Hi, I hope this is the right place for asking it. I've come across a relaxation oscillator circuit with an opamp. I've understood how it works and the influence of the various components, but i can't figure out the function of the resistor R4 at the top right. Could you give me some elucidations?? Thanks

 

Welcome to AAC!
OP1 might be a comparator e.g. LM393 with an open collector output. In that case, R4 acts a the pull-up that allows the output to go high. The pinout looks like a 741 OpAmp though so maybe not.

 

Welcome to AAC!
OP1 might be a comparator e.g. LM393 with an open collector output. In that case, R4 acts a the pull-up that allows the output to go high. The pinout looks like a 741 OpAmp though so maybe not.

It's exactly what i've thought, but it specifies a ua741 opamp. Can such resistance still provide some advantagies in that case?

 

Welcome to AAC!
OP1 might be a comparator e.g. LM393 with an open collector output. In that case, R4 acts a the pull-up that allows the output to go high. The pinout looks like a 741 OpAmp though so maybe not.

R4 is pull up resistor?

 

This guy from TI has some info about using op amps as comparators and why not to do it. Doesn't directly address the pullup R but some good info..
http://www.ti.com/lit/ml/slyt701/sl...PS FOR EVERYONE&tisearch=Search-EN-everything

 

The 741 is not a rail-to-rail output opamp, so the pull-up action of R4 will help to raise the output above what it would otherwise be.

 

In fact, through a simulation in Tina (Opamp ua741C, dual supply 10V, R1=R2=100k, R1=27k), the output changes steeply for R4 small, until it stabilizes. However I can't figure out why the opamp out can be affected by the resistor.

 

Between the Vcc rail and the output pin of the 741 output stage there are the collector-emitter path of an NPN transistor, plus a series resistor. These drop voltage. The pull-up provides a current bypass to these, reducing the drop.

 

I hasn't thought of it. Thanks.