# Integrator triangular wave

Discussion in 'The Projects Forum' started by thiagofdp, Nov 6, 2014.

1. ### thiagofdp Thread Starter New Member

Nov 6, 2014
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Guys, I need help with this project.

I have this circuit

This is the project

Here's what I want from V1

Can someone give me a hand? Thanks in advance

2. ### MikeML AAC Fanatic!

Oct 2, 2009
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1,066
What frequency, rise time, fall time, are you trying to achieve?

R3*C1 determines the rise time. R4*C1 determines the fall time. The sum of these determines period = 1/frequency.

10*R2/R1 determines amplitude.

I question why the inverting input of the second opamp is connect to +5V instead of ground?

Last edited: Nov 6, 2014
3. ### thiagofdp Thread Starter New Member

Nov 6, 2014
4
0
it's on the third picture..
rise time 500ms
fall time 1000ms

thanks

Oct 2, 2009
5,451
1,066

Nov 6, 2014
4
0

6. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
Looks like your math is good. Note that the sim and your analysis depends on the opamps being rail-to-rail output. Note V(b); see how close it must pull to the rails... Note I used repeated trials to get close to your 1500ms period.

7. ### thiagofdp Thread Starter New Member

Nov 6, 2014
4
0
Thanks a bunch!
But it would be a nice a capacitor of lower value, because the period is too damn big
I tried 500k, 1M and 10nF, doesn't work that well... Can you show me how you calculated R3, R4 and C1?
Thans a lot.

8. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
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The times are a little difficult to calculate because of the 5V bias on the Schmitt stage. I just played around with R and C until the rise time ~ 1s and fall time ~0.5s. The time constant is roughly R*C = 1e6*1e-6 or 1meg and 1uF gives one second. The input bias current of the opamps puts an upper bound on R. I wouldn't go much over ~1meg. So you are stuck with C.