# confused about opamp circuit from practical electronics for inventors

• posted

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 | |\\ | | ----/\\/\\---|-\\____|__/\\/\\_|_|+\\_ |____| ___|+/ ___|-/ | |/ | |/ | | --- --- /// ///

• posted

"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

• posted

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

• posted

"panfilero" "Phil Allison"

** 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

• posted

"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

• posted

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

• posted

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.

```--
Sue```
• posted

"panfilero"

** 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

• posted

View in a fixed-width font such as Courier.

. . . . . Vx . In1>-[R2]---+----[R3]--. . | | . | |\\ | . '---|+\\ | . | >---+-> Out1 . .---|-/ . | |/ OA . --- . gnd . . . Derived from node equation: . . Vx-In1 Vx-Out1 . ------ + ------- =0, assuming current into OA (+) is 0 . R2 R3 . . . . In1*R3 + Out1*R2 . Eq.1 Vx= ---------------- . R3 + R2 . . . . But Vx is also the differential input (+)-(-) so that . . Out1= A *Vx , where A is open loop gain of the OA, . OL OL . . and OA Out1 can only follow A *Vx up to +V in the postive . OL sat . . direction, and down to -V in the negative direction. . sat . . . The thresholds are determined by computing the value of . . In1= V that causes the OA to transition from -V to . th+ sat . . +V , and In1=V that causes the OA to transition from . sat th- . . +V to -V . . sat sat . . . Mathematically these conditions are from Eq.1: . . . . V *R3 +(-V )*R2 . th+ sat > > R2 . C1. ------------------ =Vx =0 or V = - (-V )* -- . R3 + R2 th+ sat R3 . . . R2 . Note that when In1 exactly equals - (-V )* -- , Vx=0, and . sat R3 . . Out1 starts to move from -V to 0 forcing Vx>0 which in . sat . . turn forces Out1 to head for +V . This is how positive . sat . R2 . feedback works. So we take V = - (-V )* -- . . th+ sat R3 . . . . and . . . . V *R3 +(+V )*R2 . th- sat < R2 . C2. ------------------ =Vx

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.