# Are these two OP-AMP circuits behaving different?

I am looking for DC biasing circuit for a bipolar square wave with frequency < 10 MHz. I'd like to make a square wave with high 1.5-1.7 V and low (-1.5)-(-1.7) V into a square wave with high ~ 5 V and low ~ 0 V, such that midpoint is 2.5 V

And I've found two examples.

The First one is like:

And the second one is like:

As I have little knowledge about circuit, I am wondering if they are behaving same or not.

A signal in the first circuit is fed into (+), but in the second circuit, it is fed into (-). I think both circuit will output inverted waveform respect to each other. Is it right, or do I have a wrong understanding of circuits?

• I found some article explaining OP-AMP written in Korean. It says, if there is negative feedback on OP-AMP, Vin+ and Vin- becomes same. And the output voltage can be calculated from that hint. I will try calculation.
– Jeon
Jun 27, 2015 at 8:51
– Jeon
Jun 27, 2015 at 8:56
• why not just a comparator
– user16222
Jun 27, 2015 at 9:47
• You could use a RUM002N02 MOSFET and a resistor if the output voltage is to be fixed at the supply rails. Jun 27, 2015 at 14:37

Yes they are behaving differently.

• The first one is not inverting. The second one is inverting.

To be more accurate the output signal in the second one should look like this (phase inverted; also the amplitude may be changed, depending on R1 || R2 and R4):

For most audio purposes or many other cases it doesn't matter if phase is inverted along the signal path or not.

• And of course the first circuit includes a (quite) low pass filter consisting of R? (100kOhm) and C? (100nF) at the output.
The second circuit doesn't. That's a big difference.