# Level shifting digital signals with a comparator

I'm trying to understand the voltage levels associated with a comparator used for level shifting digital signals. The comparator is the MAX995. The input voltage levels are +1.5V/-2V, and the output levels are +3.3V/0V.

Could someone explain to me what exactly is happening with the voltages at the positive comparator input when the a logic high and logic low are transmitted, respectively? Could someone explain to me what exactly is happening with the voltages at the positive comparator input when the a logic high and logic low are transmitted, respectively?

When the input is 1.5 volts, clearly and without any math, the voltage at the non-inverting input = 1.5 volts (this causes the comparator to produce a logical 1 output).

When the input is -2.0 volts use what you know about potential dividers to calculate the voltage. Basically you'll have 3.5 volts across 35k. This causes a current of 0.1mA and that current produces a volt drop of 1.5 volts across the 15k. Because one end is tied to +1.5 volts, the non-inverting input has to be at 1.5 volts minus 1.5 volts i.e. 0V. This cause a logical zero output on the comparator.

So, basically I am disagreeing with @sabirmoglad's calculation because he's saying it's the other way round but he's wrong!

this op-amp works as a comparator. when the input at the non-inverting terminal is greater the one at the inverting input( the Reference ) , the op-amp output will follow the +V (3.3 v) and if its the opposite, the output will be the -V

when your digital input is 1.5 volt, the voltage difference, between the two resistors is going to be 0 volt (1.5 -1.5 ), therefore the non-inverting input is going to be 0 volt as well which is less than the Reference voltage which makes the output of the opamp 0v

when the input is -2 volt, the voltage difference between the two resistors is going to be 3.5 (1.5 - (-2))volt, therefore, the voltage at the non-inverting input is going to be 2 volt :

(20/35)*3.5 = 2 volt (simple voltage divider)

which is higher than the reference, which makes the output of the opamp 3.3 volt (the +V terminal )

• No, see my answer - you have your thinking back-to-front. – Andy aka Jul 23 '15 at 9:49
• you are right. @Andy aka thanks for the correction – Sabir Moglad Jul 23 '15 at 9:56