+ - V_{IN}}{2} \$ so the input voltage can range from \$ \pm V_+\$
  • \$\begingroup\$ This looks interesting. Though, this would apply additional voltage on Vin. I suppose I should figure out how the input will react on it .. \$\endgroup\$
    – Art
    Apr 8, 2016 at 23:21
  • \$\begingroup\$ Edit your question to include details on the signal source and we can consider how it will be affected. Thanks for accepting my answer but if you want to "un-accept" for a day or so it may prompt a better answer. \$\endgroup\$
    – Transistor
    Apr 9, 2016 at 7:41

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