Hi all,

When I Google Triangular Generator, all circuits provided are formed from a schmitt trigger connected to the negative input of an integrating circuit. This causes the triangular output to be inverted from the square wave.
Is it possible to connect the input of the integrator circuit to the positive terminal instead? Will this work? and do I have to use a different feedback layout?

Any help would be appreciate.
Thanks.

 

Is it possible to connect the input of the integrator circuit to the positive terminal instead? Will this work?

No. If you need to invert the waveform then use an additional inverting opamp.

 

Also, in the attached 2 steps, I can understand that (R2/R2+R2)Vsat changed to -(R2/R2+R2)Vsat
but what exactly happened to -Vramp? and how did R2 in the denominator changed to -R3?

I really cannot understand what happened in these 2 steps.

 

Your circuit doesn't show a Vramp node. Presumably Vramp = V0 (the output of A2)?

I can understand why Vramp is subtracted from Vsat, but why Vramp is then added to the voltage divider equation?

R2 and R3 form a voltage-divider. The voltage at the top is V0', and at the bottom is Vo. Therefore the voltage across the divider is Vo'-Vo and so the voltage at P is V0 + (Vo'-Vo)*R2/(R2+R3).

 

Hi Alec_t,

With reference to post #3, I took these equations from the link below (Pg 248):

http://avitec13.weebly.com/uploads/7/8/6/9/7869447/lic3.pdf

I can understand that he took Vramp as common, but why was R2 replaced with R3 please?

 

With reference to post #3, I took these equations from the link below (Pg 248)

The linked article has only 43 pages . I don't see those steps anywhere, nor R2 replaced by R3. 5.6 is the relevant section.

he took Vramp as common

Ground is common; not Vramp (the A2 output). The text is confusing because the author uses '-Vramp' to denote the negative part of the triangle.

 

Hi,

Sorry I think it was another pdf. I cannot find the file, but I have this screenshot of the derivation. From this part I cannot understand how R2 changed to R3 in the numerator. Also, what happened to -Vramp?

Thanks in advance.

 

The linked article has only 43 pages . I don't see those steps anywhere, nor R2 replaced by R3. 5.6 is the relevant section.

Ground is common; not Vramp (the A2 output). The text is confusing because the author uses '-Vramp' to denote the negative part of the triangle.

Sorry forgot the attachment.

 

From this part I cannot understand how R2 changed to R3 in the numerator. Also, what happened to -Vramp?

Combine the two Vramp terms.

 

\(-Vramp+Vramp\times\frac{R2}{R2+R3} = Vramp\times(\frac{R2}{R2+R3}-\frac{R2+R3}{R2+R3})=Vramp\times\frac{-R3}{R2+R3}\)