a teacher of mine told me to play around with the potentiometer in this circuit and observer the wave forms. so the horrible drawing u see with numbers 1 en 2 are the wave forms. when im not varying the potentiometer the wave forms just look like a straight horizontal line( upper left corner drawing). now as i vary the potmeter over a wide range the wave forms were going up and down(upper right corner drawing). so what is actually happening as i vary the wave form??

 

hi sean.
I will ask you a question, your title says Inverting op-amp, what does that mean.?
E

 

Hello,

Have a look at the second half of this page of the eBook:
https://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/divided-feedback/
And this page for the analogy:
https://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/an-analogy-for-divided-feedback/

Bertus

 

hi sean.
I will ask you a question, your title says Inverting op-amp, what does that mean.?
E

it means that the output is 180 degrees out of phase with the input. if you apply a negative vin you will get a positive vout and vice versa.

 

Hello,

Have a look at the second half of this page of the eBook:
https://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/divided-feedback/
And this page for the analogy:
https://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/an-analogy-for-divided-feedback/

Bertus

thanks for the links

 

Hello,

And what is the relation of the output and input voltage in relation to R7 and R4 in your schematic?

Bertus

 

hi,
OK, thats a good start.
I guess you also know that its configured as an amplifier.?
For that type of circuit the gain is R7/R4 or more generally written as Rfeedback/Rinput, so thats ~ gain = ~ 10k/1k = 10.

So if I set the pot R6 so that the input voltage was say +0.1V, what would the output voltage measure.?
E

 

hi,
OK, thats a good start.
I guess you also know that its configured as an amplifier.?
For that type of circuit the gain is R7/R4 or more generally written as Rfeedback/Rinput, so thats ~ gain = ~ 10k/1k = 10.

So if I set the pot R6 so that the input voltage was say +0.1V, what would the output voltage measure.?
E

+1V. because the ouput is just vin*gain right?

 

No, think about it, you said its an Inverting amp.??


output is 180 degrees out of phase with the input

 

Hello,

And what is the relation of the output and input voltage in relation to R7 and R4 in your schematic?

Bertus

hmm i think ericgibbs already gave the answere i think.im new to electronics and just started messing with amplifier circuits a couple of days ago

 

No, think about it, you said its an Inverting amp.??


output is 180 degrees out of phase with the input

oh yes thats right. -1v than

 

OK.
Excellent, you have now answered you own original question..

Are there are other points you would like to ask.

BTW: on your circuit all the resistors to the left of R4 will effect the actual gain of the circuit.

E

 

OK.
Excellent, you have now answered you own original question..

Are there are other points you would like to ask.

BTW: on your circuit all the resistors to the left of R4 will effect the actual gain of the circuit.

E

thank you. no, not at the moment. im currently busy with different type of amplifier circuits that i got from my teacher to analyze.so when im done with this one i might ask bunch of more questions.because i really want to be good in electronics because is something i've found a passion for

 

hi,
A little future note:
For the Gain = R7/R4 [Rfb/Rin] to be accurate, the voltage source driving into Rin will have a zero resistance to 0V.
In your circuit you have a resistor network of R1,2,3,5 and R6 whose total parallel and series value of resistance to 0v, from the pot wiper, has to be added to R4, call it Rsum.
The Gain would be then R7/(R4+Rsum) which means the Gain will be less than 10.
E

 

hi,
A little future note:
For the Gain = R7/R4 [Rfb/Rin] to be accurate, the voltage source driving into Rin will have a zero resistance to 0V.
In your circuit you have a resistor network of R1,2,3,5 and R6 whose total parallel and series value of resistance to 0v, from the pot wiper, has to be added to R4, call it Rsum.
The Gain would be then R7/(R4+Rsum) which means the Gain will be less than 10.
E

thanks, i appreciate the help

 

You are measuring the input voltage from the left side of R4.
The voltage gain is -R7/R4.

 

hi,
In his simple point to point measurement he will show a gain of 10, but he must be made aware that the effect of source impedance of the driving voltage must be taken into consideration.

Approx equiv resistance of input network 300R
E