confused about opamp circuit from practical electronics for inventors

[panfilero]

hello, OK, I've been breaking my head over this one for a while and
just can't figure it out. It's an example from a book called
"Practical Electronics For Inventors" It's basically a "Simple
Triangle-wave/Square-wave Generator" It has a integrator opamp
circuit feeding the non-inverting input on a positive feedback....
(schmitt trigger or comparator not about the terminology here) opamp
circuit, and then that feeds the inverting input back on the
integrator...

The book then shows this equation for the threshold voltage (Vt): Vt =
Vsat / (R3 - R2) , but i can't for the life of me see how they got
this..... the circuit is as follows, anyone have any idea of how Vt is
found? The rails are at +/- 15V for each opamp

Thanks
Joshua




(view in courier font)



---------------------------------------
| |
| |
| C R3 |
| |---| |--| _/\/\_ |
| | | | | |
| R1 | |\ | R2 | |\ | |
----/\/\---|-\____|__/\/\_|_|+\_ |____|
___|+/ ___|-/
| |/ | |/
| |
--- ---
/// ///

 

[Phil Allison]

"panfilero"

The book then shows this equation for the threshold voltage (Vt): Vt =
Vsat / (R3 - R2) , but i can't for the life of me see how they got
this..... the circuit is as follows, anyone have any idea of how Vt is
found? The rails are at +/- 15V for each opamp

** Your sketch shows a " voltage follower Schmitt trigger ".

That Vt formula has a typo:

The correct one is:

Vt = Vsat x R2 / R3


The input voltage at R2 must be of the opposite sign to the op-amp's output
for it to switch over.

The threshold voltage is simply that input voltage at R2 needed to bring the
+ op-amp input to zero volts.





........ Phil

 

[panfilero]

"panfilero"


** Your sketch shows a " voltage follower Schmitt trigger ".

That Vt formula has a typo:

The correct one is:

Vt = Vsat x R2 / R3

The input voltage at R2 must be of the opposite sign to the op-amp's output
for it to switch over.

The threshold voltage is simply that input voltage at R2 needed to bring the
+ op-amp input to zero volts.

....... Phil

I still can't see it.... I guess where I get confused is I see a
triangle wave coming out of the integrator going through R2.....
so..... if the output of the 2nd opamp is saturated positively i'm
seeing a steady decreasing voltage out of the integrator.... now i
think when the non-inverting input of the 2nd opamp goes below 0 the
thing should flip (since the inverting input is grounded) ..... but i
have a hard time finding the voltage at the non-inverting input of the
2nd op-amp..... do i use superposition? with the output of the
integraor being - 1/RC * Vsat ..... ? Then I could use superposition
to find the voltage at the noninverting input....

something like.... [ (-1/RC * Vsat) * R3/(R2+R3) ] + [R2/
(R2+R3)]*Vsat....

which.... i'm not sure why it doesn't work.... but i guess it
doesn't. Maybe I'm over complicating this, could you tell me how you
got to your result?

thank you
joshua

 

[Phil Allison]

"panfilero"
"Phil Allison"

** Your sketch shows a " voltage follower Schmitt trigger ".

That Vt formula has a typo:

The correct one is:

Vt = Vsat x R2 / R3

The input voltage at R2 must be of the opposite sign to the op-amp's
output
for it to switch over.

The threshold voltage is simply that input voltage at R2 needed to bring
the
+ op-amp input to zero volts.

I still can't see it.... I guess where I get confused is I see a
triangle wave coming out of the integrator going through R2.....
so..... if the output of the 2nd opamp is saturated positively i'm
seeing a steady decreasing voltage out of the integrator.... now i
think when the non-inverting input of the 2nd opamp goes below 0 the
thing should flip (since the inverting input is grounded) ..... but i
have a hard time finding the voltage at the non-inverting input of the
2nd op-amp..... do i use superposition? with the output of the
integraor being - 1/RC * Vsat ..... ? Then I could use superposition
to find the voltage at the noninverting input....

something like.... [ (-1/RC * Vsat) * R3/(R2+R3) ] + [R2/
(R2+R3)]*Vsat....

which.... i'm not sure why it doesn't work.... but i guess it
doesn't. Maybe I'm over complicating this, could you tell me how you
got to your result?

