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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:

enter image description here

And the second one is like:

enter image description here

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?

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  • \$\begingroup\$ 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. \$\endgroup\$
    – Jeon
    Jun 27, 2015 at 8:51
  • \$\begingroup\$ If someone can read Korean, senslab.co.kr/Class/OPAMP%C6%AF%B0%AD.pdf \$\endgroup\$
    – Jeon
    Jun 27, 2015 at 8:56
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    \$\begingroup\$ why not just a comparator \$\endgroup\$
    – user16222
    Jun 27, 2015 at 9:47
  • \$\begingroup\$ You could use a RUM002N02 MOSFET and a resistor if the output voltage is to be fixed at the supply rails. \$\endgroup\$ Jun 27, 2015 at 14:37

1 Answer 1

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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):
    enter image description here

    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.

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