I'm trying to understand this circuit taken off of this application note.

LM101A Fast Integrator Circuit

To get it out of the way, I'm familiar with the basic integrator circuit and its workings. What I'm having trouble with are the extra components R2 and C2.

I've done a Laplace nodal analysis and got the equations and transfer function, but that didn't help me much. I feel that R2 and C2 have a much more "practical" role in this circuit.

So my questions are:

  1. What is the purpose of R2 and C2?
  2. Does the C1 capacitor have any influence in the circuit besides compensation?
  • \$\begingroup\$ Do you mind sharing the transfer function you allegedly have made? \$\endgroup\$ – Harry Svensson Dec 5 '17 at 22:11
  • \$\begingroup\$ If C1 was a "normal" compensation capacitor for this opamp, it would be connected to pins 1 and 8. \$\endgroup\$ – τεκ Dec 5 '17 at 22:28
  • 2
    \$\begingroup\$ Not at all. \$\endgroup\$ – jakubiak Dec 5 '17 at 23:12
  • \$\begingroup\$ @τεκ Yeah, that's part of the question. I'm not completely sure about C1. \$\endgroup\$ – jakubiak Dec 5 '17 at 23:17

The LM101 is about as old as the UA741 but is externally compensated. It has amazing but difficult properties. I would not suggest you try it, unless you really like challenges.

It has a large signal gain of >50k at 100kHz with a +/-10Vpp output swing. This would be an equivalent GBW product of 5GHz more, than a thousand times greater, except for the complexities of external compensation and bias null adjustment.

I won't even attempt to explain, other than it must be done this way to make it stable and only applies to this chip.

enter image description here

  • 2
    \$\begingroup\$ Any thoughts on R2 and C2? \$\endgroup\$ – jakubiak Dec 5 '17 at 23:14
  • \$\begingroup\$ From a Bode Analysis you could find it converts the OA into a 1st order LPF with RC=T=5us for phase margin compensation as opposed to internally compensated OA's with T<10ms and if GBW was 5e6 only gives you a gain of 50 at 100Khz instead of this one which still has high gain. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 5 '17 at 23:23
  • \$\begingroup\$ Wow, thanks a bunch! \$\endgroup\$ – jakubiak Dec 5 '17 at 23:38
  • \$\begingroup\$ Widlar has lots of fun with this? Or was the designer Gene Horni? \$\endgroup\$ – analogsystemsrf Dec 6 '17 at 5:23
  • \$\begingroup\$ Widlar after quit for more $ and joined National books.google.ca/… \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 6 '17 at 6:27

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