** It is just a simple voltage divider.

The magnitude of Vin is of no interest to the output

- only its sign is relevant.

So the formula for Vt is all you need.




......... Phil

 

[Phil Allison]

"Phil Allison"


** It is just a simple voltage divider.

The magnitude of the voltage at the + op-amp input is
of no interest to the output - only its sign is relevant.

So the formula for Vt is all you need.





........ Phil

 

[panfilero]

"Phil Allison"

** It is just a simple voltage divider.

The magnitude of the voltage at the + op-amp input is
of no interest to the output - only its sign is relevant.

So the formula for Vt is all you need.

........ Phil

Thanks for your answers, I'm still really confused.... could you
please tell me how you derived Vt = Vsat x R2 / R3 , if I take a
voltage divider at the + op-amp input, I get Vsat x R2/(R2 + R3)....
I'm ignoring the voltage coming out of the integrator..... could you
tell me how you got your answer?

thanks,
joshua

 

[Palindrome]

hello, OK, I've been breaking my head over this one for a while and
just can't figure it out. It's an example from a book called
"Practical Electronics For Inventors" It's basically a "Simple
Triangle-wave/Square-wave Generator" It has a integrator opamp
circuit feeding the non-inverting input on a positive feedback....
(schmitt trigger or comparator not about the terminology here) opamp
circuit, and then that feeds the inverting input back on the
integrator...

The book then shows this equation for the threshold voltage (Vt): Vt =
Vsat / (R3 - R2) , but i can't for the life of me see how they got
this..... the circuit is as follows, anyone have any idea of how Vt is
found? The rails are at +/- 15V for each opamp

Thanks
Joshua




(view in courier font)



---------------------------------------
| |
| |
| C R3 |
| |---| |--| _/\/\_ |
| | | | | |
| R1 | |\ | R2 | |\ | |
----/\/\---|-\____|__/\/\_|_|+\_ |____|
___|+/ ___|-/
| |/ | |/
| |
--- ---
/// ///

You might like to think about this in a different way:

At some moment during the changeover the output voltage must be zero -
and that can only be true when there is no differential input voltage.
Thus the non-inverting input must be at 0v at that time.

The input is a summing point for two currents - one from the output of
the second op amp, determined by the output voltage divided by R3. The
other the output voltage from the first op amp divided by R2.

Most of the time, the output of the second op amp is at Vsat (positive,
or negative). However, the output of the first op amp is changing with
time.

When those two currents are exactly equal and opposite, the voltage at
the non-inverting input will be at 0v and the op amp will be in its
linear region. The output from the second op amp will go to 0v, as there
is no differential input voltage. However, this will cause the input
current at the summing point, produced by the output voltage, to change,
due to the change in output voltage. The result of this positive
feedback is to drive the output of the op amp to its (opposite sign) Vsat.

Thus the only factors that determine the changeover point are the
voltage from the first op amp, divided by its resistor and the voltage
from the second op amp, divided by its resistor. At the instant that
they are equal (and opposite) change-over takes place.

 

[Phil Allison]

"panfilero"

Thanks for your answers, I'm still really confused.... could you
please tell me how you derived Vt = Vsat x R2 / R3 , if I take a
voltage divider at the + op-amp input, I get Vsat x R2/(R2 + R3)....
I'm ignoring the voltage coming out of the integrator..... could you
tell me how you got your answer?

** When the voltage at the + input = 0

Vin / R2 = Vsat / R3

( Because no current flows into or out of the + input, the currents MUST
be the same)

So it follows:


Vin = Vsat x R2 / R3



Almost too simple for words .....




........ Phil